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#### Page No 4.120:

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 4.120:

(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

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.

#### Page No 4.120:

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 4.120:

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 4.120:

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:

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,

Therefore, the altitude of the isosceles triangle is .

#### Page No 4.120:

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,

AC2 = AB2 + BC2

⇒ AC2 = (8 m)2 + (6 m)2 = (10 m)2

⇒ AC = 10 m

Now, again applying Pythagoras theorem in ΔEBD

DE2 = EB2 + BD2

⇒ (10 m)2 = (8 m)2 + BD2

⇒ BD2 = 100 m2 − 64 m2 = 36 m2

⇒ BD = 6 m

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

#### Page No 4.120:

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:

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 4.121:

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 4.121:

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, $Ar\left(∆FBE\right)=\frac{1}{2}×BF×BE$

cm2

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,

AB2 = 63 × 6

AB2 = 64

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 4.121:

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)

(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 4.121:

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 2a

Therefore, .

#### Page No 4.121:

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

Therefore, length of the side of the rhombus is .

#### Page No 4.121:

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 4.121:

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:

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.

AD2 = 144 − 36 = 108

Taking square root, we get

Therefore, the height of the equilateral triangle is.

#### Page No 4.121:

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

We need to prove that.

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:

(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:

(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 4.122:

Given: ΔABC where BAC is obtuse. PB AC and QCAB.

To prove: (i) AB × AQ = AC × AP and (ii) BC2 = 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,

⇒ BC2 = CQ2 + QB2

⇒ BC2 = CQ2 + (QA + AB)2

⇒ BC2 = CQ2 + QA2 + AB2 + 2QA × AB

⇒ BC2 = AC2 + AB2 + QA × AB + QA × AB                  [In right ΔACQ, CQ2 + QA2 = AC2]

⇒ BC2 = AC2 + AB2 + QA × AB + AC × AP                  (Using (1))

⇒ BC2 = AC (AC + AP) + AB (AB + QA)

⇒ BC2 = AC × CP + AB × BQ

#### Page No 4.122:

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 4.122:

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 4.122:

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, 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 4.122:

(i)

In and ,

$\angle ACB=\angle A=90°$

(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 AB2/AC2 = BD/DC, we will divide equation (1) by equation (2) as shown below,

Canceling BC we get,

Therefore,

#### Page No 4.122:

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 4.122:

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 4.122:

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 4.123:

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 4.124:

If , then

Hence, the value of is .

#### Page No 4.124:

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 4.124:

Given, .

$\angle ADE=\angle C$
(Corresponding angles)

$\angle A=\angle A$     (Common)

(AA Similarity)

Hence the value of and is

#### Page No 4.125:

(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 4.125:

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

In ΔAPQ and ΔABC,

So, $∆\mathrm{APQ}~∆\mathrm{ABC}$      (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 4.125:

(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 ΔAPQ and ΔABC,

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

two triangles are similar

(iii) In two triangle, we observe that

In ΔABC and ΔCDE

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

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

ΔABC $~$ 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:

Given and .

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

#### Page No 4.125:

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 4.126:

We are given ∆AMB ∼ ∆CMD

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

Given

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

By cross multiplication we get

Hence, the value of MD is .

#### Page No 4.126:

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 4.126:

(i) Three pair of similar triangles are-

$∆ABK~∆PQK$       (AAA Similarity)

$∆CBK~∆RQK$       (AAA Similarity)

$∆ACK~∆PRK$        (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 4.126:

(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:

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.

From (1) and (2), we get

Hence, the value of AD is .

#### Page No 4.126:

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:

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 = 3CD 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:

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 4.127:

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 4.127:

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 4.127:

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:

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:

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 4.127:

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:

(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:

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 4.128:

In ΔABC in which A is an acute angle with 60°.

Now apply Pythagoras theorem in triangle BCD

Hence

#### Page No 4.128:

Given: ΔABC where C is an obtuse angle, AD BC and AB2 = AC2 + 3BC2

To prove: BC = CD

Proof:

In ΔABC,  C is obtuse.

Therefore,

AB2 = AC2 + BC2 + 2BC × DC          (Obtuse angle theorem)          …(1)

AB2 = AC2 + 3BC2         (Given)                …(2)

From (1) and (2), we get

AC2 + 3BC2 = AC2 + BC2 + 2BC × DC

⇒ 3BC2 = BC2 + 2BC × DC

⇒ 2BC2 = 2BC × DC

⇒ BC = DC

#### Page No 4.128:

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 4.128:

Since is right triangle right angled at D

In right , we have

Now substitute

#### Page No 4.128:

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

To prove: AC2 + BD2 = AD2 + BC2

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,

AC2 = AO2 + OC2

BD2 = OB2 + OD2

∴ AC2 + BD2 = (AO2 + OD2) + (OC2 + OB2)

⇒ AC2 + BD2 = AD2 + BC2

This proves the given relation.

#### Page No 4.128:

Since is right triangle right angled at D

Substitute

Now, in , we have

Therefore,

Hence proved.

#### Page No 4.128:

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

To prove: BM2 + BN2 = DM2 + DN2

Proof:

Apply Pythagoras Theorem in ΔAMB and ΔCND,

AB2 = AM2 + MB2

CD2 = CN2 + ND2

Since AB = CD, AM2 + MB2 = CN2 + ND2

⇒ AM2 − CN2 = ND2 − MB2 … (i)

Again apply Pythagoras Theorem in ΔAMD and ΔCNB,

CB2 = CN2 + NB2

Since AD = BC, AM2 + MD2 = CN2 + NB2

⇒ AM2 − CN2 = NB2 − MD2 … (ii)

Equating (i) and (ii),

ND2 − MB2 = NB2 − MD2

I.e., BM2 + BN2 = DM2 + DN2

This proves the given relation.

#### Page No 4.128:

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 4.128:

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 4.128:

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 4.128:

∠B = 90°

So,    (Pythagoras theorem)

and

(Given)

So,

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

#### Page No 4.128:

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:

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

In this case the diagram will look like the following

Now The length of the new string AD = ACCD = 3.00 − 0.6 = 2.4 m

Now in triangle ADB we have

∴ Required distance =

Hence, the horizontal distance is .

#### Page No 4.129:

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 4.129:

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 4.129:

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 4.130:

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:

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

In ΔABC and ΔDEF, if

Then,

#### Page No 4.130:

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

In ΔABC and ΔDEF, if

Then,

#### Page No 4.130:

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 4.130:

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 4.130:

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 4.130:

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 4.130:

Let ∆ABC and ΔPQR are similar triangles. The area of triangles is 169cm2 and 121cm2, 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 4.130:

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 4.130:

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 4.130:

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 4.130:

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 4.130:

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 4.130:

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:

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 $∆\mathrm{LMK}$ and $∆\mathrm{PNK}$,

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

Hence we got the result as .

#### Page No 4.131:

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

Now, we know that the areas of two similar triangles are in the ratio of the squares of the corresponding sides. Therefore,

#### Page No 4.131:

Given:

AK = 10 cm

BC = 3.5 cm

HK = 7 cm

To find: AC

Since, so their corresponding sides are proportional.

#### Page No 4.132:

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

To find: DE

So,

#### Page No 4.132:

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

To find: DE

#### Page No 4.132:

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 4.132:

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 4.133:

Given: Areas of two similar triangles are 9cm2 and 16cm2.

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 4.133:

Given: Areas of two similar triangles ΔABC and ΔDEF are 144cm2 and 81cm2.

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

= 27 cm

#### Page No 4.133:

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.

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 4.133:

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,

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

Hence,

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

Hence we got the result as

#### Page No 4.133:

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 4.133:

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 4.133:

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 4.133:

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,

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

It is given that AB = 4BX.

Let AB = 4x and BX = x.

Then, AX = 3x

Hence the correct answer is .

#### Page No 4.133:

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

AC2 = AE2 + EC2

AC2 = 52 + 122                  (since CDEB forms a rectangle and opposite sides of rectangle are equal)

AC2 = 25 + 144

AC2 = 169

Thus, the distance between their tops is 13m.

#### Page No 4.133:

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 4.133:

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 4.133:

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 4.133:

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°.

#### Page No 4.133:

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.

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 option is .

#### Page No 4.133:

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

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

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

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

∴ ∆ACB $~$∆ADC     (AA Similarity)

We got the result as

#### Page No 4.134:

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

#### Page No 4.134:

Given: In an equilateral ΔABC,

Applying Pythagoras theorem,

In ΔABD,

We got the result as .

#### Page No 4.134:

Given: In an equilateral ΔABC, .

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

Applying Pythagoras theorem,

We got the result as

#### Page No 4.134:

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

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 4.134:

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.

We know that

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

Now ,

#### Page No 4.134:

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 4.134:

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 4.134:

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

#### Page No 4.134:

Given: In ΔABC, and.

To find: BD: DC

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

#### Page No 4.134:

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 4.134:

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 4.134:

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

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

We got the result as

#### Page No 4.135:

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 4.135:

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 4.135:

Given:

To find: measure of EF

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

#### Page No 4.135:

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

#### Page No 4.135:

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

#### Page No 4.135:

Given: In Δ ABC and Δ DEF

To find: Ar(Δ DEF)

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

#### Page No 4.135:

Given: In Δ ABC and ΔPQR

To find: Measure of QR

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

#### Page No 4.135:

Given: The area of two similar triangles is 121cm2 and 64cm2 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

Hence the correct answer is .

#### Page No 4.135:

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 4.135:

In an equilateral ΔABC, .

In ΔADC, applying Pythagoras theorem, we get

Hence, the correct option is (c).

#### Page No 4.135:

∆ABC is an equilateral triangle and .

In ∆ABD, applying Pythagoras theorem, we get

We got the result as .

#### Page No 4.135:

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 4.135:

Given: In Isosceles ΔABC, AC = BC and AB2 = 2AC2.

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 4.136:

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 4.136:

In ΔABC and ΔDEF

∴ ΔABC and ΔDEF are similar triangles.

Hence

Hence the correct answer is (b).

#### Page No 4.136:

Given: In an isosceles triangle ABC, AB = AC = 25 cm and BC = 14cm.

To find: Measure of altitude from A on BC

Since AD ⊥ BC, so BD = CD =

In right ∆ABD,

We got the result as .

#### Page No 4.136:

ΔABC and ΔDEF,

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

ΔABC $~$ ΔEFD   (SAS Similarity)

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

#### Page No 4.136:

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 4.136:

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 4.137:

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

To find: the value of x and y respectively.

This relation is satisfied by option (d).

Hence, x = 16 cm and y = 8cm

Hence the result is .

#### Page No 4.137:

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

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 4.19:

(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

$AE=\frac{4×2.5}{5}=2$ 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 = 8x − 7, DB = 5x − 3, AE = 4x − 3 and EC = 3x − 1.

We have to find the value of .

So   (by Thales theorem)

Then,

Hence,

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

(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:

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

#### Page No 4.19:

It is given that ,, and .

We have to find BD and CE.

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

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

∠AED = ∠ACB   (corresponding angles)

∠AED = ∠ACB  (proved above)

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

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

Hence, BD = 3.6 cm  and  CE = 4.8 cm.

#### Page No 4.19:

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:

(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 4.2:

(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 4.20:

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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

#### Page No 4.30:

(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,

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

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:

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:

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:

We have to prove that .
In ∆ABC,

(Given)

So, AD is the bisector of .

Therefore,

#### Page No 4.31:

It is given that , and .

We have to find , and BD.

Sinceis bisector of

So

Then,

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

So,

Since is the bisector of .

So,

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

So

Now since is the bisector of

So

So

Hence

And

#### Page No 4.31:

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

(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

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

(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:

It is given that bisect . Also, , and .

We have to find .

Since is the bisector of

Then

Hence

#### Page No 4.37:

(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:

It is given that .

,, and.

We have to find and .

Since

So

Similarly

Hence, and

#### Page No 4.74:

We have to find the height of .

Now,

∆ABC $~$∆PQR      (AA Similarity)

Hence

#### Page No 4.74:

It is given that

,and PR = 6 cm

We have to find .

Since

So

Hence,

#### Page No 4.74:

It is given that .

,and

We have to find .

Since

()

So

Hence,

#### Page No 4.74:

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:

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:

It is given that,,, and

We have to find .

Since

So

Hence,

#### Page No 4.74:

It is given that, and .

Since

So

Hence,

#### Page No 4.74:

It is given that , and are each perpendicular to .

We have to prove that

In we have

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

Now in , we have

…… (2)

Adding (1) and (2) we have

Hence, .

#### Page No 4.75:

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:

#### Page No 4.75:

Comparing ΔCAB and ΔCED,

CAB = CED                [Given]

ACB = ECD                [Common]

∴ ΔCAB ΔCED

#### Page No 4.75:

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.

Hence

#### Page No 4.75:

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:

It is given that,, and .

We have to prove that

Since clearly

Also, is common in and

So     (SAS Similarity)

#### Page No 4.75:

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 $\mathrm{BP}×\mathrm{DQ}=\mathrm{AB}×\mathrm{BC}$

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.

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

$⇒\frac{\mathrm{AB}}{\mathrm{QC}}=\frac{\mathrm{BP}}{\mathrm{CP}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{AB}×\mathrm{CP}=\mathrm{QC}×\mathrm{BP}$

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

#### Page No 4.75:

It is given that .

,and

We have to calculate the values of and .

In and , we have

(Vertically opposite angles)

(Alternate interior angles)

So

Similarly in we have

Hence and

#### Page No 4.75:

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)
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 4.76:

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 4.76:

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)

$⇒\frac{AP}{AC}=\frac{BC}{BQ}$
Also,

Hence,            (SAS Similarity)

#### Page No 4.76:

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

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

Hence the length of her shadow after is 1.6 m.

#### Page No 4.76:

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 4.76:

(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 4.76:

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,

Hence the height of the tower is .

#### Page No 4.76:

It is given that is right angle triangle and

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

So by similarly criterion, we have

Since

So

And

Hence, and

#### Page No 4.95:

Given: ΔABC and ΔDEF are similar triangles

To find:

(i) If area of ΔABC = 16cm2, area of ΔDEF = 25cm2 and BC = 2.3 cm, Find EF.

(ii) If area of ΔABC = 9cm2, area of ΔDEF = 64cm2 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 = 36cm2, area of ΔDEF = 64cm2 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)

#### Page No 4.95:

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.

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

#### Page No 4.96:

Given: The area of two similar triangles is 81cm2 and 49cm2 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

#### Page No 4.96:

Given: The area of two similar triangles is 169cm2 and 121cm2 respectively.The longest side of the larger triangle is 26cm.

To find: Longest side of the smaller triangle

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

= 22 cm

Hence, the longest side of the smaller triangle is .

#### Page No 4.96:

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.

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

$\therefore \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{PQR}\right)}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{36}{25}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AD}}{\mathrm{PS}}=\frac{6}{5}$

Hence, the ratio of their corresponding heights is 6 : 5.

#### Page No 4.96:

Given: The area of two similar triangles is 25cm2 and 36cm2 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

Hence, the corresponding altitude of the other is 2.88 cm.

#### Page No 4.96:

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.

#### Page No 4.96:

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.

#### Page No 4.96:

In the given figure, we have DE || BC.

So, $∆ADE~∆ABC$      (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 = 9x sq. units and Area of ΔABC = 25x sq. units

Now,

#### Page No 4.96:

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, $\mathrm{DE}=\frac{1}{2}\mathrm{BC}$

So, $∆\mathrm{ADE}~∆\mathrm{ABC}$      (AA Similarity)

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

#### Page No 4.96:

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)

#### Page No 4.96:

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

(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.

#### Page No 4.96:

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

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

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

Let Area of ΔAPQ= 1 sq. units and Area of ΔABC = 9x sq. units

Now,

#### Page No 4.96:

Given: The area of two similar triangles is 100cm2 and 49cm2 respectively. If the altitude of bigger triangle is 5 cm

To find: their corresponding altitude of 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 side

#### Page No 4.97:

Given: The area of two similar triangles is 121cm2 and 64cm2 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.

#### Page No 4.97:

Given: The area of two similar ΔABC = 20cm2, ΔDEF = 45cm2 respectively and AB = 5cm.

To find: measure of DE

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

#### Page No 4.97:

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,

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

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

Hence

#### Page No 4.97:

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.

$\frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{PQR}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^{2}$

#### Page No 4.97:

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,

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

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

Hence

#### Page No 4.97:

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,

So, $∆\mathrm{BDE}~∆\mathrm{ABC}$      (AA Similarity)

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

Let AD = 2x and BD = 3x.

Hence

#### Page No 4.97:

Given: In ΔABC and ΔBDE are equilateral triangles. D is the midpoint of BC.

To find: $\frac{\mathrm{Ar}\left(∆\mathrm{ABC}\right)}{\mathrm{Ar}\left(∆\mathrm{BDE}\right)}$

In ΔABC and ΔBDE

Since D is the midpoint of BC, BD : DC = 1.

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

Let DC = x, and BD = x

Therefore BC = BD + DC = 2x

Hence

#### Page No 4.97:

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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