RD Sharma 2017 Solutions for Class 10 Math Chapter 4 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among class 10 students for Math Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2017 Book of class 10 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2017 Solutions. All RD Sharma 2017 Solutions for class 10 Math are prepared by experts and are 100% accurate.

#### Page No 4.119:

#### Question 1:

If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is a right-angled triangle.

#### Answer:

We have,

In order to prove that the triangle is a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.

Here, the larger side is .

Hence, we have to prove that.

Let solve the left hand side of the above equation.

Now we will solve the right hand side of the equation,

Here we can observe that left hand side is not equal to the right hand side.

Therefore, the given triangle is not a right angled triangle.

#### Page No 4.119:

#### Question 2:

The sides of certain triangles are given below. Determine which of them are right triangles.

(i) *a* = 7 cm, *b* = 24 cm and *c* = 25 cm

(ii) *a* = 9 cm, *b* = 16 cm and *c* = 18 cm

(iii) *a* = 1.6 cm, *b* = 3.8 cm and *c* = 4 cm

(iv) *a* = 8 cm, *b* = 10 cm and *c* = 6 cm

#### Answer:

(i) Let

In order to prove that the given sides of a certain triangle forms a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.

Here, the larger side is.

Hence, we have to prove that.

Let solve the left hand side of the above equation.

Now we will solve the right hand side of the equation,

Here we can observe that left hand side is equal to the right hand side that is .

Therefore, the given sides of a certain triangle form a right angled triangle.

(ii) Let

In order to prove that the given sides of a certain triangle forms a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.

Here, the larger side is .

Hence, we have to prove that.

Let solve the left hand side of the above equation.

Now we will solve the right hand side of the equation,

Here we can observe that left hand side is not equal to the right hand side.

Therefore, the given sides of a certain triangle do not form a right angled triangle.

(iii) Let

In order to prove that the given sides of a certain triangle forms a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.

Here, the larger side is .

Hence, we have to prove that.

Let solve the left hand side of the above equation.

Now we will solve the right hand side of the equation,

Here we can observe that left hand side is not equal to the right hand side.

Therefore, the given sides of a certain triangle do not form a right angled triangle.

(iv) Let

Here, the larger side is .

Hence, we have to prove that .

Let solve the left hand side of the above equation.

Now we will solve the right hand side of the equation,

Here we can observe that left hand side is equal to the right hand side that is .

Therefore, the given sides of a certain triangle form a right angled triangle.

#### Page No 4.119:

#### Question 3:

A man goes 15 meters due west and then 8 meters due north. How far is he from the starting point?

#### Answer:

Let us draw the diagram. Let A be the starting point. From point B he goes to the north.

Therefore, we obtained the following drawing.

Now we have to find how far is he from the starting point that is we have to find .

Now we will use Pythagoras theorem to find the length of AC.

………(1)

Let us substituting the values of AB and BC in equation (1) we get,

Let us take the square root we get,

Since AC is the distance therefore it should be positive.

Therefore, he is from the starting point.

#### Page No 4.119:

#### Question 4:

A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

#### Answer:

Let us draw the diagram from the given information we get a right angled triangle ABC as shown below,

Let the window be at the point A. We know that angle formed between the building and ground is always 90°.

Given: AB = 15 m and CA = 17 m

Now we will use Pythagoras theorem to find .

Let us substitute the values we get,

Subtracting 225 from both the sides of the equation we get,

Let us take the square root we get,

Therefore, the distance of the foot of the ladder from the building is.

#### Page No 4.119:

#### Question 5:

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

#### Answer:

Let us draw the diagram from the given information.

Let us draw a perpendicular from B on CD which meets CD at P.

It is clear that BP = 12 m because it is given that distance between feet of the two poles is 12 m.

After drawing the perpendicular we get a rectangle BACP such that AB = PC and BP = AC.

Because of this construction we also obtained a right angled triangle BPD.

Now we will use Pythagoras theorem,

Let us substitute the values of BP and PD we get,

Taking the square root we get,

Therefore, distance between the top of the two poles is .

#### Page No 4.120:

#### Question 6:

In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm, Calculate the altitude from A on BC.

#### Answer:

We know that altitude that is a perpendicular drawn on the unequal side of the isosceles triangle bisects that side.

Therefore, BD = DC = 7 cm.

Let us use the Pythagoras theorem in right angled triangle ADB we get,

Substituting the values we get,

Subtracting 49 from both the sides we get,

Let us take the square root we get,

AD = 24 cm

Therefore, the altitude of the isosceles triangle is .

#### Page No 4.120:

#### Question 7:

The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?

#### Answer:

The given information can be represented as follows.

Here, A is the position of the window and AC is the ladder.

Also, DE is the same ladder when it is shifted.

C and E are the original and final position of the foot of the ladder.

Now, applying Pythagoras theorem in ΔABC,

AC^{2} = AB^{2} + BC^{2}

⇒ AC^{2} = (8 m)^{2} + (6 m)^{2} = (10 m)^{2}

⇒ AC = 10 m

Now, again applying Pythagoras theorem in ΔEBD

DE^{2} = EB^{2} + BD^{2}

⇒ (10 m)^{2} = (8 m)^{2} + BD^{2}

⇒ BD^{2}^{ }= 100 m^{2} − 64 m^{2} = 36 m^{2}

⇒ BD = 6 m

Thus, the tip of the ladder is now at the height of 6 m above the ground.

#### Page No 4.120:

#### Question 8:

Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

#### Answer:

Let us draw the diagram from the given information.

As we are given that distance between their feet is 12 m

.

Now we get a right angled triangle DCE.

Let us applying the Pythagoras theorem we get,

Substituting the values we get,

Let us take the square root we get,

Therefore, distance between their top is .

#### Page No 4.120:

#### Question 9:

Using Pythagoras theorem determine the length of AD in terms of b and c shown in the given figure.

#### Answer:

In ∆*ABC* and ∆*DBA*,

(90º each)

(Common)

Therefore, by AA-criterion for similarity, we have .

Now we will substitute the values of AC and AB

We are finding the value of AD therefore; we will use the following ratios,

Now we will multiple both sides of the equation by .

We will simplify the above equation as below,

..…(1)

But we know that substituting the value of BC in equation (1) we get,

Therefore, the value of AD in terms of *b* and *c* is.

#### Page No 4.120:

#### Question 10:

A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm.

#### Answer:

Since we obtained two right angled triangles, and .

In and

(Common angle)

So, by AA-criterion

Now we will multiply both sides of the equation by.

.....(1)

Let us simplify the equation (1) as given below,

Now we will substitute the values of BC, AB and AC.

Therefore, the length of the altitude is.

#### Page No 4.120:

#### Question 11:

ABCD is a square. F is the mid-point of AB. BE is one third of BC. If the area of ∆FBE = 108 cm^{2}, find the length of AC.

#### Answer:

It is given that F is the midpoint of AB. Therefore, we have AF = FB.

It is also given that .....(1)

Now look at the figure. Quadrilateral ABCD is a square and hence all angles are of 90º.

In , and hence it is a right angle triangle.

We know that the area of the right angle triangle is $\frac{1}{2}$ × base × height

Therefore, $Ar\left(\u2206FBE\right)=\frac{1}{2}\times BF\times BE$

cm^{2}

Now we will multiple both sides of the equation by 2 we get, .....(2)

But we know that and.

Let us substitute these values in equation (2) we get,

Let us simplify the above equation as below,

But we know that ABCD is a square and hence AB = BC = CD = AD.

.....(3)

We know that 216 is the cube of 6 therefore we can write the equation (3) as below,

AB^{2} = 6^{3} × 6

AB^{2} = 6^{4}

Therefore, side of the square ABCD is 36 cm.

Now we are going to find the diagonal AC.

Diagonal of the square can be calculate by using the formula given below,

Diagonal = $\sqrt{2}$ Side

.....(4)

We know that

Let us substitute the value of in equation (3).

Therefore, the length of AC is .

#### Page No 4.120:

#### Question 12:

In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.

#### Answer:

We have given an isosceles triangle and we know that the altitude drawn on the unequal side of the isosceles triangle bisects that side.

Therefore, in and

$\angle B=\angle C$ (Equal sides have equal angles opposite to them)

$\angle ADB=\angle ADC\left(90\xb0\mathrm{each}\right)$

(AAS congruence theorem)

Now we will use Pythagoras theorem in right angled triangle ADB.

Let us substitute the values of AB and AD we get,

Subtracting 25 from both sides we get,

Since

Therefore, length of BC is.

#### Page No 4.120:

#### Question 13:

In a ∆ABC, AB = BC = CA = 2*a* and AD ⊥ BC. Prove that

(i) $\mathrm{AD}=a\sqrt{3}$

(ii) $Area\left(\u2206ABC\right)=\sqrt{3}{a}^{2}$

#### Answer:

In and

$\angle B=\angle C$ (60º each)

$\angle ADB=\angle ADC$ (90º each)

(AAS congruence theorem)

But therefore, we get,

………(1)

Now we will divide both sides of the equation (1) by 2, we get,

Now we will use Pythagoras theorem in right angled triangle ADB.

Now we will substitute the values of AB and BD we get,

Therefore, .

We have given an equilateral triangle and we know that the area of the equilateral triangle is.

Here, side is 2*a*

Therefore, .

#### Page No 4.120:

#### Question 14:

The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.

#### Answer:

We know that a quadrilateral is said to a rhombus if all sides of the quadrilateral are equal. Diagonals of a rhombus bisect each other at right angles.

Quadrilateral ABCD is a rhombus and diagonals AC and BD intersect at point O.

As we defined above, we get , , and angle .

We are given that AC = 10 cm and BD = 24 cm.

Therefore, we get, AO = OC = 5 cm and BO = OD = 12 cm.

Now we will use Pythagoras theorem in the right angled triangle AOD as below,

…......(1)

Now we will substitute the values of AO and OD in equation (1) we get,

Let us take the square root

AD = 13

Therefore, length of the side of the rhombus is .

#### Page No 4.120:

#### Question 15:

Each side of a rhombus is 10 cm. If one its diagonals is 16 cm find the length of the other diagonal.

#### Answer:

We know that a quadrilateral is said to a rhombus if all sides of the quadrilateral are equal. Diagonals of a rhombus bisect each other at right angles.

Quadrilateral ABCD is a rhombus and diagonals AC and BD intersect at point O.

As we defined above, we get , , and angle .

We are given that AB = 10 cm and AC = 16 cm. Now we will find length of BD.

As we know the definition of rhombus we get AB = BC = CD = AD.

Therefore, we get, AB = BC = CD = AD = 10 cm

Also we know that diagonals of rhombus bisect each other at right angles therefore, we get,

,

and

Here, we know the length of AC therefore, we get, .

Now we will use Pythagoras theorem in the right angled triangle AOD as below,

…......(1)

Now we will substitute the values of AD and AO in equation (1) we get,

…......(2)

Now we will subtract 64 from both sides of the equation (2)

Now we will take the square root.

OD = 6

We know that

Therefore, length of the other diagonal is .

#### Page No 4.120:

#### Question 16:

Calculate the height of an equilateral triangle each of whose sides measures 12 cm.

#### Answer:

We are asked to find the height of the equilateral triangle.

Let us draw the figure. Let us draw the altitude AD. We know that altitude is also median of the equilateral triangle.

Therefore,.

In right angled triangle ADB, we will Pythagoras theorem, as shown below,

Now we will substitute the values.

144 = *AD*^{2 }+ 36

*AD*^{2 }= 144 − 36 = 108

*AD* = 10.39 cm

Therefore, the height of the equilateral triangle is.

#### Page No 4.120:

#### Question 17:

In the given figure, ∠B < 90° and segment AD ⊥ BC, show that

(i) ${b}^{2}={h}^{2}+{a}^{2}+{x}^{2}-2ax$

(ii) ${b}^{2}={a}^{2}+{c}^{2}-2ax$

#### Answer:

(i) Since AD perpendicular to BC we obtained two right angled triangles, triangle ADB and triangle ADC.

We will use Pythagoras theorem in the right angled triangle ADC

…......(1)

Let us substitute AD = *h*, AC = *b* and DC = (*a − x*) in equation (1) we get,

…......(2)

(ii) Let us use Pythagoras theorem in the right angled triangle ADB as shown below,

…......(3)

Let us substitute AB = *c*, AD = *h* and BD = *x* in equation (3) we get,

Let us rewrite the equation (2) as below,

…......(4)

Now we will substitute in equation (4) we get,

Therefore, .

#### Page No 4.120:

#### Question 18:

In an equilateral ∆ABC, AD ⊥ BC prove that AD^{2} = 3BD^{2}.

#### Answer:

We have to prove that .

In right angled, using Pythagoras theorem we get,

….....(1)

We know that in an equilateral triangle every altitude is also median.

Therefore, AD bisects BC.

Therefore, we have

Since is an equilateral triangle,

Therefore, we can write equation (1) as

…......(2)

But

Therefore, equation (2) becomes,

Simplifying the equation we get,

Therefore, .

#### Page No 4.121:

#### Question 19:

∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that

(i) AB^{2} = BC . BD

(ii) AC^{2} = BC . DC

(iii) AD^{2} = BD . CD

(iv) $\frac{{\mathrm{AB}}^{2}}{{\mathrm{AC}}^{2}}=\frac{\mathrm{BD}}{\mathrm{DC}}$

#### Answer:

(i)

In and ,

$\angle ACB=\angle A=90\xb0$

(Common angle)

So, by AA criterion

.....(1)

(ii) In and ,

(Common angle)

So, by AA criterion

.....(2)

(iii) We have shown that is similar to and is similar to therefore, by the property of transitivity is similar to .

.....(3)

(iv) Now to obtained AB^{2}/AC^{2} = BD/DC, we will divide equation (1) by equation (2) as shown below,

Canceling BC we get,

Therefore,

#### Page No 4.121:

#### Question 20:

A guy wire attached to a vertical pole of height 18 m is 24 m long has a stake attached to the other end. How far from the base of pole should the stake be driven so that the wire will be taut?

#### Answer:

We will draw the figure from the given information as below,

Let AB be the vertical pole of length 18 m and let the stake be at the point C so the wire will be taut.

Therefore, we have , and we have to find BC.

Now we will use Pythagoras theorem,

Let us substitute the values we get,

Subtracting 324 from both sides of the equation we get,

We can rewrite the 252 as , therefore, our equation becomes,

Now we will take the square root,

Therefore, the stake should be far from the base of the pole so that the wire will be taut.

#### Page No 4.121:

#### Question 21:

Determine whether the triangle having sides (*a* − 1) cm, $2\sqrt{a}\mathrm{cm}$ and (*a* + 1) cm is a right angled triangle.

#### Answer:

Let

Larger side is

We know that any number plus 1 is always greater than that number minus 1 and product of 2 and its square root.

For example : If *a* = 36

If a = 5

In order to prove that the given sides forms a right angled triangle we have to prove that .

Let us solve the left hand side first.

Now we will simplify the right hand side as shown below,

We can see that left hand side is equal to right hand side.

Therefore, the given sides determined the right angled triangle.

#### Page No 4.121:

#### Question 22:

In an acute-angled triangle, express a median in terms of its sides.

#### Answer:

Let ΔABC be acute angled triangle where AD is its median with respect side BC.

It is known that in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.

This is the required expression.

#### Page No 4.121:

#### Question 23:

In right-angled triangle ABC is which ∠C = 90°, if D is the mid-point of BC, prove that AB^{2} = 4 AD^{2} − 3AC^{2}.

#### Answer:

∆ABC is a right-angled triangle with ∠C = 90°. D is the mid-point of BC.

We need to prove that.

Join AD.

Since D is the midpoint of the side BC, we get

BD = DC

∴

Using Pythagoras theorem in triangles right angled triangle ABC

.....(1)

Again using Pythagoras theorem in the right angled triangle ADC

.....(2)

From (1) and (2), we get

Hence, .

#### Page No 4.121:

#### Question 24:

In the given figure, D is the mid-point of side BC and AE ⊥ BC. If BC = *a*, AC = *b*, AB = *c*, ED = *x*, AD = *p* and AE = *h*, prove that :

(i) ${b}^{2}={p}^{2}+ax+\frac{{a}^{2}}{4}$

(ii) ${c}^{2}={p}^{2}-ax+\frac{{a}^{2}}{4}$

(iii) ${b}^{2}+{c}^{2}=2{p}^{2}+\frac{{a}^{2}}{2}$

#### Answer:

(i) It is given that D is the midpoint of BC and.

Therefore, …......(1)

Using Pythagoras theorem in the right angled triangle AED,

…......(2)

Let us substitute , and in equation (2), we get

Let us take another right angled triangle that is triangle AEC.

Using Pythagoras theorem,

…......(3)

Let us substitute and in equation (3) we get,

Here we know that and .

Substituting , and we get

…......(4)

From equation (1) we can substitute in equation (4),

…......(5)

(ii) Using Pythagoras theorem in right angled triangle AEB,

…...... (6)

We know that AB = *c* and AE = *h* now we will find BE.

Therefore,

We know that and substituting these values in we get,

Now we will substitute AB = c, AE = h and in equation (6) we get,

…......(7)

Let us rewrite the equation (7) as below,

…......(8)

From equation (1) we can substitute in equation (8),

…......(9)

(iii) Now we will add equations (5) and (9) as shown below,

Therefore, .

#### Page No 4.121:

#### Question 25:

In ∆ABC, ∠A is obtuse, PB ⊥ AC and QC ⊥ AB. Prove that:

(i) AB ✕ AQ = AC ✕ AP

(ii) BC^{2} = (AC ✕ CP + AB ✕ BQ)

#### Answer:

Given: ΔABC where ∠BAC is obtuse. PB ⊥AC and QC⊥AB.

To prove: (i) AB × AQ = AC × AP and (ii) BC^{2} = AC × CP + AB × BQ

Proof: In ΔACQ and ΔABP,

∠CAQ = ∠BAP (Vertically opposite angles)

∠Q = ∠P (= 90°)

∴ ΔACQ ∼ ΔABP [AA similarity test]

In right ΔBCQ,

⇒ BC^{2} = CQ^{2} + QB^{2}

⇒ BC^{2} = CQ^{2} + (QA + AB)^{2}

⇒ BC^{2} = CQ^{2} + QA^{2} + AB^{2} + 2QA × AB

⇒ BC^{2} = AC^{2} + AB^{2} + QA × AB + QA × AB [In right ΔACQ, CQ^{2} + QA^{2} = AC^{2}]

⇒ BC^{2} = AC^{2} + AB^{2} + QA × AB + AC × AP (Using (1))

⇒ BC^{2} = AC (AC + AP) + AB (AB + QA)

⇒ BC^{2} = AC × CP + AB × BQ

#### Page No 4.121:

#### Question 26:

In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that BC^{2} = 4(AD^{2} − AC^{2}).

#### Answer:

It is given that ∆ABC is a right-angled at C and D is the mid-point of BC.

In the right angled triangle ADC, we will use Pythagoras theorem,

…......(1)

Since D is the midpoint of BC, we have

Substituting in equation (1), we get

#### Page No 4.121:

#### Question 27:

In a quadrilateral ABCD, ∠B = 90°, AD^{2} = AB^{2} + BC^{2} + CD^{2}, prove that ∠ACD = 90°.

#### Answer:

In order to prove angle it is enough to prove that .

Given,

.....(1)

Since, so applying Pythagoras theorem in the right angled triangle ABC, we get

.....(2)

From (1) and (2), we get

Therefore, angle . (Converse of pythagoras theorem)

#### Page No 4.121:

#### Question 28:

An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after $1\frac{1}{2}\mathrm{hours}$?

#### Answer:

Let us draw the figure first.

An aeroplane which flies due north at a speed of 1000 km/hr covers the distance AB after hr and another aeorplane that flies due west at the speed of 1200 km/hr covers the distance BC after hr.

We know that

Let us calculate AB first as shown below,

Similarly we can calculate BC.

Now we have find AC. To find AC we will use Pythagoras theorem,

Taking square root we get,

Therefore, after hrs the aeroplanes will be approximately far apart.

#### Page No 4.123:

#### Question 1:

In each of the figures [(i)-(iv)] given below, a line segment is drawn parallel to one side of the triangle and the lengths of certain line-segment are marked. Find the value of *x* in each of the following :

(i)

(ii)

(iii)

(iv)

#### Answer:

In each of the figure, we have to find the value of *x*

(*i*)

By cross multiplication on both sides, we get

Hence the value of is .

(*ii*)

By cross multiplication on both sides, we get

Hence the value of is .

(*iii*)

By cross multiplication on both sides, we get

Hence the value of is .

(*iv*)

By cross multiplication on both sides, we get

Hence the value of is .

#### Page No 4.123:

#### Question 2:

What values of *x* will make DE || AB in the given figure?

#### Answer:

If , then

Hence, the value of is .

#### Page No 4.123:

#### Question 3:

In ∆ABC, points P and Q are on CA and CB, respectively such that CA = 16 cm, CP = 10 cm, CB = 30 cm and CQ = 25 cm. Is PQ || AB?

#### Answer:

Given: and, we get

We will check whether or not to conclude whether.

$\frac{CP}{AC}=\frac{10\mathrm{cm}}{16\mathrm{cm}}=\frac{5}{8}\phantom{\rule{0ex}{0ex}}\frac{CQ}{CB}=\frac{25\mathrm{cm}}{30\mathrm{cm}}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}\therefore \frac{CP}{AC}\ne \frac{CQ}{CB}\phantom{\rule{0ex}{0ex}}$

Hence,*PQ*is not parallel to

*AB*.

#### Page No 4.123:

#### Question 4:

In the given figure, DE || BD. Determine AC and AE.

#### Answer:

Given, .

In ∆*ABC *and ∆*ADE
$\angle ADE=\angle C$ *(Corresponding angles)

*$\angle A=\angle A$*(Common)

(AA Similarity)

Hence the value of and is

#### Page No 4.124:

#### Question 5:

In the given figure, given that ∆ABC ∼ ∆PQR and quad ABCD ∼ quad PQRS. Determine the value of *x*, *y*,* z* in each case.

#### Answer:

(i) We have,

So the ratio of the sides of the triangles will be proportional to each other.

Therefore put the values of the known terms in the above equation to get,

On solving these simultaneous equations, we get

(ii) We have,

So the ratio of the sides of the quadrilaterals will be proportional to each other.

Therefore put the values of the known terms in the above equation to get,

On solving these simultaneous equations, we get

#### Page No 4.124:

#### Question 6:

In ∆ABC, P and Q are points on sides AB and AC respectively such that PQ || BC. If AP = 4 cm, PB = 6 cm and PQ = 3 cm, determine BC.

#### Answer:

In triangle *ABC*, *P* and *Q* are points on sides *AB* and *AC* respectively such that.

In Δ*APQ* and Δ*ABC*,

$\angle APQ=\angle B\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle PAQ=\angle BAC\left(\mathrm{Common}\right)$

So, $\u2206\mathrm{APQ}~\u2206\mathrm{ABC}$ (AA Similarity)

$\frac{AP}{AB}=\frac{PQ}{BC}$

Substituting value and , we get

By cross multiplication we get

Hence, the value of *BC* is .

#### Page No 4.124:

#### Question 7:

In each of the following figures, you find who triangles. Indicate whether the triangles are similar. Give reasons in support of your answer.

(i)

(ii)

(iii)

(iv)

(v)

#### Answer:

(*i*) In two triangles, we observe that

In similar triangle corresponding sides are proportional to each other.

Therefore, by SSS-criterion of similarity,

two triangles are similar

(*ii*)

PQ || BC (Corresponding angles formed are equal)

In Δ*APQ* and Δ*ABC*,

$\angle APQ=\angle B\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle PAQ=\angle BAC\left(\mathrm{Common}\right)$

So, $\u2206\mathrm{APQ}~\u2206\mathrm{ABC}$ (AA Similarity)

two triangles are similar

(*iii*) In two triangle, we observe that

In Δ*ABC* and ΔCDE

$\frac{CD}{CE}=\frac{CB}{CA}$

$\angle ACB=\angle DCE$ (Vertically opposite angles)

Δ*ABC* $~$ ΔCDE (SAS Similarity)

two triangles are similar

(*iv*) In two triangle, we observe that

In two triangles corresponding sides are not proportional to each other.

two triangles are not similar.

(v) In two triangle, we observe that

In two triangles corresponding sides are not proportional to each other.

two triangles are not similar.

#### Page No 4.125:

#### Question 8:

In ∆PQR, M and N are points on sides PQ and PR respectively such that PM = 15 cm and NR = 8 cm. If PQ = 25 cm and PR = 20 cm state whether MN || QR.

#### Answer:

Given and .

$\frac{PM}{PQ}=\frac{15\mathrm{cm}}{25\mathrm{cm}}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}\frac{PN}{PR}=\frac{12\mathrm{cm}}{20\mathrm{cm}}=\frac{3}{5}\left(PN=PR-NR=20-8=12\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\therefore \frac{PM}{PQ}=\frac{PN}{PR}$

So, by the converse of basic proportionality theorem MN || QR.

#### Page No 4.125:

#### Question 9:

In ∆ABC, P and Q are points on sides AB and AC respectively such that PQ || BC. If AP = 3 cm, PB = 5 cm and AC = 8 cm, find AQ.

#### Answer:

In, *P* and *Q* are points on sides *AB* and *AC* respectively such that

Then we have

and

Hence the value of *AQ* is

#### Page No 4.125:

#### Question 10:

In the given figure, ∆AMB ∼ ∆CMD; determine MD in terms of *x*, *y* and *z*.

#### Answer:

We are given ∆AMB ∼ ∆CMD

We have to determine the value of *MD* in terms of *x, y* and *z.*

Given

$\Rightarrow \frac{BM}{MD}=\frac{AM}{CM}\phantom{\rule{0ex}{0ex}}\frac{x}{MD}=\frac{y}{z}$

By cross multiplication we get

Hence, the value of *MD* is .

#### Page No 4.125:

#### Question 11:

In ∆ABC, the bisector of ∠A intersects BC in D. If AB = 18 cm, AC = 15 cm and BC = 22 cm, find BD.

#### Answer:

We have to find the value of *BD.*

Given and .

In , *AD* the bisector of .

On cross multiplication, we get

*BD *= 12 cm

Hence, the value of *BD* is .

#### Page No 4.125:

#### Question 12:

In the given figure, *l* || *m*

(i) Name three pairs of similar triangles with proper correspondence; write similarities.

(ii) Prove that $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{RQ}}$

#### Answer:

(i) Three pair of similar triangles are-

$\u2206ABK~\u2206PQK$ (AAA Similarity)

$\u2206CBK~\u2206RQK$ (AAA Similarity)

$\u2206ACK~\u2206PRK$ (AAA Similarity)

(ii) Since the pair of similar triangles mentioned above can give us the desired result. The ratios of the corresponding side of the similar triangle are equal.

So,

Therefore,

……equation (1)

Similarly in ,

……equation (2)

Similarly ,

……equation (3)

From the above equations 1 and 2 we have,

=

……eqaution (4)

Combining it with equation (4)

hence proved

#### Page No 4.125:

#### Question 13:

In the given figure,

AB || DC prove that

(ii) DM × BV = BM ✕ DU

#### Answer:

(*i*) Given

In triangle DMU and BMV, we have

Each angle is equal to 90°

Each are vertically opposite angles.

Therefore, by AA-criterion of similarity

(*ii*)Since

By cross multiplication, we get

Hence proved that

#### Page No 4.126:

#### Question 14:

ABCD is a trapezium in which AB || DC. P and Q are points on sides AD and BC such that PQ || AB. If PD = 18, BQ = 35 and QC = 15, find AD.

#### Answer:

In trapezium *ABCD*, AB || DC. P and Q are points on sides AD and BC such that PQ || AB.

Join AC. Suppose AC intersects PQ in O.

$\mathrm{In}\u2206ACD,OP\left|\right|CD\phantom{\rule{0ex}{0ex}}\therefore \frac{AP}{PD}=\frac{AO}{OC}.....\left(1\right)\left(\mathrm{BPT}\right)\phantom{\rule{0ex}{0ex}}\mathrm{In}\u2206ABC,OQ\left|\right|AB\phantom{\rule{0ex}{0ex}}\therefore \frac{BQ}{QC}=\frac{AO}{OC}.....\left(2\right)\left(\mathrm{BPT}\right)$

From (1) and (2), we get

Hence, the value of *AD* is .

#### Page No 4.126:

#### Question 15:

In ∆ABC, D and E are points on sides AB and AC respectively such that AD ✕ EC = AE ✕ DB. Prove that DE || BC.

#### Answer:

Given: In, *D* and *E* are points on sides *AB* and *AC* such that

To Prove:

Proof:

Since

∴ *DE *||* BC *(Converse of basic proportionality theorem)

#### Page No 4.126:

#### Question 16:

ABCD is a trapezium having AB || DC. Prove that O, the point of intersection of diagonals, divides the two diagonals in the same ratio. Also prove that $\frac{\mathrm{ar}\left(\u2206\mathrm{OCD}\right)}{\mathrm{ar}\left(\u2206\mathrm{OAB}\right)}=\frac{1}{9}$, if AB = 3 CD.

#### Answer:

We are given *ABCD* is a trapezium with *AB*||*DC*

Consider the triangles *AOB* and *COD* in which

Therefore,

Hence we have proved that *O*, the point of intersection of diagonals, divides the two diagonals in the same ratio.

We are given *AB* = 3*CD* and we have to prove that

We already have proved that *AOB* and *COD* are similar triangles

So

Hence, Prove that

#### Page No 4.126:

#### Question 17:

Corresponding sides of two triangles are in the ratio 2 : 3. If the area of the smaller triangle is 48 cm^{2}, determine the area of the larger triangle.

#### Answer:

The ratio of the areas of two similar triangles is equal to the ratio of the square of any two corresponding sides.

.

Hence the area of the larger triangle is .

#### Page No 4.126:

#### Question 18:

The area of two similar triangles are 36 cm^{2} and 100 cm^{2}. If the length of a side of the smaller triangle in 3 cm, find the length of the corresponding side of the larger triangle.

#### Answer:

Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

(Corresponding side of larger triangle)^{2} =

(Corresponding side of larger triangle)^{2} =

(Corresponding side of larger triangle)^{2} = 25

⇒ Corresponding side of larger triangle = 5

Hence, the length of the corresponding side of the larger triangle is

#### Page No 4.126:

#### Question 19:

Corresponding sides of two similar triangles are in the ratio 1 : 3. If the area of the smaller triangle in 40 cm^{2}, find the area of the larger triangle.

#### Answer:

Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Hence the area of the larger triangle is

#### Page No 4.126:

#### Question 20:

In the given figure, each of PA, QB, RC and SD is perpendicular to *l*. If AB = 6 cm, BC = 9 cm, CD = 12 cm and PS = 36 cm, then determine PQ, QR and RS.

#### Answer:

Given and

*PA*, *QB, RC* and *SD* is perpendicular to *l*,

Therefore, by the corollory of basic proportionality theorem, we have

Now for QR

Again for *RS*

Hence, the values of *PQ, QR* and *RS* are respectively.

#### Page No 4.126:

#### Question 21:

In each of the figures given below, an altitude is drawn to the hypotenuse by a right-angled triangle. The length of different line-segment are marked in each figure. Determine *x*, *y*, *z* in each case.

(i)

(ii)

#### Answer:

(*i*) is right angled triangle right angled at B

is right triangle right angled at D

is right triangle right angled at D

By canceling equation and by elimination method, we get

*y* canceling and by elimination method we get

Now, substituting in equation (*iv*) we get

Now, substituting in equation (*ii*) we get

Hence the values of *x, y, z* is

(*ii*) is a right triangle, right angled at Q

…… (*i*)

is a right triangle right angled at S

is a right triangle right angled at S

…… (*ii*)

Now substituting in equation (*i*) we get

Now substituting in equation (*iii*) we get

Now substituting and in equation (*ii*) we get

Hence the value of *x, y* and *z* are

#### Page No 4.127:

#### Question 22:

Prove that in an equilateral triangle, three times the square of a side is equal to four times the square of its altitudes.

#### Answer:

Let *ABC* be an equilateral triangle and let.

In and we have

And

Since is a right triangle right-angled at *D*. So

Hence proved.

#### Page No 4.127:

#### Question 23:

In ∆ABC, AD and BE are altitude. Prove that $\frac{\mathrm{ar}\left(\u2206\mathrm{DEC}\right)}{\mathrm{ar}\left(\u2206\mathrm{ABC}\right)}=\frac{{\mathrm{DC}}^{2}}{{\mathrm{AC}}^{2}}$.

#### Answer:

**Given: **ΔABC in which AD and BE are altitudes on sides BC and AC respectively.

Since ∠ADB = ∠AEB = 90°, there must be a circle passing through point D and E having AB as diameter.

We also know that, angle in a semi-circle is a right angle.

Now, join DE.

So, ABDE is a cyclic quadrilateral with AB being the diameter of the circle.

∠A + ∠BDE = 180° [Opposite angles in a cyclic quadrilateral are supplementary]

⇒ ∠A + (∠BDA + ∠ADE) = 180°

⇒ ∠BDA + ∠ADE = 180° − ∠A ..... (1)

Again,

∠BDA + ∠ADC = 180° [Linear pair]

⇒ ∠BDA + ∠ADE + ∠EDC = 180°

⇒ ∠BDA + ∠ADE = 180° − ∠EDC ..... (2)

Equating (1) and (2), we get

180° − ∠A = 180° − ∠EDC

⇒ ∠A = ∠EDC

Similarly, ∠B = ∠CED

Now, in ΔABC and ΔDEC, we have

∠A = ∠EDC

∠B = ∠CED

∠C = ∠C

∴ ΔABC ∼ ΔDEC

#### Page No 4.127:

#### Question 24:

The diagonals of quadrilateral ABCD intersect at O. Prove that $\frac{\mathrm{ar}\left(\u2206\mathrm{ACB}\right)}{\mathrm{ar}\left(\u2206\mathrm{ACD}\right)}=\frac{\mathrm{BO}}{\mathrm{DO}}$.

#### Answer:

We are given the following quadrilateral with *O* as the intersection point of diagonals

To Prove:

Given *ACB* and *ACD* are two triangles on the same base *AC*

Consider *h* as the distance between two parallel sides

Now we see that the height of these two triangles *ACB* and *ACD* are same and are equal to *h*

So

Now consider the triangles *AOB* and *COD* in which

Therefore,

From equation (1) and (2) we get

Hence prove that

#### Page No 4.127:

#### Question 25:

In ∆ABC, ray AD bisects ∠A and intersects BC in D. If BC = *a*, AC = *b* and AC = *c*, prove that

(i) $\mathrm{BD}=\frac{ac}{b+c}$

(ii) $\mathrm{DC}=\frac{ab}{b+c}$

#### Answer:

Given: In ray *AD* bisects angle *A* and intersects *BC* in *D*, If and

To Prove:

(i)

(ii)

(i) The corresponding figure is as follows

Proof: In triangle *ABC, AD* is the bisector of

Therefore

Substitute and we get,

By cross multiplication we get.

We proved that

(ii) Since *BC = CD + BD*

#### Page No 4.127:

#### Question 26:

There is a staircase as shown in the given figure, connecting points A and B. Measurements of steps are marked in the figure. Find the straight line distance between A and B.

#### Answer:

We are given the following figure with the related information

In the above figure complete the triangle *ABC* with right angled at *C*

So

*AC* = 2 + 2 + 2 + 2 = 8 and

*BC* = 1 + 1.6 + 1.6 + 1.8 = 6

Using Pythagoras theorem for triangle *ABC* to find

Hence the distance between *A* and *B* is .

#### Page No 4.127:

#### Question 27:

In ∆ABC, ∠A = 60°. Prove that BC^{2} = AB^{2} + AC^{2} − AB . AC.

#### Answer:

In Δ*ABC* in which ∠*A* is an acute angle with 60°.

Now apply Pythagoras theorem in triangle *BCD*

Hence

#### Page No 4.127:

#### Question 28:

In ∆ABC, ∠C is an obtuse angle. AD ⊥ BC and AB^{2} = AC^{2} + 3 BC^{2}. Prove that BC = CD.

#### Answer:

**Given**: ΔABC where ∠C is an obtuse angle, AD ⊥ BC and AB^{2} = AC^{2} + 3BC^{2}

**To prove**: BC = CD

**Proof: **

In ΔABC, ∠C is obtuse.

Therefore,

AB^{2} = AC^{2} + BC^{2} + 2BC × DC (Obtuse angle theorem) …(1)

AB^{2} = AC^{2} + 3BC^{2} (Given) …(2)

From (1) and (2), we get

AC^{2} + 3BC^{2} = AC^{2} + BC^{2} + 2BC × DC

⇒ 3BC^{2} = BC^{2} + 2BC × DC

⇒ 2BC^{2} = 2BC × DC

⇒ BC = DC

#### Page No 4.127:

#### Question 29:

A point D is on the side BC of an equilateral triangle ABC such that $\mathrm{DC}=\frac{1}{4}\mathrm{BC}$. Prove that AD^{2} = 13 CD^{2}.

#### Answer:

We are given *ABC* is an equilateral triangle with

We have to prove

Draw

In and we have

So by right side criterion of similarity we have

Thus we have

and

Since, therefore

We know that *AB* = *BC* = *AC*

We know that

Substitute in we get

Hence we have proved that

#### Page No 4.127:

#### Question 30:

In ∆ABC, if BD ⊥ AC and BC^{2} = 2 AC . CD, then prove that AB = AC.

#### Answer:

Since is right triangle right angled at *D*

In right , we have

Now substitute

#### Page No 4.127:

#### Question 31:

In a quadrilateral ABCD, given that ∠A + ∠D = 90°. Prove that AC^{2} + BD^{2} = AD^{2} + BC^{2}.

#### Answer:

Given: A quadrilateral ABCD where ∠A + ∠D = 90°.

To prove: AC^{2} + BD^{2} = AD^{2} + BC^{2}

Construction: Extend AB and CD to intersect at O.

Proof:

In ΔAOD, ∠A + ∠O + ∠D = 180°

⇒ ∠O = 90° [∠A + ∠D = 90°]

Apply Pythagoras Theorem in ΔAOC and ΔBOD,

AC^{2} = AO^{2} + OC^{2}

BD^{2} = OB^{2} + OD^{2}

∴ AC^{2} + BD^{2} = (AO^{2} + OD^{2}) + (OC^{2} + OB^{2})

⇒ AC^{2} + BD^{2} = AD^{2} + BC^{2}

This proves the given relation.

#### Page No 4.127:

#### Question 32:

In ∆*ABC*, given that *AB *= *AC *and *BD *⊥ *AC*. Prove that *BC*^{2} = 2 *AC*. *CD*

#### Answer:

Since is right triangle right angled at *D*

Substitute

Now, in , we have

Therefore,

Hence proved.

#### Page No 4.127:

#### Question 33:

ABCD is a rectangle. Points M and N are on BD such that AM ⊥ BD and CN ⊥ BD. Prove that* *BM^{2} + BN^{2}^{ }= DM^{2}^{ }+ DN^{2}.

#### Answer:

Given: A rectangle ABCD where AM ⊥ BD and CN ⊥ BD.

To prove: BM^{2} + BN^{2} = DM^{2} + DN^{2}

Proof:

Apply Pythagoras Theorem in ΔAMB and ΔCND,

AB^{2} = AM^{2} + MB^{2}

CD^{2} = CN^{2} + ND^{2}

Since AB = CD, AM^{2} + MB^{2} = CN^{2} + ND^{2}

⇒ AM^{2} − CN^{2} = ND^{2} − MB^{2} … (i)

Again apply Pythagoras Theorem in ΔAMD and ΔCNB,

AD^{2} = AM^{2} + MD^{2}

CB^{2} = CN^{2} + NB^{2}

Since AD = BC, AM^{2} + MD^{2} = CN^{2} + NB^{2}

⇒ AM^{2} − CN^{2} = NB^{2} − MD^{2} … (ii)

Equating (i) and (ii),

ND^{2} − MB^{2} = NB^{2} − MD^{2}

I.e., BM^{2} + BN^{2} = DM^{2} + DN^{2}

This proves the given relation.

#### Page No 4.127:

#### Question 34:

In ∆ABC, AD is a median. Prove that AB^{2} + AC^{2} = 2AD^{2} + 2DC^{2}.

#### Answer:

We have the following figure.

Since triangle *ABM* and *ACM* are right triangles right angled at *M*

…… (*i*)

…… (*ii*)

Adding (*i*) and (*ii*), we get

Since in triangle *ADM* we have

So,

*BM + CM = BC*

So,

Now we have

So,

Hence proved

#### Page No 4.127:

#### Question 35:

In ∆ABC, ∠ABC = 135°. Prove that AC^{2} = AB^{2} + BC^{2} + 4* ar* (∆ABC)

#### Answer:

We have the following figure.

Here is a right triangle right angled at *D*. Therefore by Pythagoras theorem we have

Again is a right triangle right angled at *D*.

Therefore, by Pythagoras theorem, we have

Since angle ABD is 45°and therefore angle *BAD* is also 45°.

Hence *AB = DB*

So,

Since

So,

Hence we have proved that

#### Page No 4.127:

#### Question 36:

In a quadrilateral ABCD, ∠B = 90°. If AD^{2} = AB^{2} + BC^{2} + CD^{2} then prove that ∠ACD = 90°.

#### Answer:

In quadrilateral *ABCD*, we have

∠B = 90°

So, (Pythagoras theorem)

and

(Given)

So,

Hence, ∠*ACD* = 90° (Converse of Pythagoras theorem)

#### Page No 4.127:

#### Question 37:

In a triangle ABC, N is a point on AC such that BN ⊥ AC. If BN^{2} = AN . NC, prove that ∠B = 90°.

#### Answer:

In , BN ⊥ AC.

Also,

We have to prove that .

In triangles *ABN* and *BNC*, we have

Adding above two equations, we get

So,

Hence

#### Page No 4.128:

#### Question 38:

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the road. Assuming that her string (from the tip of her road to the fly) is taut, how much string does she have out (in the given figure)? If she pulls the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds.

#### Answer:

Let us take

*AB* = Tip of fishing rod above the surface of the water

*BC* = The string rest from directly under the tip of the rod.

*AC *= The length of string

In *ABC* right triangle right angled at *B*

Hence 3 m string is out

The string pulled in 12 seconds is equal to to point *D *

In this case the diagram will look like the following

Now The length of the new string *AD* = *AC* − *CD* = 3.00 − 0.6 = 2.4 m

Now in triangle *ADB* we have

∴ Required distance =

Hence, the horizontal distance is .

#### Page No 4.128:

#### Question 1:

State basic proportionality theorem and its converse.

#### Answer:

TO STATE: The basic proportionality theorem and its converse.

BASIC PROPORTIONALITY THEOREM: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

CONVERSE OF BASIC PROPORTIONALITY THEOREM: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

#### Page No 4.128:

#### Question 2:

In the adjoining figure, find AC.

#### Answer:

**GIVEN: **In the figure we are given AD = 6cm, BD = 9cm, AE = 8cm

**TO FIND:** AC

According to** **BASIC PROPORTIONALITY THEOREM, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ∆ABC, DE || BC. So,

Now,

#### Page No 4.128:

#### Question 3:

In the adjoining figure, if AD is the bisector of ∠A, what is AC?

#### Answer:

**GIVEN: **AB = 6cm, BD = 3cm and DC = 2cm. Also, AD is the bisector of .

**TO FIND:** AC

**SOLUTION: **We know that the internal bisector of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Therefore,

#### Page No 4.128:

#### Question 4:

State *AAA* similarity criterion.

#### Answer:

AAA Similarity Criterion: If two triangles are equiangular, then they are similar.

In ΔABC and ΔDEF, if

Then,

#### Page No 4.128:

#### Question 5:

State SSS similarity criterion.

#### Answer:

SSS Similarity Criterion: If the corresponding sides of two triangles are proportional, then they are similar.

In ΔABC and ΔDEF, if

Then,

#### Page No 4.128:

#### Question 6:

State SAS similarity criterion.

#### Answer:

SAS Similarity Criterion: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then two triangles are similar.

In ΔABC and ΔDEF, if

Then,

#### Page No 4.128:

#### Question 7:

In the adjoining figure, DE is parallel to BC and AD = 1 cm, BD = 2 cm. What is the ratio of the area of ∆ABC to the area of ∆ADE?

#### Answer:

**GIVEN: **DE is parallel to BC, AD = 1cm and BD = 2cm.

**TO FIND:** Ratio of ΔABC to area of ΔADE

According to** **BASIC PROPORTIONALITY THEOREM, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ΔABC, DE || BC.

So

#### Page No 4.128:

#### Question 8:

In the figure given below DE || BC. If AD = 2.4 cm, DB = 3.6 cm, AC = 5 cm. Find AE.

#### Answer:

**GIVEN: **AD = 2.4cm, BD = 3.6cm and AC = 5cm.

**TO FIND:** AE

According to** **BASIC PROPORTIONALITY THEOREM If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ∆ABC, DE || BC.

#### Page No 4.128:

#### Question 9:

If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm, what is the length of QR?

#### Answer:

Given: ΔABC and ΔPQR are similar triangles. Area of ΔABC: Area of ΔPQR = 9:16 and BC = 4.5cm.

To find: Length of QR

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Hence,

#### Page No 4.128:

#### Question 10:

The areas of two similar triangles are 169 cm^{2} and 121 cm^{2} respectively. If the longest side of the larger triangle is 26 cm, what is the length of the longest side of the smaller triangle?

#### Answer:

Let ∆ABC and ΔPQR are similar triangles. The area of triangles is 169cm^{2} and 121cm^{2}, respectively.

Longest side of the larger triangle is 26cm

TO FIND: length of longest side of the smaller side.

Suppose longest side of the larger triangle is BC and longest side of the smaller triangle is QR.

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Hence,

#### Page No 4.128:

#### Question 11:

If ABC and DEF are similar triangles such that ∠A = 57° and ∠E = 73°, what is the measure of ∠C?

#### Answer:

GIVEN: There are two similar triangles ΔABC and ΔDEF.

^{, }

TO FIND: measure of

SAS Similarity Criterion: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then two triangles are similar.

In ΔABC and ΔDEF if

Then,

So,

Similarly

Now we know that sum of all angles of a triangle is equal to 180°,

#### Page No 4.128:

#### Question 12:

If the altitude of two similar triangles are in the ratio 2 : 3, what is the ratio of their areas?

#### Answer:

GIVEN: Altitudes of two similar triangles are in ratio 2:3.

TO FIND: Ratio of the areas of two similar triangles.

Let first triangle be ΔABC and the second triangle be ΔPQR

We know that the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.

#### Page No 4.128:

#### Question 13:

If ∆ABC and ∆DEF are two triangles such that $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}=\frac{3}{4}$, then write Area (∆ABC) : Area (∆DEF)

#### Answer:

GIVEN: ΔABC and ΔDEF are two triangles such that .

TO FIND:

We know that two triangles are similar if their corresponding sides are proportional.

Here, ΔABC and ΔDEF are similar triangles because their corresponding sides are given proportional, i.e.

Since the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.

#### Page No 4.128:

#### Question 14:

If ∆ABC and ∆DEF are similar triangles such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm and EF = 4 cm, write the perimeter of ∆DEF.

#### Answer:

GIVEN: ΔABC and ΔDEF are similar triangles such that AB = 3cm, BC = 2cm, CA = 2.5cm and EF = 4cm.

TO FIND: Perimeter of ΔDEF.

We know that if two triangles are similar then their corresponding sides are proportional.

Hence,

Substituting the values, we get

Similarly,

#### Page No 4.128:

#### Question 15:

State Pythagoras theorem and its converse.

#### Answer:

TO STATE: Pythagoras theorem and its converse.

PYTHAGORAS THEOREM: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

CONVERSE OF PYTHAGORAS THEOREM: In a triangle, if the square of one side is equal to the sum of the square of the other two sides then the angle opposite to the greatest side is a right angle.

#### Page No 4.128:

#### Question 16:

The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus.

#### Answer:

GIVEN: the lengths of the diagonals of a rhombus are 30 cm and 40 cm.

TO FIND: side of the rhombus.

Let the diagonals AC and CD of the rhombus ABCD meet at point O.

We know that the diagonals of the rhombus bisect each other perpendicularly.

Hence in right triangle AOD, by Pythagoras theorem

Hence the side of the rhombus is

#### Page No 4.130:

#### Question 17:

In the given figure, PQ || BC and AP : PB = 1 : 2. Find $\frac{\mathrm{area}\left(\u2206\mathrm{APQ}\right)}{\mathrm{area}\left(\u2206\mathrm{ABC}\right)}$.

#### Answer:

GIVEN: In the given figure PQ || BC, and AP: PB = 1:2

TO FIND:

We know that according to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other side, then it divides the two sides in the same ratio.

Since triangle APQ and ABC are similar

Hence,

Now, it is given that .

So;

Since the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.

$\frac{\mathrm{Area}\left(\mathrm{APQ}\right)}{\mathrm{Area}\left(\mathrm{ABC}\right)}={\left(\frac{\mathrm{AP}}{\mathrm{AB}}\right)}^{2}={\left(\frac{1}{3}\right)}^{2}=\frac{1}{9}$

Hence we got the result

#### Page No 4.130:

#### Question 18:

In the given figure, S and T are points on the sides PQ and PR respectively of ∆PQR such that PT = 2 cm, TR = 4 cm and ST is parallel to QR. Find the ratio of the areas of ∆PST and ∆PQR.

#### Answer:

Given: In ΔPQR, S and T are the points on the sides PQ and PR respectively such that PT = 2cm, TR = 4cm and ST is parallel to QR.

To find: Ratio of areas of ΔPST and ΔPQR

$\mathrm{In}\u2206\mathrm{PST}\mathrm{and}\u2206\mathrm{PQR},\phantom{\rule{0ex}{0ex}}\angle \mathrm{PST}=\angle \mathrm{Q}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{P}=\angle \mathrm{P}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\therefore \u2206\mathrm{PST}~\u2206\mathrm{PQR}\left(\mathrm{AA}\hspace{0.17em}\mathrm{Similarity}\right)\phantom{\rule{0ex}{0ex}}$Now, we know that the areas of two similar triangles are in the ratio of the squares of the corresponding sides. Therefore,

#### Page No 4.130:

#### Question 19:

In the given figure, ∆AHK is similar to ∆ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC.

#### Answer:

Given:

AK = 10 cm

BC = 3.5 cm

HK = 7 cm

To find: AC

Since, so their corresponding sides are proportional.

#### Page No 4.130:

#### Question 20:

In the given figure, DE || BC in ∆ABC such that BC = 8 cm, AB = 6 cm and DA = 1.5 cm. Find DE.

#### Answer:

Given: In ∆ABC, DE || BC. BC = 8 cm, AB = 6 cm and DA = 1.5 cm.

To find: DE

In ∆ABC and ∆ADE

$\angle \mathrm{B}=\angle \mathrm{ADE}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}=\angle \mathrm{A}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\therefore \u2206\mathrm{ABC}~\u2206\mathrm{ADE}\left(\mathrm{AA}\mathrm{Similarity}\right)$

So,

#### Page No 4.131:

#### Question 21:

In the given figure, DE || BC and $\mathrm{AD}=\frac{1}{2}\mathrm{BD}$. If BC = 4.5 cm, find DE.

#### Answer:

Given: In ∆ABC, DE || BC. and BC = 4.5 cm.

To find: DE

In ∆ABC and ∆ADE$\angle \mathrm{B}=\angle \mathrm{ADE}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}=\angle \mathrm{A}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\therefore \u2206\mathrm{ABC}~\u2206\mathrm{ADE}\left(\mathrm{AA}\mathrm{Similarity}\right)$

#### Page No 4.131:

#### Question 22:

In the given figure, LM = LN = 46°. Express *x* in terms of *a*, *b *and *c* where *a*, *b*, *c* are lengths of LM, MN and NK respectively.

#### Answer:

Given: In the given figure

TO EXPRESS: *x* in terms of *a*, *b*, *c* where *a*,* b*, and *c* are the lengths of LM, MN and NK respectively.

Here we can see that. It forms a pair of corresponding angles.

Hence, LM || PN

In $\u2206\mathrm{LMK}$ and $\u2206\mathrm{PNK}$,

$\angle \mathrm{LMK}=\angle \mathrm{PNK}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{LKM}=\angle \mathrm{PKN}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\therefore \u2206\mathrm{LMK}~\u2206\mathrm{PNK}\left(\mathrm{AA}\mathrm{Similarity}\right)$

$\frac{\mathrm{ML}}{\mathrm{NP}}=\frac{\mathrm{MK}}{\mathrm{NK}}\phantom{\rule{0ex}{0ex}}\frac{a}{x}=\frac{b+c}{c}\phantom{\rule{0ex}{0ex}}x=\frac{ac}{b+c}$

Hence we got the result as .

#### Page No 4.131:

#### Question 1:

Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.

(a) 2 : 3

(b) 4 : 9

(c) 81 : 16

(d) 16 : 81

#### Answer:

Given: Sides of two similar triangles are in the ratio 4:9

To find: Ratio of area of these triangles

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence the correct answer is option

#### Page No 4.131:

#### Question 2:

The areas of two similar triangles are in respectively 9 cm^{2} and 16 cm^{2}. The ratio of their corresponding sides is

(a) 3 : 4

(b) 4 : 3

(c) 2 : 3

(d) 4 : 5

#### Answer:

Given: Areas of two similar triangles are 9cm^{2} and 16cm^{2}.

To find: Ratio of their corresponding sides.

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Taking square root on both sides, we get

$\frac{\mathrm{side}1}{\mathrm{side}2}=\frac{3}{4}$

So, the ratio of their corresponding sides is 3 : 4.

Hence the correct answer is .

#### Page No 4.131:

#### Question 3:

The areas of two similar triangles ∆ABC and ∆DEF are 144 cm^{2} and 81 cm2 respectively. If the longest side of larger ∆ABC be 36 cm, then the longest side of the smaller triangle ∆DEF is

(a) 20 cm

(b) 26 cm

(c) 27 cm

(d) 30 cm

#### Answer:

Given: Areas of two similar triangles ΔABC and ΔDEF are 144cm^{2} and 81cm^{2}.

If the longest side of larger ΔABC is 36cm

To find: the longest side of the smaller triangle ΔDEF

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Taking square root on both sides, we get

$\frac{12}{9}=\frac{36}{\mathrm{longest}\mathrm{side}\mathrm{of}\mathrm{smaller}\u2206\mathrm{DEF}}$

= 27 cm

Hence the correct answer is

#### Page No 4.132:

#### Question 4:

∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangle ABC and BDE is

(a) 2 : 1

(b) 1 : 2

(c) 4 : 1

(d) 1 : 4

#### Answer:

Given: ΔABC and ΔBDE are two equilateral triangles such that D is the midpoint of BC.

To find: Ratio of areas of ΔABC and ΔBDE.

ΔABC and ΔBDE are equilateral triangles; hence they are similar triangles.

Since D is the midpoint of BC, BD = DC.

Hence the correct answer is .

#### Page No 4.132:

#### Question 5:

If ∆ABC and ∆DEF are similar such that 2AB = DE and BC = 8 cm, then EF =

(a) 16 cm

(b) 12 cm

(c) 8 cm

(d) 4 cm

#### Answer:

Given: ΔABC and ΔDEF are similar triangles such that 2AB = DE and BC = 8 cm.

To find: EF

We know that if two triangles are similar then there sides are proportional.

Hence, for similar triangles ΔABC and ΔDEF

Hence the correct answer is .

#### Page No 4.132:

#### Question 6:

If ∆ABC and ∆DEF are two triangles such that $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}=\frac{2}{5}$, then Area (∆ABC) : Area (∆DEF) =

(a) 2 : 5

(b) 4 : 25

(c) 4 : 15

(d) 8 : 125

#### Answer:

Given: ΔABC and ΔDEF are two triangles such that .

To find:

We know that if the sides of two triangles are proportional, then the two triangles are similar.

Since , therefore, ΔABC and ΔDEF are similar.

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence the correct answer is .

#### Page No 4.132:

#### Question 7:

XY is drawn parallel to the base BC of a ∆ABC cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY =

(a) 2 cm

(b) 4 cm

(c) 6 cm

(d) 8 cm

#### Answer:

Given: XY is drawn parallel to the base BC of a ΔABC cutting AB at X and AC at Y. AB = 4BX and YC = 2 cm.

To find: AY

In ΔAXY and ΔABC,

$\angle \mathrm{AXY}=\angle \mathrm{B}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}=\angle \mathrm{A}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\therefore \u2206\mathrm{AXY}~\u2206\mathrm{ABC}\left(\mathrm{AA}\mathrm{similarity}\right)$

We know that if two triangles are similar, then their sides are proportional.

It is given that AB = 4BX.

Let AB = 4*x* and BX = *x*.

Then, AX = 3*x*

Hence the correct answer is .

#### Page No 4.132:

#### Question 8:

Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, the distance between their tops is

(a) 12 m

(b) 14 m

(c) 13 m

(d) 11 m

#### Answer:

Given: Two poles of heights 6m and 11m stand vertically upright on a plane ground. Distance between their foot is 12 m.

To find: Distance between their tops.

Let CD be the pole with height 6m.

AB is the pole with height 11m, distance between their foot i.e. DB is 12 m.

Let us assume a point E on the pole AB which is 6m from the base of AB.

Hence

AE = AB − 6 = 11 − 6 = 5 m

Now in right triangle AEC, Applying Pythagoras theorem

AC^{2} = AE^{2} + EC^{2}

AC^{2} = 5^{2} + 12^{2} (since CDEB forms a rectangle and opposite sides of rectangle are equal)

AC^{2} = 25 + 144

AC^{2} = 169

Thus, the distance between their tops is 13m.

Hence correct answer is .

#### Page No 4.132:

#### Question 9:

In ∆ABC, D and E are points on side AB and AC respectively such that DE || BC and AD : DB = 3 : 1. If EA = 3.3 cm, then AC =

(a) 1.1 cm

(b) 4 cm

(c) 4.4 cm

(d) 5.5 cm

#### Answer:

Given: In ΔABC, D and E are points on the side AB and AC respectively such that DE || BC and AD : DB = 3 : 1. Also, EA = 3.3cm.

To find: AC

In ∆ABC, DE || BC.

Using corollory of basic proportionality theorem, we have

Hence the correct answer is .

#### Page No 4.132:

#### Question 10:

In triangles ABC and DEF, ∠A = ∠E = 40°, AB : ED = AC : EF and ∠F = 65°, then ∠B =

(a) 35°

(b) 65°

(c) 75°

(d) 85°

#### Answer:

Given: In ΔABC and ΔDEF

To find: Measure of angle B.

In ΔABC and ΔDEF

Hence in similar triangles ΔABC and ΔDEF

We know that sum of all the angles of a triangle is equal to 180°.

Hence the correct answer is .

#### Page No 4.132:

#### Question 11:

If ABC and DEF are similar triangles such that ∠A = 47° and ∠E = 83°, then ∠C =

(a) 50°

(b) 60°

(c) 70°

(d) 80°

#### Answer:

Given: If ΔABC and ΔDEF are similar triangles such that

To find: Measure of angle C

In similar ΔABC and ΔDEF,

We know that sum of all the angles of a triangle is equal to 180°.

Hence the correct answer is

#### Page No 4.132:

#### Question 12:

If D, E, F are the mid-points of sides BC, CA and AB respectively of ∆ABC, then the ratio of the areas of triangles DEF and ABC is

(a) 1 : 4

(b) 1 : 2

(c) 2 : 3

(d) 4 : 5

#### Answer:

GIVEN: In ΔABC, D, E and F are the midpoints of BC, CA, and AB respectively.

TO FIND: Ratio of the areas of ΔDEF and ΔABC

Since it is given that D and, E are the midpoints of BC, and AC respectively.

Therefore DE || AB, DE || FA……(1)

Again it is given that D and, F are the midpoints of BC, and, AB respectively.

Therefore, DF || CA, DF || AE……(2)

From (1) and (2) we get AFDE is a parallelogram.

Similarly we can prove that BDEF is a parallelogram.

Now, in ΔADE and ΔABC

Hence the correct option is .

#### Page No 4.132:

#### Question 13:

In a ∆ABC, ∠A = 90°, AB = 5 cm and AC = 12 cm. If AD ⊥ BC, then AD =

(a) $\frac{13}{2}\mathrm{cm}$

(b) $\frac{60}{13}\mathrm{cm}$

(c) $\frac{13}{60}\mathrm{cm}$

(d) $\frac{2\sqrt{15}}{13}\mathrm{cm}$

#### Answer:

Given: In ΔABC,, AC = 12cm, and AB = 5cm.

To find: AD

In ∆ACB and ∆ADC,

$\angle \mathrm{C}=\angle \mathrm{C}$ (Common)

$\angle \mathrm{A}=\angle \mathrm{ADC}=90\xb0$

∴ ∆ACB $~$∆ADC (AA Similarity)

We got the result as

#### Page No 4.132:

#### Question 14:

If ∆ABC is an equilateral triangle such that AD ⊥ BC, then AD^{2} =

(a) $\frac{3}{2}{\mathrm{DC}}^{2}$

(b) 2 DC^{2}

(c) 3 CD^{2}

(d) 4 DC^{2}

#### Answer:

Given: In an equilateral ΔABC, .

Since , BD = CD = $\frac{\mathrm{BC}}{2}$

Applying Pythagoras theorem,

In ΔADC

We got the result as

#### Page No 4.132:

#### Question 15:

In a ∆ABC, AD is the bisector of ∠BAC. If AB = 6 cm, AC = 5 cm and BD = 3 cm, then DC =

(a) 11.3 cm

(b) 2.5 cm

(c) 3 : 5 cm

(d) None of these

#### Answer:

Given: In a ΔABC, AD is the bisector of . AB = 6cm and AC = 5cm and BD = 3cm.

To find: DC

We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

Hence,

Hence we got the result

#### Page No 4.132:

#### Question 16:

In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. Find AC

(a) 4 cm

(b) 6 cm

(c) 3 cm

(d) 8 cm

#### Answer:

Given: In a ΔABC, AD is the bisector of angle BAC. AB = 8cm, and DC = 3cm and BD = 6cm.

To find: AC

We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

Hence,

Hence we got the result

#### Page No 4.132:

#### Question 17:

ABCD is a trapezium such that BC || AD and AD = 4 cm. If the diagonals AC and BD intersect at O such that $\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{DO}}{\mathrm{OB}}=\frac{1}{2}$, then BC =

(a) 7 cm

(b) 8 cm

(c) 9 cm

(d) 6 cm

#### Answer:

Given: ABCD is a trapezium in which BC||AD and AD = 4 cm

The diagonals AC and BD intersect at O such that

To find: DC

In ΔAOD and ΔCOB

$\angle \mathrm{OAD}=\angle \mathrm{OCB}\left(\mathrm{Alternate}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{ODA}=\angle \mathrm{OBC}\left(\mathrm{Alternate}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{AOD}=\angle \mathrm{BOC}\left(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\mathrm{So},\u2206\mathrm{AOD}~\u2206\mathrm{COB}\left(\mathrm{AAA}\mathrm{similarity}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{correponding}\mathrm{sides}\mathrm{of}\mathrm{similar}\u2206\text{'}\mathrm{s}\mathrm{are}\mathrm{proportional}.\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{AD}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{\mathrm{AD}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{4}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=8\mathrm{cm}$

Hence the correct answer is

#### Page No 4.133:

#### Question 18:

If ABC is a right triangle right-angled at B and M, N are the mid-points of AB and BC respectively, then 4(AN^{2} + CM^{2}) =

(a) 4 AC^{2}

(b) 5 AC^{2}

(c) $\frac{5}{4}{\mathrm{AC}}^{2}$

(d) 6 AC^{2}

#### Answer:

M is the mid-point of AB.

∴ $\mathrm{BM}=\frac{\mathrm{AB}}{2}$

N is the mid-point of BC.

∴ $\mathrm{BN}=\frac{\mathrm{BC}}{2}$

Now,

Hence option (b) is correct.

#### Page No 4.133:

#### Question 19:

If in ∆ABC and ∆DEF, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}}$, then ∆ABC ∼ ∆DEF when

(a) ∠A = ∠F

(b) ∠A = ∠D

(c) ∠B = ∠D

(d) ∠B = ∠E

#### Answer:

Given: In ΔABC and ΔDEF, .

We know that if in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.

Then,Hence, ΔABC is similar to ΔDEF, we should have.

Hence the correct answer is .

#### Page No 4.133:

#### Question 20:

If in two triangles ABC and DEF, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FE}}=\frac{\mathrm{CA}}{\mathrm{FD}}$, then

(a) ∆FDE ∼ ∆CAB

(b) ∆FDE ∼ ∆ABC

(c) ∆CBA ∼ ∆FDE

(d) ∆BCA ∼ ∆FDE

#### Answer:

We know that if two triangles are similar if their corresponding sides are proportional.

It is given that ΔABC and ΔDEF are two triangles such that .

$\angle A=\angle D\phantom{\rule{0ex}{0ex}}\angle B=\angle E\phantom{\rule{0ex}{0ex}}\angle C=\angle F\phantom{\rule{0ex}{0ex}}$

∴ ΔCAB $~$ΔFDE

Hence the correct answer is .

#### Page No 4.133:

#### Question 21:

∆ABC ∼ ∆DEF, ar(∆ABC) = 9 cm^{2}, ar(∆DEF) = 16 cm^{2}. If BC = 2.1 cm, then the measure of EF is

(a) 2.8 cm

(b) 4.2 cm

(c) 2.5 cm

(d) 4.1 cm

#### Answer:

Given:

To find: measure of EF

Hence the correct answer is

#### Page No 4.133:

#### Question 22:

The length of the hypotenuse of an isosceles right triangle whose one side is $4\sqrt{2}\mathrm{cm}$ is

(a) 12 cm

(b) 8 cm

(c) $8\sqrt{2}\mathrm{cm}$

(d) $12\sqrt{2}\mathrm{cm}$

#### Answer:

**Given: **One side of isosceles right triangle is 4√2cm

**To find:** Length of the hypotenuse.

We know that in isosceles triangle two sides are equal.

In isosceles right triangle ABC, let AB and AC be the two equal sides of measure 4√2cm.

Applying Pythagoras theorem, we get

Hence correct answer is .

#### Page No 4.133:

#### Question 23:

A man goes 24 m due west and then 7 m due north. How far is he from the starting point?

(a) 31 m

(b) 17 m

(c) 25 m

(d) 26 m

#### Answer:

A man goes 24m due to west and then 7m due north.

Let the man starts from point B and goes 24 m due to west and reaches point A, then walked 7m north and reaches point C.

Now we have to find the distance between the starting point and the end point i.e. BC.

In right triangle ABC, applying Pythagoras theorem, we get

Hence correct answer is .

#### Page No 4.133:

#### Question 24:

∆ABC ∼ ∆DEF. If BC = 3 cm, EF = 4 cm and ar(∆ABC) = 54 cm^{2}, then ar(∆DEF) =

(a) 108 cm^{2}

(b) 96 cm^{2}

(c) 48 cm^{2}

(d) 100 cm^{2}

#### Answer:

Given: In Δ ABC and Δ DEF

To find: Ar(Δ DEF)

Hence the correct answer is

#### Page No 4.133:

#### Question 25:

∆ABC ∼ ∆PQR such that ar(∆ABC) = 4 ar(∆PQR). If BC = 12 cm, then QR =

(a) 9 cm

(b) 10 cm

(c) 6 cm

(d) 8 cm

#### Answer:

Given: In Δ ABC and ΔPQR

To find: Measure of QR

Hence the correct answer is

#### Page No 4.133:

#### Question 26:

The areas of two similar triangles are 121 cm^{2} and 64 cm^{2} respectively. If the median of the first triangle is 12.1 cm, then the corresponding median of the other triangle is

(a) 11 cm

(b) 8.8 cm

(c) 11.1 cm

(d) 8.1 cm

#### Answer:

Given: The area of two similar triangles is 121cm^{2} and 64cm^{2} respectively. The median of the first triangle is 2.1cm.

To find: Corresponding medians of the other triangle

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians.

Taking square root on both side, we get

$\frac{11}{8}=\frac{12.1\mathrm{cm}}{\mathrm{median}2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{median}2=8.8\mathrm{cm}$

Hence the correct answer is .

#### Page No 4.133:

#### Question 27:

In an equilateral triangle ABC if AD ⊥ BC, then AD^{2} =

(a) CD^{2}

(b) 2CD^{2}

(c) 3CD^{2}

(d) 4CD^{2}

#### Answer:

In an equilateral ΔABC, .

In ΔADC, applying Pythagoras theorem, we get

#### Page No 4.133:

#### Question 28:

In an equilateral triangle ABC if AD ⊥ BC, then

(a) 5AB^{2} = 4AD^{2}

(b) 3AB^{2} = 4AD^{2}

(c) 4AB^{2} = 3AD^{2}

(d) 2AB^{2} = 3AD^{2}

#### Answer:

∆ABC** **is an equilateral triangle and** **.

In ∆ABD, applying Pythagoras theorem, we get

We got the result as .

#### Page No 4.133:

#### Question 30:

∆ABC is an isosceles triangle in which ∠C = 90. If AC = 6 cm, then AB =

(a) $6\sqrt{2}\mathrm{cm}$

(b) 6 cm

(c) $2\sqrt{6}\mathrm{cm}$

(d) $4\sqrt{2}\mathrm{cm}$

#### Answer:

Given: In an isosceles ΔABC, , AC = 6 cm.

To find: AB

In an isosceles ΔABC,.

Therefore, BC = AC = 6 cm

Applying Pythagoras theorem in ΔABC, we get

We got the result as .

#### Page No 4.134:

#### Question 31:

If in two triangle ABC and DEF, ∠A = ∠E, ∠B = ∠F, then which of the following is not true?

(a) $\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{AC}}{\mathrm{DE}}$

(b) $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{DF}}$

(c) $\frac{\mathrm{AB}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DE}}$

(d) $\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{AB}}{\mathrm{EF}}$

#### Answer:

In ΔABC and ΔDEF

Hence

Hence the correct answer is (b).

#### Page No 4.134:

#### Question 32:

In the given figure the measure of ∠D and ∠F are respectively

(a) 50°, 40°

(b) 20°, 30°

(c) 40°, 50°

(d) 30°, 20°

#### Answer:

ΔABC and ΔDEF,

$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{EF}}{\mathrm{ED}}\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}=\angle \mathrm{E}=130\xb0$

ΔABC $~$ ΔEFD (SAS Similarity)

$\therefore \angle \mathrm{F}=\angle \mathrm{B}=30\xb0\phantom{\rule{0ex}{0ex}}\angle \mathrm{D}=\angle \mathrm{C}=20\xb0$

Hence the correct answer is

#### Page No 4.134:

#### Question 33:

In the given figure, the value of *x* for which DE || AB is

(b) 1

(c) 3

(d) 2

#### Answer:

Given: In ∆ABC, DE || AB.

To find: the value of *x*

According to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ∆ABC, DE || AB

* x* = 2

Hence we got the result .

#### Page No 4.134:

#### Question 34:

In the given figure, if ∠ADE = ∠ABC, then CE =

(b) 5

(c) 9/2

(d) 3

#### Answer:

Given:

To find: The value of CE

Since

∴ DE || BC (Two lines are parallel if the corresponding angles formed are equal)

According to basic proportionality theorem if a line is parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ∆ABC, DE || BC

Hence we got the result .

#### Page No 4.134:

#### Question 35:

In the given figure, RS || DB || PQ. If CP = PD = 11 cm and DR = RA = 3 cm. Then the values of *x* and* y* are respectively.

(a) 12, 10

(b) 14, 6

(c) 10, 7

(d) 16, 8

#### Answer:

Given: RS || DB || PQ. CP = PD = 11cm and DR = RA = 3cm

To find: the value of *x* and y respectively.

$\mathrm{In}\u2206\mathrm{ASR}\mathrm{and}\u2206\mathrm{ABD},\phantom{\rule{0ex}{0ex}}\angle \mathrm{ASR}=\angle \mathrm{ABQ}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}=\angle \mathrm{A}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\therefore \u2206\mathrm{ASR}~\u2206\mathrm{ABD}\left(\mathrm{AA}\hspace{0.17em}\mathrm{Similarity}\right)\phantom{\rule{0ex}{0ex}}$

This relation is satisfied by option (d).

Hence, *x* = 16 cm and *y* = 8cm

Hence the result is .

#### Page No 4.135:

#### Question 36:

In the given figure, if PB || CF and DP || EF, then $\frac{\mathrm{AD}}{\mathrm{DE}}=$

(a) $\frac{3}{4}$

(b) $\frac{1}{3}$

(c) $\frac{1}{4}$

(d) $\frac{2}{3}$

#### Answer:

Given: PB||CF and DP||EF. AB = 2 cm and AC = 8 cm.

To find: AD: DE

According to** **BASIC PROPORTIONALITY THEOREM, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ∆ACF, PB || CF.

.....(1)

Again, DP||EF.

Hence we got the result .

#### Page No 4.135:

#### Question 37:

A chord of a circle of radius 10 cm subtends a right angle at the centre. The length of the chord (in cm) is

(a) $5\sqrt{2}$ (b) $10\sqrt{2}$ (c) $\frac{5}{\sqrt{2}}$ (d) $10\sqrt{3}$ [CBSE 2014]

#### Answer:

In right ∆OAB,

${\mathrm{AB}}^{2}={\mathrm{OA}}^{2}+{\mathrm{OB}}^{2}\left(\mathrm{Pythagoras}\mathrm{Theorem}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AB}}^{2}={\left(10\right)}^{2}+{\left(10\right)}^{2}\left(\mathrm{OA}=\mathrm{OB}=10\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AB}}^{2}=100+100=200\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}=\sqrt{200}=10\sqrt{2}\mathrm{cm}$

Thus, the length of the chord is $10\sqrt{2}$ cm.

Hence, the correct answer is option B.

#### Page No 4.135:

#### Question 38:

A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. The height of the tower is

(a) 100 m

(b) 120 m

(c) 25 m

(d) 200 m

#### Answer:

Given: Vertical stick 20m long casts a shadow 10m long on the ground. At the same time a tower casts the shadow 50 m long on the ground.

To determine: Height of the tower

Let AB be the vertical stick and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow.

Join BC and EF.

In ΔABC and ΔDEF, we have

We know that in any two similar triangles, the corresponding sides are proportional. Hence,

Hence the correct answer is option .

#### Page No 4.135:

#### Question 39:

Two isosceles triangles have equal angles and their areas are in the ratio 16 : 25. The ratio of their corresponding heights is

(a) 4 : 5

(b) 5 : 4

(c) 3 : 2

(d) 5 : 7

#### Answer:

Given: Two isosceles triangles have equal vertical angles and their areas are in the ratio of 16:25.

To find: Ratio of their corresponding heights.

Let ∆ABC and ∆PQR be two isosceles triangles such that $\angle \mathrm{A}=\angle \mathrm{P}$. Suppose AD ⊥ BC and PS ⊥ QR .

In ∆ABC and ∆PQR,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}=\angle \mathrm{P}\phantom{\rule{0ex}{0ex}}\therefore \u2206\mathrm{ABC}~\u2206\mathrm{PQR}\left(\mathrm{SAS}\mathrm{similarity}\right)$

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Hence,

$\frac{\mathrm{Ar}\left(\u2206\mathrm{ABC}\right)}{\mathrm{Ar}\left(\u2206\mathrm{PQR}\right)}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{16}{25}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{AD}}{\mathrm{PS}}=\frac{4}{5}$

Hence we got the result as

#### Page No 4.135:

#### Question 40:

∆ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ∆DEF ∼ ∆ABC and EF = 4 cm, then perimeter of ∆DEF is

(a) 7.5 cm

(b) 15 cm

(c) 22.5 cm

(d) 30 cm

#### Answer:

Given: In ΔABC, AB = 3cm, BC = 2cm, CA = 2.5cm. and EF = 4cm.

To find: Perimeter of ΔDEF.

We know that if two triangles are similar, then their sides are proportional

Since ΔABC and ΔDEF are similar,

From (1) and (2), we get

Perimeter of ΔDEF = DE + EF + FD = 6 + 4 +5 = 15 cm

Hence the correct answer is .

#### Page No 4.136:

#### Question 41:

In ∆ABC, a line XY parallel to BC cuts AB at X and AC at Y. If BY bisects ∠XYC, then

(a) BC = CY

(b) BC = BY

(c) BC ≠ CY

(d) BC ≠ BY

#### Answer:

Given: XY||BC and BY is bisector of $\angle $XYC.

Since XY||BC

So $\angle $YBC = $\angle $BYC (Alternate angles)

Now, in triangle BYC two angles are equal. Therefore, the two corresponding sides will be equal.

Hence, BC = CY

Hence option (a) is correct.

#### Page No 4.136:

#### Question 43:

In a ∆ABC, perpendicular AD from A and BC meets BC at D. If BD = 8 cm, DC = 2 cm and AD = 4 cm, then

(a) ∆ABC is isosceles

(b) ∆ABC is equilateral

(c) AC = 2AB

(d) ∆ABC is right-angled at A

#### Answer:

Given: In ΔABC,, BD = 8cm, DC = 2 cm and AD = 4cm.

In ΔADC,

Similarly, in ΔADB

Now, In ΔABC

${\mathrm{BC}}^{2}={\left(\mathrm{CD}+\mathrm{DB}\right)}^{2}={\left(2+8\right)}^{2}={\left(10\right)}^{2}=100$

and

${\mathrm{AB}}^{2}+{\mathrm{CA}}^{2}=80+20=100$

$\therefore {\mathrm{AB}}^{2}+{\mathrm{CA}}^{2}={\mathrm{BC}}^{2}$

Hence, triangle ABC is right angled at A.

We got the result as

#### Page No 4.136:

#### Question 44:

In a ∆ABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium. If DE : BC = 3 : 5, then Area (∆ ADE) : Area (◻BCED) =

(a) 3 : 4

(b) 9 : 16

(c) 3 : 5

(d) 9 : 25

#### Answer:

Given: In ΔABC, D is on side AB and point E is on side AC, such that BCED is a trapezium. DE: BC = 3:5.

To find: Calculate the ratio of the areas of ΔADE and the trapezium BCED.

In ΔADE and ΔABC,

$\angle \mathrm{ADE}=\angle \mathrm{B}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}=\angle \mathrm{A}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\therefore \u2206\mathrm{ADE}~\u2206\mathrm{ABC}\left(\mathrm{AA}\mathrm{Similarity}\right)$

We know that

Let Area of ΔADE = 9*x* sq. units and Area of ΔABC = 25*x* sq. units

Now ,

Hence the correct answer is.

#### Page No 4.136:

#### Question 45:

If ABC is an isosceles triangle and D is a point of BC such that AD ⊥ BC, then

(a) AB^{2} − AD^{2} = BD.DC

(b) AB^{2} − AD^{2} = BD^{2} − DC^{2}

(c) AB^{2} + AD^{2} = BD.DC

(d) AB^{2} + AD^{2} = BD^{2} − DC^{2}

#### Answer:

Given: ΔABC is an isosceles triangle, D is a point on BC such that

We know that in an isosceles triangle the perpendicular from the vertex bisects the base.

∴ BD = DC

Applying Pythagoras theorem in ΔABD

Since

Hence correct answer is .

#### Page No 4.136:

#### Question 46:

∆ABC is a right triangle right-angled at A and AD ⊥ BC. Then, $\frac{\mathrm{BD}}{\mathrm{DC}}=$

(a) ${\left(\frac{\mathrm{AB}}{\mathrm{AC}}\right)}^{2}$

(b) $\frac{\mathrm{AB}}{\mathrm{AC}}$

(c) ${\left(\frac{\mathrm{AB}}{\mathrm{AD}}\right)}^{2}$

(d) $\frac{\mathrm{AB}}{\mathrm{AD}}$

#### Answer:

Given: In ΔABC, and.

To find: BD: DC

$\angle \mathrm{CAD}+\angle \mathrm{BAD}=90\xb0.....\left(1\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{BAD}+\angle \mathrm{ABD}=90\xb0.....\left(2\right)\left(\angle \mathrm{ADB}=90\xb0\right)\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\phantom{\rule{0ex}{0ex}}\angle \mathrm{CAD}=\angle \mathrm{ABD}$

In ΔADB and ΔADC,

$\angle \mathrm{ADB}=\angle \mathrm{ADC}\left(90\xb0\mathrm{each}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{ABD}=\angle \mathrm{CAD}\left(\mathrm{Proved}\right)\phantom{\rule{0ex}{0ex}}\therefore \u2206\mathrm{ADB}~\u2206\mathrm{ADC}\left(\mathrm{AA}\mathrm{Similarity}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\mathrm{AD}}{\mathrm{BD}}\left(\mathrm{Corresponding}\mathrm{sides}\mathrm{are}\mathrm{proportional}\right)$

**Disclaimer**: The question is not correct. The given ratio cannot be evaluated using the given conditions in the question.

#### Page No 4.136:

#### Question 47:

If E is a point on side CA of an equilateral triangle ABC such that BE ⊥ CA, then AB^{2} + BC^{2} + CA^{2} =

(a) 2 BE^{2}

(b) 3 BE^{2}

(c) 4 BE^{2}

(d) 6 BE^{2}

#### Answer:

In triangle ABC, E is a point on AC such that .

We need to find .

Since , *CE = AE *= $\frac{\mathit{A}\mathit{C}}{\mathit{2}}$ (In a equilateral triangle, the perpendicular from the vertex bisects the base.)

In triangle ABE, we have

Since AB = BC = AC

Therefore,

Since in triangle BE is an altitude, so

Hence option (c) is correct.

#### Page No 4.136:

#### Question 48:

In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then

(a) AQ^{2} + CP^{2} = 2(AC^{2}^{ }+ PQ^{2})

(b) 2(AQ^{2} + CP^{2}) = AC^{2} + PQ^{2}

(c) AQ^{2}^{ }+ CP^{2} = AC^{2} + PQ^{2}

(d) $\mathrm{AQ}+\mathrm{CP}=\frac{1}{2}\left(\mathrm{AC}+\mathrm{PQ}\right)$

#### Answer:

**Disclaimer**: There is mistake in the problem. The question should be "In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and BC respectively, then"

Given: In the right ΔABC, right angled at B. P and Q are points on the sides AB and BC respectively.

Applying Pythagoras theorem,

In ΔAQB,

${\mathrm{AQ}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BQ}}^{2}$ .....(1)

In ΔPBC

** **.....(2)

Adding (1) and (2), we get

${\mathrm{AQ}}^{2}+{\mathrm{CP}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BQ}}^{2}+{\mathrm{PB}}^{2}+{\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}$ .....(3)

In ΔABC,

.....(4)

In ΔPBQ,

${\mathrm{PQ}}^{2}={\mathrm{PB}}^{2}+{\mathrm{BQ}}^{2}$ .....(5)

From (3), (4) and (5), we get

${\mathrm{AQ}}^{2}+{\mathrm{CP}}^{2}={\mathrm{AC}}^{2}+{\mathrm{PQ}}^{2}$

#### Page No 4.136:

#### Question 49:

If ∆ABC ∼ ∆DEF such that DE = 3 cm, EF = 2 cm, DF = 2.5 cm, BC = 4 cm, then perimeter of ∆ABC is

(a) 18 cm

(b) 20 cm

(c) 12 cm

(d) 15 cm

#### Answer:

Given: ΔABC and ΔDEF are similar triangles such that DE = 3cm, EF = 2cm, DF = 2.5cm and BC = 4cm.

To find: Perimeter of ΔABC.

We know that if two triangles are similar then their corresponding sides are proportional.

Hence,

Substituting the values we get

Similarly,

Hence the correct option is

#### Page No 4.136:

#### Question 50:

If ∆ABC ∼ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25 cm, then the perimeter of ∆ABC is

(a) 36 cm

(b) 30 cm

(c) 34 cm

(d) 35 cm

#### Answer:

Given: ΔABC is similar to ΔDEF such that AB= 9.1cm, DE = 6.5cm. Perimeter of ΔDEF is 25cm.

To find: Perimeter of ΔABC.

We know that the ratio of corresponding sides of similar triangles is equal to the ratio of their perimeters.

Hence,

Hence the correct answer is .

#### Page No 4.136:

#### Question 51:

In an isosceles triangle ABC if AC = BC and AB^{2} = 2AC^{2}, then ∠C =

(a) 30°

(b) 45°

(c) 90°

(d) 60°

#### Answer:

Given: In Isosceles ΔABC, AC = BC and AB^{2} = 2AC^{2}.

To find: Measure of angle C

In Isosceles ΔABC,

AC = BC

$\angle \mathrm{B}=\angle \mathrm{A}$ (Equal sides have equal angles opposite to them)

Hence the correct answer is .

#### Page No 4.19:

#### Question 1:

In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC.

(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, find AC.

(ii) If $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{4}$ and AC = 15 cm, find AE.

(iii) If $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{2}{3}$ and AC = 18 cm, find AE.

(iv) If AD = 4, AE = 8, DB = *x* − 4, and EC = 3*x* − 19, find *x*.

(v) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.

(vi) if AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.

(vii) If AD = 2 cm, AB = 6 cm, and AC = 9 cm, find AE.

(viii) If $\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{4}{5}$ and EC = 2.5 cm, find AE.

(ix) If AD = *x*, DB = *x* − 2, AE = *x* + 2 and EC = *x* − 1, find the value of *x*.

(x) If AD = 8*x* − 7, DB = 5*x* − 3, AE = 4*x* − 3 and EC = (3*x* − 1), find the value of *x*.

(xi) If AD = 4*x* − 3, AE = 8*x* − 7, BD = 3*x* − 1 and CE = 5*x* − 3. find the volume *x*.

(xii) If AD = 2.5 cm, BD = 3.0 cm and AE = 3.75 cm find the length of AC.

#### Answer:

(i) It is given that and *DE || BC*

We have to find the

Since

So (by Thales theorem)

Then

Hence

(ii)It is given that and

We have to find

Let

So (by Thales theorem)

Then

Hence

(iii)It is given that and

We have to find

Let and

So (by Thales theorem)

Then

Hence

(iv)It is given that ,, and .

We have to find

So (by Thales theorem)

Then

Hence

(v) It is given that , and .

We have to find CE.

So (by Thales theorem)

Then

Hence

(vi) It is given that , and .

We have to find .

So (by Thales theorem)

Then

Hence

(vii) It is given that , and .

We have to find .

Now

*DB = *6 − 2 = 4 cm

So (by Thales theorem)

Then (Let)

Hence

(viii) It is given that and *EC *= 2.5 cm

We have to find .

So (by Thales theorem)

Then

$AE=\frac{4\times 2.5}{5}=2$ cm

Hence

(ix) It is given that ,, and .

We have to find the value of .

So (by Thales theorem)

Then

Hence

(x) It is given that *AD* = 8*x* − 7, *DB* = 5*x* − 3, *AE* = 4*x* − 3 and *EC* = 3*x *− 1.

We have to find the value of .

So (by Thales theorem)

Then,

$\frac{8x-7}{5x-3}=\frac{4x-3}{3x-1}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(8x-7\right)\left(3x-1\right)=\left(5x-3\right)\left(4x-3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 24{x}^{2}-29x+7=20{x}^{2}-27x+9\phantom{\rule{0ex}{0ex}}\Rightarrow 4{x}^{2}-2x-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left[2{x}^{2}-x-1\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}-x-1=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}-2x+x-1=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x\left(x-1\right)+1\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-1\right)\left(2x+1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-1=0\mathrm{or}2x+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=1\mathrm{or}x=-\frac{1}{2}\left(\mathrm{rejected}\right)$

Hence,

(xi) It is given that *AD* = 4*x* − 3, *BD* = 3*x* − 1, *AE* = 8*x *− 7 and *EC* = 5*x *− 3.

We have to find the value of .

So (by Thales theorem)

Then

$\left(4x-3\right)\left(5x-3\right)=\left(3x-1\right)\left(8x-7\right)$

Then

Hence

(xii) It is given that , and .

So (by Thales theorem)

Then

Now

#### Page No 4.19:

#### Question 2:

In a ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC :

(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.

(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 and AE = 1.8 cm.

(iii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.

(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm

#### Answer:

(i) It is given that and are point on sides *AB* and *AC*.

We have to prove that *DE || BC.*

According to Thales theorem we have

(Proportional)

Hence, *DE || BC.*

(ii) It is given that and are point on sides *AB* and *AC*.

We have to prove that *DE || BC.*

According to Thales theorem we have

(Proportional)

Hence, *DE || BC.*

(iii) It is given that and are point on sides *AB* and *AC*.

We have to prove that *DE || BC.*

According to Thales theorem we have

So

$AD=AB-DB=10.8-4.5=6.3$

And

$EC=AC-AE=4.8-2.8=2$

Now

Hence, *DE || BC*.

(iv) It is given that and are point on sides *AB* and *AC*.

We have to prove that *DE || BC.*

According to Thales theorem we have

(Proportional)

Hence, *DE || BC*.

#### Page No 4.19:

#### Question 3:

In a ∆ABC, P and Q are points on sides AB and AC respectively, such that PQ || BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find the AB and PQ.

#### Answer:

It is given that ,, and .

We have to find *AB* and .

So (by Thales theorem)

Then

Now

Since PQ$\parallel $BC, AB is a transversal, then∠APQ = ∠ABC (corresponding angles)

Since PQ$\parallel $BC, AC is a transversal, then

∠AQP = ∠ACB (corresponding angles)

In ∆APQ and ∆ABC,

∠APQ = ∠ABC (proved above)

∠AQP = ∠ACB (proved above)

so, ∆ APQ ∼ ∆ ABC (Angle Angle Similarity)

Since the corresponding sides of similar triangles are proportional, then

$\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AQ}}{\mathrm{AC}}$

$\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\frac{2.4}{6}=\frac{\mathrm{PQ}}{6}\phantom{\rule{0ex}{0ex}}\mathrm{so},\mathrm{PQ}=2.4\mathrm{cm}$

#### Page No 4.19:

#### Question 4:

In a ∆ABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.

#### Answer:

It is given that ,, and .

We have to find BD and CE.

Since DE$\parallel $BC, AB is transversal, then

∠ADE = ∠ABC (corresponding angles)

Since DE$\parallel $BC, AC is a transversal, then

∠AED = ∠ACB (corresponding angles)

In ∆ADE and ∆ABC,

∠ADE = ∠ABC (proved above)

∠AED = ∠ACB (proved above)

so, ∆ADE ∼ ∆ABC (Angle Angle similarity)

Since, the corresponding sides of similar triangles are proportional, then

$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{DE}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2.4}{2.4+\mathrm{DB}}=\frac{2}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 2.4+\mathrm{DB}=6\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{DB}=6-2.4\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{DB}=3.6\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3.2}{3.2+\mathrm{EC}}=\frac{2}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 3.2+\mathrm{EC}=8\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{EC}=8-3.2\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{EC}=4.8\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Hence, BD = 3.6 cm and CE = 4.8 cm.

#### Page No 4.19:

#### Question 5:

In Fig. 4.35, state if PQ || EF.

#### Answer:

It is given that *EP = *3 cm, *PG* = 3.9 cm, *FQ = *3.6 cm and *FQ* = 2.4 cm.

We have to check that or not.

According to Thales theorem we have

$\frac{PG}{GE}=\frac{GQ}{FQ}$

Now,

$\frac{3.9}{3}\ne \frac{3.6}{2.4}$

Hence, it is not proportional.

So, PQ ∦ EF.

#### Page No 4.19:

#### Question 6:

M and N are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether MN || QR.

(i) PM = 4 cm, QM = 4.5 cm, PN = 4 cm, NR = 4.5 cm

(ii) PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, PN = 0.32 cm

#### Answer:

(1)It is given that , , and .

We have to check that or not.

According to Thales theorem we have

(Proportional)

Hence,

(2) It is given that *PQ = *1.28 cm, *PR* = 2.56 cm, *PM* = 0.16 cm and *PN* = 0.32 cm.

We have to check that or not.

According to Thales theorem we have

Now,

$\frac{PM}{MQ}=\frac{0.16}{1.12}=\frac{1}{7}\phantom{\rule{0ex}{0ex}}\frac{PN}{NR}=\frac{0.32}{2.24}=\frac{1}{7}\phantom{\rule{0ex}{0ex}}\therefore \frac{0.16}{1.12}=\frac{0.32}{2.24}$

Hence,

#### Page No 4.20:

#### Question 7:

In three line segments OA, OB and OC points L, M, N respectively are so chosen that LM || AB and MN || BC but neither of L, M, N nor of A, B, C are collinear. Show that LN || AC.

#### Answer:

$\mathrm{In}\u2206\mathrm{OAB},\mathrm{since}\mathrm{LM}\parallel \mathrm{AB},\mathrm{then}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{OL}}{\mathrm{LA}}=\frac{\mathrm{OM}}{\mathrm{MB}}\left(\mathrm{By}\mathrm{BPT}\right)........\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\u2206\mathrm{OBC},\mathrm{since}\mathrm{MN}\parallel \mathrm{BC},\mathrm{then}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{OM}}{\mathrm{MB}}=\frac{\mathrm{ON}}{\mathrm{NC}}\left(\mathrm{By}\mathrm{BPT}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{ON}}{\mathrm{NC}}=\frac{\mathrm{OM}}{\mathrm{MB}}.........\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{from}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{OL}}{\mathrm{LA}}=\frac{\mathrm{ON}}{\mathrm{NC}}.........\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\u2206\mathrm{OCA},\mathrm{we}\mathrm{have},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{OL}}{\mathrm{LA}}=\frac{\mathrm{ON}}{\mathrm{NC}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{LN}\parallel \mathrm{AC}\left(\mathrm{By}\mathrm{converse}\mathrm{of}\mathrm{BPT}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 4.20:

#### Question 8:

If D and E are points on sides AB and AC respectively of a ∆ABC such that DE || BC and BD = CE. Prove that ∆ABC is isosceles.

#### Answer:

It is given that in , *DE* || *BC* and *BD* = *CE*.

We have to prove that ∆ABC is isosceles.

By Thales theorem we have

Now and

So

Hence

So, ∆ABC is isosceles

#### Page No 4.2:

#### Question 1:

Fill in the blanks using the correct word given in brackets :

(i) All circles are ......... (congruent, similar).

(ii) All squares are ........(similar, congruent).

(iii) All .......... triangles are similar (isosceles, equilateral):

(iv) Two triangles are similar, if their corresponding angles are .......... (proportional, equal)

(v) Two triangles are similar, if their corresponding sides are ........... (proportional, equal)

(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are ...........(equal, proportional).

#### Answer:

(i) Since all circles have centre and circumference, therefore all circles are similar.

Hence

(ii) Since all squares have each angle and sides are proportional, therefore all squares are similar.

Hence

(iii) In equilateral triangle each angle is therefore all equilateral triangles are similar.

Hence

(iv) Two triangles are similar, if their corresponding angles are _{.}

(v) Two triangles are similar, if their corresponding sides are.

(vi) Two polygons of same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are .

#### Page No 4.3:

#### Question 2:

Write the truth value (T/F) of each of the following statements:

(i) Any two similar figures are congruent.

(ii) Any two congruent figures are similar.

(iii) Two polygons are similar, if their corresponding sides are proportional.

(iv) Two polygons are similar, if their corresponding angles are proportional.

(v) Two triangles are similar if their corresponding sides are proportional.

(vi) Two triangles are similar if their corresponding angles are proportional.

#### Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

#### Page No 4.31:

#### Question 1:

If a ∆ ABC, AD is the bisector of ∠A, meeting side BC at D.

(i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC.

(ii) If BD = 2 cm, AB = 5 cm and DC = 3 cm, find AC.

(iii) If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD.

(iv) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

(v) If AC = 4.2 cm, DC = 6 cm and BC = 10 cm. find AB.

(vi) If AB = 5.6 cm, AC = 6 cm and DC = 3 cm, find BC.

(vii) If AD = 5.6 cm, BC = 6 cm and BD = 3.2 cm, find AC.

(viii) If AB = 10 cm, AC = 6 cm and BC = 12 cm, find BD and DC.

#### Answer:

(i) It is given that , and .

In , is the bisector of, meeting side at .

We have to find .

Since is bisector

Then

Hence

(ii) It is given that , and.

In , is the bisector of, meeting side at .

We have to find .

Since is bisector

So (is bisector of and side)

Then

Hence

(iii) It is given that , and .

In , is the bisector of, meeting side at .

We have to find *BD.*

Since is bisector

So (is bisector of and side)

Then

Hence

(iv) It is given that , and .

In , is the bisector of, meeting side at .

We have to find *BD *and .

Since is bisector

So (is bisector of and side)

Let BD = *x* cm. Then CD = (6 − *x*) cm

Then,

$\frac{10}{14}=\frac{x}{6-x}\phantom{\rule{0ex}{0ex}}\Rightarrow 14x=60-10x\phantom{\rule{0ex}{0ex}}\Rightarrow 24x=60\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{60}{24}=2.5$

Hence, BD = 2.5 cm and DC = 6 − 2.5 = 3.5 cm.

(v) It is given that , and .

In , is the bisector of, meeting side at .

We have to find .

Since is bisector

So

Then

Hence

(vi) It is given that , and .

In , is the bisector of, meeting side at .

We have to find .

Since is bisector

So

Then

So

Hence

(vii) If it is given that *AB = *5.6 cm, and _{.}

In, is the bisector of, meeting side at

(viii) It is given that , and .

In , is the bisector of, meeting side at .

We have to find *BD* and .

Since is bisector

So

Let *BD* = *x* cm

Then

⇒

Now

Hence and

#### Page No 4.31:

#### Question 2:

In Fig. 4.57, AE is the bisector of the exterior ∠CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, find CE.

#### Answer:

It is given that is the bisector of the exterior

Meeting produced and, and

Since is the bisector of the exterior

So

Hence

#### Page No 4.31:

#### Question 3:

In Fig. 4.58, ∆ABC is a triangle such that $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}},\angle \mathrm{B}=70\xb0,\angle \mathrm{C}=50\xb0$. Find the ∠BAD.

#### Answer:

It is given that in , , and .

We have to find .

In ,

Since , therefore, *AD* is the bisector of $\angle A$.

Hence,

#### Page No 4.31:

#### Question 4:

In Fig. 4.60, check whether AD is the bisector of ∠A of ∆ABC in each of the following:

(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm

(iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm

(iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm

(v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm

#### Answer:

(i) It is given that,, and .

We have to check whether is bisector of .

First we will check proportional ratio between sides.

Now

$\frac{AB}{AC}=\frac{5}{10}=\frac{1}{2}$

$\frac{BD}{CD}=\frac{1.5}{3.5}=\frac{3}{7}$

Since $\frac{AB}{AC}\ne \frac{BD}{CD}$

Hence is not the bisector of .

(ii) It is given that,,, and .

We have to check whether is bisector of .

First we will check proportional ratio between sides.

So

$\Rightarrow \frac{4}{6}=\frac{1.6}{2.4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{3}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}$

(It is proportional)

Hence, is bisector of .

(iii) It is given that,,, and .

We have to check whether is bisector of .

First we will check proportional ratio between sides.

Now

So

(It is proportional)

Hence, is bisector of .

(iv) It is given that,* AB* = 6 cm, *AC* = 8 cm, *BD* = 1.5 cm and *CD* = 2 cm.

We have to check whether is bisector of .

First we will check proportional ratio between sides.

So

$\Rightarrow \frac{6}{8}=\frac{1.5}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{4}=\frac{3}{4}$

(It is proportional)

Hence is bisector of .

(v) It is given that *AB* = 5 cm, *AC* = 12 cm, *BD* = 2.5 cm and *BC* = 9 cm

We have to check whether is bisector of .

First we will check proportional ratio between sides.

Now

$\frac{AB}{AC}=\frac{5}{12}$

$\frac{BD}{CD}=\frac{2.5}{9}=\frac{5}{18}$

Since $\frac{AB}{AC}\ne \frac{BD}{CD}$

Hence is not the bisector of .

#### Page No 4.31:

#### Question 5:

In Fig. 4.60, AD bisects ∠A, AB = 12 cm, AC = 20 cm and BD = 5 cm, determine CD.

#### Answer:

It is given that bisect . Also, , and .

We have to find .

Since is the bisector of

Then

Hence

#### Page No 4.32:

#### Question 6:

In ∆ABC (Fig. 4.59), if ∠1 = ∠2, prove that $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}$.

#### Answer:

We have to prove that .

In ∆ABC,

(Given)

So, *AD* is the bisector of .

#### Page No 4.32:

#### Question 7:

D, E and F are the points on sides BC, CA and AB respectively of ∆ABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C. If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AF, CE and BD.

#### Answer:

It is given that , and .

We have to find _{,} and *BD.*

Sinceis bisector of

So

Then,

$\frac{5}{4}=\frac{BD}{BC-BD}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5}{4}=\frac{BD}{8-BD}\phantom{\rule{0ex}{0ex}}\Rightarrow 40-5BD=4BD\phantom{\rule{0ex}{0ex}}\Rightarrow 9BD=40$

So,

Since is the bisector of .

So,

$\frac{AB}{BC}=\frac{AE}{EC}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{AB}{BC}=\frac{AC-EC}{EC}$

So

Now since is the bisector of

So

So

Hence

And

#### Page No 4.37:

#### Question 1:

(i) In the given figure, if AB || CD, find the value of *x*.

*x*. (iii) In the given figure, AB || CD, If OA = 3

*x*− 19, OB =

*x*− 4, OC =

*x*− 3 and OD = 4, find

*x*.

#### Answer:

(i) It is given that .

We have to find the value of .

Diagonals of the para

Now

So

Therefore and

Hence

(ii) It is given that

We have to find the value of

Now

So

Therefore or

Hence *x* = $\frac{1}{2}$ or *x* = 2.

For *x* = $\frac{1}{2}$, OD is negative.

Hence

(iii) It is given that.

And, and

We have to find the value of

Now

So

Therefore and

Hence or _{.}

#### Page No 4.73:

#### Question 1:

In the given figure, ∆ACB ∼ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.

#### Answer:

It is given that .

,, and.

We have to find and .

Since

So

Similarly

Hence, and

#### Page No 4.73:

#### Question 2:

In the given figure, AB || QR. Find the length of PB.

#### Answer:

It is given that

,and *PR* = 6 cm

We have to find .

Since

So

Hence,

#### Page No 4.73:

#### Question 3:

In the given figure, XY || BC. Find the length of XY.

#### Answer:

It is given that .

,and

We have to find .

Since

()

So

Hence,

#### Page No 4.73:

#### Question 4:

In a right angled triangle with sides *a* and* b* and hypotenuse *c*, the altitude drawn on the hypotenuse is *x*. Prove that *ab* = *cx*.

#### Answer:

Let ∆ABC be a right angle triangle having sides and ; and hypotenuse . BD is the altitude drawn on the hypotenuse AC.

We have to find to prove .

Since the altitude is perpendicular on the hypotenuse, both the triangles are similar

Hence, .

#### Page No 4.74:

#### Question 5:

In the given figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, find CD.

#### Answer:

It is given that and _{. }

When ,we have to find the.

Since is right angle triangle and is perpendicular on , so

(AA similarity)

Hence,

#### Page No 4.74:

#### Question 6:

In the given figure, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

#### Answer:

It is given that,,, and

We have to find .

Since

So

Hence,

#### Page No 4.74:

#### Question 7:

In the given figure, DE || BC such that AE = (1/4) AC. If AB = 6 cm, find AD.

#### Answer:

It is given that, and .

We have to find *AD.*

Since

So

Hence,

#### Page No 4.74:

#### Question 8:

In Fig. 4.145, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, prove that ∆CED ∼ ∆ABC.

#### Answer:

It is given that , and .

We have to prove that .

Now,

,, so.

In ∆ABC and ∆CED,

(Given)

$\angle A=\angle ECD$ ( Alternate angles)

So, (*AA* similarly rule)

#### Page No 4.74:

#### Question 9:

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion for two triangles, show that $\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$.

#### Answer:

It is given that trapezium with . *O* is the point of intersection of *AC *and *BD*.

We have to prove that

Now, in and

$\angle AOB=\angle COD$ (Vertically opposite angles)

$\angle OAB=\angle OCD$ (Alternate angles)

∴ $~$ (AA Similarity)

Hence, (Corresponding sides are proportional)

#### Page No 4.74:

#### Question 10:

If ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively such that ∠MAP = ∠BAC. Prove that

(i) ∆ABC ∼ ∆AMP

(ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$

#### Answer:

(1) It is given that and are two right angle triangles.

Now, in and , we have(Given)

(AA Similarity)

(2)

So, (Corresponding sides are proportional)

#### Page No 4.74:

#### Question 11:

A vertical stick 10 cm long casts a shadow 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower.

#### Answer:

We have to find the height of _{.}

Now,

∆ABC $~$∆PQR (AA Similarity)

Hence

#### Page No 4.75:

#### Question 12:

In the given figure, ∠A = ∠CED, prove that ∆CAB ∼ ∆CED. Also, find the value of *x*.

#### Answer:

Comparing Δ*CAB* and Δ*CED*,

∠*CAB* = ∠*CED** *[Given]

∠*ACB* = ∠*ECD** *[Common]

∴ Δ*CAB* ∼ Δ*CED*

#### Page No 4.75:

#### Question 13:

The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, what is the corresponding side of the other triangle?

#### Answer:

It is given that perimeter of two similar triangle are and and one side.

We have to find the other side.

Let the corresponding side of the other triangle be *x* cm.

$\frac{25\mathrm{cm}}{15\mathrm{cm}}=\frac{9\mathrm{cm}}{x}$

Hence

#### Page No 4.75:

#### Question 14:

In ∆ABC and ∆DEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL ⊥ BC and DM ⊥ EF, find AL : DM.

#### Answer:

It is given that ,,,, and .

We have to find

Since both triangle are similar

So,

Here, we use the result that in similar triangle the ratio of corresponding altitude is same as the ratio of the corresponding sides.

∴ *AL *: *DM *= 1 : 2

#### Page No 4.75:

#### Question 15:

D and E are the points on the sides AB and AC respectively of a ∆ABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm, Prove that BC = 5/2 DE.

#### Answer:

It is given that,, and .

We have to prove that

Since clearly

Also, is common in and

So (SAS Similarity)

$\Rightarrow \frac{BC}{DE}=\frac{AB}{AD}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{BC}{DE}=\frac{1}{\left({\displaystyle \frac{AD}{AB}}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{BC}{DE}=\frac{1}{\left({\displaystyle \frac{2}{5}}\right)}\left({\displaystyle \frac{AD}{AB}=\frac{2}{5}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{BC}{DE}=\frac{5}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)BC=\frac{5}{2}DE}$

#### Page No 4.75:

#### Question 16:

D is the mid-point of side BC of a ∆ABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE = EX = 3 : 1

#### Answer:

**Given: **ABC is a triangle in which D is the mid point of BC, E is the mid point of AD. BE produced meets AC at X.

**To Prove:** BE : EX = 3:1.

**Construction: **We draw a line DY parallel to BX.

**Proof**:

$\mathrm{In}\u2206\mathrm{BCX}\mathrm{and}\u2206\mathrm{DCY},\phantom{\rule{0ex}{0ex}}\angle \mathrm{CBX}=\angle \mathrm{CDY}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{CXB}=\angle \mathrm{CYD}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{BCX}~\u25b3\mathrm{DCY}\left(\mathrm{AA}\mathrm{similarity}\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{corresponding}\mathrm{sides}\mathrm{of}\mathrm{similar}\mathrm{triangles}\mathrm{are}\mathrm{proportional}.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\frac{\mathrm{BC}}{\mathrm{DC}}=\frac{\mathrm{BX}}{\mathrm{DY}}=\frac{\mathrm{CX}}{\mathrm{CY}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{BX}}{\mathrm{DY}}=\frac{\mathrm{BC}}{\mathrm{DC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{BX}}{\mathrm{DY}}=\frac{2\mathrm{DC}}{\mathrm{DC}}\left(\mathrm{As}\mathrm{D}\mathrm{is}\mathrm{the}\mathrm{mid}\mathrm{point}\mathrm{of}\mathrm{BC}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{BX}}{\mathrm{DY}}=\frac{2}{1}....\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{In}\u2206\mathrm{AEX}\mathrm{and}\u2206\mathrm{ADY},\phantom{\rule{0ex}{0ex}}\angle \mathrm{AEX}=\angle \mathrm{ADY}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{AXE}=\angle \mathrm{AYD}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{AEX}~\u25b3\mathrm{ADY}\left(\mathrm{AA}\mathrm{similarity}\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{corresponding}\mathrm{sides}\mathrm{of}\mathrm{similar}\mathrm{triangles}\mathrm{are}\mathrm{proportional}.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\frac{\mathrm{AE}}{\mathrm{AD}}=\frac{\mathrm{EX}}{\mathrm{DY}}=\frac{\mathrm{AX}}{\mathrm{AY}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{EX}}{\mathrm{DY}}=\frac{\mathrm{AE}}{\mathrm{AD}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{EX}}{\mathrm{DY}}=\frac{\mathrm{AE}}{2\mathrm{AE}}\left(\mathrm{As}\mathrm{D}\mathrm{is}\mathrm{the}\mathrm{mid}\mathrm{point}\mathrm{of}\mathrm{BC}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{EX}}{\mathrm{DY}}=\frac{1}{2}....\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\left(1\right)\mathrm{by}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{BX}}{\mathrm{EX}}=4\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BX}=4\mathrm{EX}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BE}+\mathrm{EX}=4\mathrm{EX}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BE}=3\mathrm{EX}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BE}:\mathrm{EX}=3\hspace{0.17em}:1$

#### Page No 4.75:

#### Question 17:

ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC.

#### Answer:

**Given: **

ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q.

**To Prove: **

The rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC. We need to prove that $\mathrm{BP}\times \mathrm{DQ}=\mathrm{AB}\times \mathrm{BC}$

**Proof: **

In ∆ABP and ∆QCP, we have

∠ABP = ∠QCP (Alternate angles as AB || DC)

∠BPA = ∠QPC ( Vertically opposite angles)

By AA similarity, we get

∆ABP ~ ∆QCP

We know that corresponding sides of similar triangles are proportional.

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{QC}}=\frac{\mathrm{BP}}{\mathrm{CP}}=\frac{\mathrm{AP}}{\mathrm{QP}}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{QC}}=\frac{\mathrm{BP}}{\mathrm{CP}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}\times \mathrm{CP}=\mathrm{QC}\times \mathrm{BP}$

Adding $\mathrm{AB}\times \mathrm{BP}$ in both sides, we get

$\Rightarrow \mathrm{AB}\times \mathrm{CP}+\mathrm{AB}\times \mathrm{BP}=\mathrm{QC}\times \mathrm{BP}+\mathrm{AB}\times \mathrm{BP}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}\times \left(\mathrm{CP}+\mathrm{BP}\right)=\left(\mathrm{QC}+\mathrm{AB}\right)\times \mathrm{BP}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}\times \left(\mathrm{CP}+\mathrm{BP}\right)=\left(\mathrm{QC}+\mathrm{CD}\right)\times \mathrm{BP}\left(\mathrm{ABCD}\hspace{0.17em}\mathrm{is}\mathrm{a}\mathrm{parallelogram},\mathrm{AB}=\mathrm{CD}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}\times \mathrm{BC}=\mathrm{DQ}\times \mathrm{BP}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BP}\times \mathrm{DQ}=\mathrm{AB}\times \mathrm{BC}$

#### Page No 4.75:

#### Question 18:

In ∆ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that:

(i) ∆OMA ∼ ∆OLC

(ii) $\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OM}}{\mathrm{OL}}$

#### Answer:

$\left(\mathrm{i}\right).\mathrm{In}\u2206\mathrm{OMA}\mathrm{and}\u2206\mathrm{OLC},\phantom{\rule{0ex}{0ex}}\angle \mathrm{AOM}=\angle \mathrm{COL}\left[\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}\right]\phantom{\rule{0ex}{0ex}}\angle \mathrm{OMA}=\angle \mathrm{OLC}\left[90\xb0\mathrm{each}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206\mathrm{OMA}~\u2206\mathrm{OLC}\left[\mathrm{AA}\mathrm{similarity}\right]\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right).\mathrm{Since}\u2206\mathrm{OMA}~\u2206\mathrm{OLC}\mathrm{by}\mathrm{AA}\mathrm{similarity},\mathrm{then}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{OM}}{\mathrm{OL}}=\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{MA}}{\mathrm{LC}}\left[\mathrm{Corresponding}\mathrm{sides}\mathrm{of}\mathrm{similar}\mathrm{triangles}\mathrm{are}\mathrm{proportional}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OM}}{\mathrm{OL}}$

#### Page No 4.75:

#### Question 19:

ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

#### Answer:

**Given:**

ABCD is quadrilateral in which AD = BC and P, Q, R, S are the mid points of AB, AC, CD, BD respectively.

**To Prove:**

PQRS is a rhombus.

**Proof:**

In ∆ABC, P and Q are the midpoints of the sides AB and AC respectively.

By the Mid point theorem, we get

PQ || BC and PQ = $\frac{1}{2}$BC ...(1)

In ∆ADC, Q and R are the midpoints of the sides AC and DC respectively.

By the Mid point theorem, we get

QR || AD and QR = $\frac{1}{2}$AD = $\frac{1}{2}$BC (Since AD = BC) ...(2)

Similarly, in ∆BCD, we have

RS || BC and RS = $\frac{1}{2}$BC ...(3)

In ∆BAD, we have

PS || AD and PS = $\frac{1}{2}$AD = $\frac{1}{2}$BC (Since AD = BC) ...(4)

From the equations (1), (2), (3), (4), we get

PQ = QR = RS = RS

Thus, PQRS is a rhombus.

#### Page No 4.75:

#### Question 20:

In an isosceles ∆ABC, the base AB is produced both the ways to P and Q such that AP ✕ BQ = AC^{2}. Prove that ∆APC ∼ ∆BCQ.

#### Answer:

It is given that is isosceles and _{.}

We have to prove that .

It is given that is an isosceles triangle, so AC = BC.

Now,

(Given)

$\Rightarrow \frac{AP}{AC}=\frac{BC}{BQ}$

Also,

$\angle CAB=\angle CBA\left(\mathrm{Equal}\mathrm{sides}\mathrm{have}\mathrm{equal}\mathrm{angles}\mathrm{opposite}\mathrm{to}\mathrm{them}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 180\xb0-\angle CAP=180\xb0-\angle CBQ\phantom{\rule{0ex}{0ex}}\Rightarrow \angle CAP=\angle CBQ\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, (SAS Similarity)

#### Page No 4.75:

#### Question 21:

A girl of heigh 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

#### Answer:

It is given that, girl height, speed and height of lamp.

We have to find the length of her shadow after

Let be the lamp post and be the girl.

Suppose is the length of her shadow.

Let* DE *= *x*

And

Now in and we have

and

So by similarly criterion

$\frac{BE}{DE}=\frac{AB}{CD}\phantom{\rule{0ex}{0ex}}\frac{4.8+x}{x}=\frac{3.6}{0.9}=4\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=4.8\phantom{\rule{0ex}{0ex}}\Rightarrow x=1.6\phantom{\rule{0ex}{0ex}}$

Hence the length of her shadow after is 1.6 m.

#### Page No 4.75:

#### Question 22:

A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

#### Answer:

It is given that length of vertical stick

We have to find the height of the tower.

Suppose is the height of the tower and is its shadow.

Now, $\left(\angle B=\angle Q\mathrm{and}\angle A=\angle P\right)$

Hence the height of the tower is _{. }

#### Page No 4.75:

#### Question 23:

In the given figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ∼ ∆ADE and hence find the length of AE and DE.

#### Answer:

It is given that is right angle triangle and

We have to prove that and find the lengths of and .

In ∆*ABC* ∼ ∆*ADE*,

$\angle A=\angle A\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\angle C=\angle E\left(90\xb0\right)$

So by similarly criterion, we have

Since

So

And

Hence, and

#### Page No 4.76:

#### Question 24:

In the given figure, PA, QB and RC are each perpendicular to AC. Prove that $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.

#### Answer:

It is given that _{,} and are each perpendicular to .

We have to prove that

In we have

$\Rightarrow \frac{y}{x}=\frac{CB}{CA}$ ......(1)

Now in _{,} we have

…… (2)

Adding (1) and (2) we have

Hence, .

#### Page No 4.76:

#### Question 25:

In the given figure, we have AB || CD || EF. If AB = 6 cm, CD =* x* cm, EF = 10 cm, BD = 4 cm and DE = *y* cm, Calculate the values of *x* and *y*.

#### Answer:

It is given that .

,and

We have to calculate the values of and .

In and , we have

(Vertically opposite angles)

(Alternate interior angles)

So

$\frac{EF}{AB}=\frac{DE}{DB}$

Similarly in we have

$\frac{DC}{AB}=\frac{DE}{BE}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{6}=\frac{y}{4+y}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{6y}{4+y}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{6\times 6.67}{4+6.67}\phantom{\rule{0ex}{0ex}}\Rightarrow x=3.75$

Hence, $x=3.75\mathrm{cm}$ and

#### Page No 4.94:

#### Question 1:

Triangles ABC and DEF are similar.

(i) If area (∆ABC) = 16 cm^{2}, area (∆DEF) = 25 cm^{2} and BC = 2.3 cm, find EF.

(ii) If area (∆ABC) = 9 cm^{2}, area (∆DEF) = 64 cm^{2} and DE = 5.1 cm , find AB.

(iii) If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles.

(iv) If area (∆ABC) = 36 cm^{2}, area (∆DEF) = 64 cm^{2} and DE = 6.2 cm, find AB.

(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of ∆ABC and ∆DEF.

#### Answer:

**Given: **ΔABC and ΔDEF are similar triangles

**To find:**

(i) If area of ΔABC = 16cm^{2}, area of ΔDEF = 25cm^{2} and BC = 2.3 cm, Find EF.

(ii) If area of ΔABC = 9cm^{2}, area of ΔDEF = 64cm^{2} and DE = 5.1 cm, Find AB.

(iii) If AC = 19cm and DF = 8cm, find the ratio of the area of two triangles.

(iv) If area of ΔABC = 36cm^{2}, area of ΔDEF = 64cm^{2} and DE = 6.2 cm, Find AB.

(v) If AB = 1.2cm and DE = 1.4cm, find the ratio of the area of two triangles.

(i) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

(ii)

(iii)

(iv)

(v)

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#### Question 2:

In the given figure, ∆ACB ∼ ∆APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Also, find the area (∆ACB) : area (∆APQ).

#### Answer:

Given: ΔACB is similar to ΔAPQ.

BC = 10 cm, PQ = 5cm, BA = 6.5cm and AP = 2.8 cm

TO FIND:

(1) CA and AQ

(2) Area of ΔACB : Area of ΔAPQ

(1) It is given that ΔACB $~$ ΔAPQ.

We know that for any two similar triangles the sides are proportional. Hence

$\frac{\mathrm{AB}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{AP}}$

Similarly,

(2) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{\mathrm{ar}\left(\u2206\mathrm{ACQ}\right)}{\mathrm{ar}\left(\u2206\mathrm{APQ}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{PQ}}\right)}^{2}={\left(\frac{10}{5}\right)}^{2}={\left(\frac{2}{1}\right)}^{2}=\frac{4}{1}$

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#### Question 3:

The areas of two similar triangles are 81 cm^{2} and 49 cm^{2}^{ }respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians?

#### Answer:

Given: The area of two similar triangles is 81cm^{2} and 49cm^{2} respectively.

To find:

(1) Ratio of their corresponding heights.

(2) Ratio of their corresponding medians.

(1) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Taking square root on both sides, we get

(2) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians.

Taking square root on both sides, we get

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#### Question 4:

The areas of two similar triangles are 169 cm^{2} and 121 cm^{2} respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.

#### Answer:

Given: The area of two similar triangles is 169cm^{2} and 121cm^{2} respectively.The longest side of the larger triangle is 26cm.

To find: Longest side of the smaller triangle

Taking square root on both sides, we get

= 22 cm

Hence, the longest side of the smaller triangle is .

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#### Question 5:

The areas of two similar triangles are 25 cm^{2} and 36 cm^{2} respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.

#### Answer:

Given: The area of two similar triangles is 25cm^{2} and 36cm^{2} respectively. If the altitude of first triangle is 2.4cm

To find: The altitude of the other triangle

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Taking square root on both sides, we get

$\frac{5}{6}=\frac{2.4}{\mathrm{altitude}2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{altitude}2=2.88\mathrm{cm}$

Hence, the corresponding altitude of the other is 2.88 cm.

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#### Question 6:

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

#### Answer:

Given: The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively.

To find: Ratio of areas of triangle.

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Hence, the ratio of areas of two triangles is 4 : 9.

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#### Question 7:

ABC is a triangle in which ∠A = 90°, AN ⊥ BC, BC = 12 cm and AC = 5 cm. Find the ratio of the area of ∆ANC and ∆ABC.

#### Answer:

Given: In ΔABC,, , BC = 12cm and AC = 5cm.

TO FIND: Ratio of the triangles ∆*ANC *and ∆*ABC*.

In ∆*ANC *and ∆*ABC*,

$\angle ACN=\angle ACB$ (Common)

$\angle A=\angle ANC$ (90º each)

∴ ∆*ANC *$~$ ∆*ABC *(AA Similarity)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\therefore \frac{Ar\left(\u2206ANC\right)}{Ar\left(\u2206ABC\right)}={\left(\frac{AC}{BC}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{Ar\left(\u2206ANC\right)}{Ar\left(\u2206ABC\right)}={\left(\frac{5\mathrm{cm}}{12\mathrm{cm}}\right)}^{2}$

⇒

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#### Question 8:

In the given figure, DE || BC

(i) If DE = 4 cm, BC = 6 cm and Area (∆ADE) = 16 cm^{2}, find the area of ∆ABC.

(ii) If DE = 4 cm, BC = 8 cm and Area (∆ADE) = 25 cm^{2}, find the area of ∆ABC.

(iii) If DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED.

#### Answer:

In the given figure, we have DE || BC.

In Δ*ADE*and Δ

*ABC*

$\angle ADE=\angle B\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle DAE=\angle BAC\left(\mathrm{Common}\right)$

So, $\u2206ADE~\u2206ABC$ (AA Similarity)

(i) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence

(ii) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence,

(iii) We know that

Let Area of ΔADE = 9*x* sq. units and Area of ΔABC = 25*x* sq. units

Now,

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#### Question 9:

In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC.

#### Answer:

Given: In ΔABC, D and E are the midpoints of AB and AC respectively.

To find: Ratio of the areas of ΔADE and ΔABC.

Since it is given that D and E are the midpoints of AB and AC, respectively.

Therefore, DE || BC (Converse of mid-point theorem)

Also, $\mathrm{DE}=\frac{1}{2}\mathrm{BC}$

$\angle \mathrm{ADE}=\angle \mathrm{B}\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle \mathrm{DAE}=\angle \mathrm{BAC}\left(\mathrm{Common}\right)$

So, $\u2206\mathrm{ADE}~\u2206\mathrm{ABC}$ (AA Similarity)

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#### Question 10:

The areas of two similar triangles are 100 cm^{2} and 49 cm^{2} respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other.

#### Answer:

Given: The area of two similar triangles is 100cm^{2} and 49cm^{2} respectively. If the altitude of bigger triangle is 5 cm

To find: their corresponding altitude of other triangle

Taking square root on both side

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#### Question 11:

The areas of two similar triangles are 121 cm^{2} and 64 cm^{2} respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.

#### Answer:

Given: The area of two similar triangles is 121cm^{2} and 64cm^{2} respectively. IF the median of the first triangle is 12.1cm

To find: corresponding medians of the other triangle

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians.

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#### Question 12:

If ∆ABC ∼ ∆DEF such that AB = 5 cm, area (∆ABC) = 20 cm^{2} and area (∆DEF) = 45 cm^{2}, determine DE.

#### Answer:

Given: The area of two similar ΔABC = 20cm^{2}, ΔDEF = 45cm^{2} respectively and AB = 5cm.

To find: measure of DE

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#### Question 13:

In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ∆ABC into two parts equal in area. Find $\frac{\mathrm{BP}}{\mathrm{AB}}$.

#### Answer:

Given: In ΔABC, PQ is a line segment intersecting AB at P, and AC at Q such that PQ || BC and PQ divides ΔABC in two parts equal in area.

To find:

We have PQ || BC

And

Now, PQ || BC and BA is a transversal.

In ΔAPQ and ΔABC,

$\angle \mathrm{APQ}=\angle \mathrm{B}\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle \mathrm{PAQ}=\angle \mathrm{BAC}\left(\mathrm{Common}\right)$

So, $\u2206\mathrm{APQ}~\u2206\mathrm{ABC}$ (AA Similarity)

Hence

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#### Question 14:

The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.

#### Answer:

Given: The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. BC = 4.5cm.

To find: length of QR

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{\mathrm{ar}\left(\u2206\mathrm{ABC}\right)}{\mathrm{ar}\left(\u2206\mathrm{PQR}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^{2}$

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#### Question 15:

ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 m, prove that the area of ∆APQ is one-sixteenth of the area of ∆ABC.

#### Answer:

Given: In ΔABC, PQ is a line segment intersecting AB at P, and AC at Q. AP = 1cm, PB = 3cm, AQ = 1.5cm and QC = 4.5cm.

To find:

In ΔABC,

According to converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Hence, PQ || BC

In ΔAPQ and ΔABC,

$\angle \mathrm{APQ}=\angle \mathrm{B}\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle \mathrm{PAQ}=\angle \mathrm{BAC}\left(\mathrm{Common}\right)$

So, $\u2206\mathrm{APQ}~\u2206\mathrm{ABC}$ (AA Similarity)

Hence

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#### Question 16:

If D is a point on the side AB of ∆ABC such that AD : DB = 3.2 and E is a point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE.

#### Answer:

Given: In ΔABC, D is a point on side AB such that AD : DB= 3 : 2. E is a point on side BC such that DE || AC.

To find:

In ΔABC,

According to converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Hence DE || AC

In ΔBDE and ΔABC,

$\angle \mathrm{BDE}=\angle \mathrm{A}\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle \mathrm{DBE}=\angle \mathrm{ABC}\left(\mathrm{Common}\right)$

So, $\u2206\mathrm{BDE}~\u2206\mathrm{ABC}$ (AA Similarity)

Let AD = 2*x* and BD = 3*x.*

Hence

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#### Question 17:

If ∆ABC and ∆BDE are equilateral triangles, where D is the mid point of BC, find the ratio of areas of ∆ABC and ∆BDE.

#### Answer:

Given: In ΔABC and ΔBDE are equilateral triangles. D is the midpoint of BC.

To find: $\frac{\mathrm{Ar}\left(\u2206\mathrm{ABC}\right)}{\mathrm{Ar}\left(\u2206\mathrm{BDE}\right)}$

In ΔABC and ΔBDE

Since D is the midpoint of BC, BD : DC = 1.

Let DC = *x*, and BD = *x*

Therefore BC = BD + DC = 2*x*

Hence

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#### Question 18:

Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.

#### Answer:

Given: Two isosceles triangles have equal vertical angles and their areas are in the ratio of 36:25.

To find: Ratio of their corresponding heights.

Suppose ∆ABC and ∆PQR are two isosceles triangles with $\angle \mathrm{A}=\angle \mathrm{P}$.

Now, AB = AC and PQ = PR

$\therefore \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{PQ}}{\mathrm{PR}}$

In ∆ABC and ∆PQR,

$\angle \mathrm{A}=\angle \mathrm{P}$

$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{PQ}}{\mathrm{PR}}$

∴ ∆ABC $~$∆PQR (SAS Similarity)

Let AD and PS be the altitudes of ∆ABC and ∆PQR, respectively.

$\therefore \frac{\mathrm{ar}\left(\u2206\mathrm{ABC}\right)}{\mathrm{ar}\left(\u2206\mathrm{PQR}\right)}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{36}{25}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{AD}}{\mathrm{PS}}=\frac{6}{5}$

Hence, the ratio of their corresponding heights is 6 : 5.

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#### Question 19:

In the given figure, ∆ABC and ∆DBC are on the same base BC. If AD and BC intersect at O, prove that $\frac{\mathrm{Area}\left(\u2206\mathrm{ABC}\right)}{\mathrm{Area}\left(\u2206\mathrm{DBC}\right)}=\frac{\mathrm{AO}}{\mathrm{DO}}$.

#### Answer:

Given: ΔABC and ΔDBC are on the same base BC. AD and BC intersect at O.

Prove that:

Construction: Draw and .

Now, in ΔALO and ΔDMO, we have

$\therefore \frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}$ (Corresponding sides are proportional)

$\frac{\mathrm{Ar}\left(\u2206\mathrm{ABC}\right)}{\mathrm{Ar}\left(\u2206\mathrm{BCD}\right)}=\frac{{\displaystyle \frac{1}{2}}\times \mathrm{BC}\times \mathrm{AL}}{{\displaystyle \frac{1}{2}}\times \mathrm{BC}\times \mathrm{DM}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \mathrm{AL}}}{\mathrm{DM}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{AO}}{\mathrm{DO}}$

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#### Question 20:

ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that : (i) ∆AOB ∼ ∆COD (ii) If OA = 6 cm, OC = 8 cm,

Find:

(a) $\frac{\mathrm{Area}\left(\u2206\mathrm{AOB}\right)}{\mathrm{Area}\left(\u2206\mathrm{COD}\right)}$

(b) $\frac{\mathrm{Area}\left(\u2206\mathrm{AOD}\right)}{\mathrm{Area}\left(\u2206\mathrm{COD}\right)}$

#### Answer:

Given: ABCD is a trapezium in which AB || CD.

The diagonals AC and BD intersect at O.

To prove:

(i)

(ii) If OA = 6 cm, OC = 8 cm

To find:

(*a*)

(*b*)

Construction: Draw a line MN passing through O and parallel to AB and CD

(i) Now in ΔAOB and ΔCOD

$\angle \mathrm{OAB}=\angle \mathrm{OCD}\left(\mathrm{Alternate}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{OBA}=\angle \mathrm{ODC}\left(\mathrm{Alternate}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}$

(*ii*) (*a*)We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

(*b*)We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{\mathrm{ar}\left(\u2206\mathrm{AOD}\right)}{\mathrm{ar}\left(\u2206\mathrm{COD}\right)}={\left(\frac{\mathrm{OA}}{\mathrm{OC}}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{6\mathrm{cm}}{8\mathrm{cm}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{9}{16}$

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#### Question 21:

In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.

#### Answer:

GIVEN: In ΔABC, P divides the side AB such that AP : PB = 1 : 2, Q is a point on AC such that PQ || BC.

TO FIND: The ratio of the areas of ΔAPQ and the trapezium BPQC.

In ΔAPQ and ΔABC

$\angle \mathrm{APQ}=\angle \mathrm{B}\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle \mathrm{PAQ}=\angle \mathrm{BAC}\left(\mathrm{Common}\right)$

So, $\u2206\mathrm{APQ}~\u2206\mathrm{ABC}$ (AA Similarity)

Let Area of ΔAPQ= 1 sq. units and Area of ΔABC = 9*x* sq. units

Now,

$\frac{\mathrm{ar}\left(\u2206\mathrm{APQ}\right)}{\mathrm{ar}\left(\mathrm{trapBCED}\right)}=\frac{x\mathrm{sq}\mathrm{units}}{8x\mathrm{sq}\mathrm{units}}=\frac{1}{8}\phantom{\rule{0ex}{0ex}}$

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#### Question 22:

AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed.

Prove that Area (∆ADE) : Area (∆ABC) = 3 : 4.

#### Answer:

We have an equilateral triangle in which AD is altitude. An equilateral triangle is drawn using AD as base. We have to prove that,

Since the two triangles are equilateral, the two triangles will be similar also.

We know that according to the theorem, the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

…… (1)

Nowis an equilateral triangle. So,

.

Therefore,

So,

We will now use this in equation (1). So,

Hence, proved.

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