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#### Question 1:

D and E are points on the sides AB and AC, respectively, of a $∆ABC$, such that $DE\parallel BC.$

(i) If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC.
(ii) If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.
(iii) If find AE.
(iv) If find AE.

(i)

Applying Thales' theorem, we get:

$\because$ AD = 3.6 cm , AB = 10 cm, AE = 4.5 cm​
$\therefore$ DB = 10 $-$ 3.6 = 6.4 cm

(ii)

(iii)

(iv)

#### Question 2:

D and E are points on the sides AB and AC respectively of a $∆ABC$ such that $DE\parallel BC.$ Find the value of x, when

(i)
(ii)
(iii)

(i)

(ii)

(iii)

#### Question 3:

D and E are points on the sides AB and AC respectively of a $∆ABC$. In each of the following cases, determine whether $DE\parallel BC$ or not.

(i)
(ii)
(iii)
(iv)

(i) We have:

(ii)
We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 $-$ 6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 $-$ 4.2 = 7 cm

Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.

(iii)
We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,
DB = 10.8 $-$ 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4 cm
Therefore,
AE = 9.6 $-$ 4 = 5.6 cm
Now,

(iv)
We have:
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 $-$ 7.2 =  4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 $-$ 6.4 = 3.6 cm
Now,

#### Question 4:

In a  is the bisector or $\angle A.$

(i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.
(ii) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.
(iii) If AB = 5.6 cm, BD = 3.8 cm and BC = 6 cm, find AC.
(iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

(i)

(ii)

(iii)

(iv)

Hence,  BC = 3 + 4.2 = 7.2 cm

#### Question 5:

The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm respectively. If PQ = 12 cm. find AB.

It is given that triangles ABC and PQR are similar.
Therefore,

#### Question 6:

The corresponding sides of two similar triangles ABC and DEF are BC = 9.1 cm and EF = 6.5 cm. If the perimeter of ∆ DEF is 25 cm, find the perimeter of ∆ABC.

Therefore, their corresponding sides will be proportional.
Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides.

Let the perimeter of ∆ABC be x cm.

Therefore,
$\frac{x}{25}=\frac{9.1}{6.5}\phantom{\rule{0ex}{0ex}}⇒x=\frac{9.1×25}{6.5}=35$

Thus, the perimeter of ∆ABC is 35 cm.

#### Question 7:

In the adjoining figure, ABCD is a trapezium in which CDAB and its diagonals intersect at O. If AO = (5x − 7) cm, OC = (2x + 1) cm, DO = (7x − 5) cm and OB = (7x + 1) cm, find the value of x.

In trapezium ABCD, $AB\parallel CD$ and the diagonals AC and BD intersect at O.
Therefore,

#### Question 8:

In ∆ABC, the bisector of ∠B meets AC at D. A line PQAC meets AB, BC and BD at P, Q and R respectively.
Show that PR × BQ = QR × BP.

In triangle ​BQP, BR bisects angle B.
Applying angle bisector theorem, we get:

This completes the proof.

#### Question 9:

Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides.

Let the trapezium ​be ABCD with E and F as the mid points of AD and BC, respectively.
Produce AD and BC to meet at P.

In
Applying Thales' theorem, we get:

Thus. EF is parallel to both AB and DC.
This completes the proof.

#### Question 10:

ABC and ∆DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQAB and PR BD are drawn, meeting AC at Q and CD at R, respectively. Prove that QRAD.

Similarly, applying Thales' theorem in , we get:

Applying the converse of Thales' theorem, we conclude that .
This completes the proof.

#### Question 11:

In the given figure, side BC of ∆ABC is bisected at D and O is any point on AD. BO and CO produced meet AC and AB at E and F, respectively, and AD is produced to X, so that D is the midpoint of OX. Prove that AO : AX = AF : AB and show that EFBC.

It is given that BC is bisected at D.
BD = DC ​
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX  bisect each other.
Therefore, BOCX is a parallelogram.
∴ BO

and
Applying Thales' theorem in $∆$ABX, we get:

From (1) and (2), we have:

Applying the converse of Thales' theorem in .
This completes the proof.

#### Question 12:

ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that $CQ=\frac{1}{4}AC.$ If PQ produced meets BC at R, prove that R is the midpoint BC.

We know that the diagonals of a parallelogram bisect each other.
Therefore,
CS = $\frac{1}{2}$AC      ...(i)
Also, it is given that CQ = $\frac{1}{4}$AC        ...(ii)
Dividing equation (ii) by (i), we get:

or, CQ = $\frac{1}{2}$CS
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in $△CSD$
$PQ\parallel DS$

In .
Applying converse of midpoint theorem ,we conclude that R is the midpoint of CB.
This completes the proof.

#### Question 13:

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

P and Q are midpoints of AB and AC, respectively.
According to the midpoint theorem,
$PQ\parallel BC$ and PQ = $\frac{1}{2}$BC = $\frac{1}{2}$DA  (It is given that AD = BC)

In $△$CDA, R and Q are midpoints of DC and AC, respectively
According to the midpoint theorem,
RQ$\parallel$DA and RQ = $\frac{1}{2}$DA

In $△$BDAP and S are midpoints of AB and BD, respectively.
According to the midpoint theorem,
SP$\parallel$DA and SP = $\frac{1}{2}$DA

Similarly, in $△$CDB, R and S are midpoints of DC and BD, respectively.
According to the midpoint theorem,
SR$\parallel$BC and SR = $\frac{1}{2}$BC = $\frac{1}{2}$DA

Therefore, SP$\parallel$RQ, PQ$\parallel$SR  and PQ = RQ = SP = SR

Hence, PQRS is a rhombus with all sides equal and opposite sides parallel to each other.

#### Question 14:

In the adjoining figure, ABC is a triangle in which AB = AC. If D and E are points on AB and AC respectively such that AD = AE, show that the points B, C, E and D are concyclic.

Given:
AB = AC     ...(ii)
Subtracting AD from both sides, we get:
$⇒$ AB $-$ AD = AC $-$ AD
$⇒$ AB $-$ AD = AC $-$ AE  (Since, AD = AE)
$⇒$BD = EC    ...(iii)
Dividing  equation  (i) by equation (iii), we get:

$\frac{AD}{DB}=\frac{AE}{EC}$

Applying the converse of Thales' theorem, DE$\parallel$BC

Therefore, B,C,E and D are concyclic points.

#### Question 1:

Find the pair of similar triangles among the given pairs. State the similarity criterion and write the similarity relation in symbolic form.

 (i)
 (ii)
 (iii)
 (iv)
 (v)

(i)
We have:

Therefore, by AAA similarity theorem,

(ii)
We have:

But, $\angle ABC \ne \angle EDF$ (Included angles are not equal)
Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.

(iii)
We have:

Also,
Therefore, by SAS similarity theorem, .

(iv)
We have
$\frac{DE}{QR}=\frac{2.5}{5}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{EF}{PQ}=\frac{2}{4}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{DF}{PR}=\frac{3}{6}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{DE}{QR}=\frac{EF}{PQ}=\frac{DF}{PR}$

Therefore, by SSS similarity theorem, $△FED~△PQR$

(v)

Therefore, by AA similarity theorem, $△\mathrm{ABC}~△\mathrm{MNR}$

#### Question 2:

In the given figure Find

(i) ∠DOC
(ii) ∠DCO
(iii) ∠OAB
(iv) ∠OBA.

(i)
It is given that DB is a straight line.
Therefore,

(ii)
In $△DOC$, we have:

(iii)
It is given that
Therefore,

(iv)
Again,
Therefore,

#### Question 3:

In the given figure Find
(i) OA
(ii) DO.

(i) Let OA be x cm.
$\because$

Hence, OA = 5.6 cm

(ii)  Let OD be y cm

$\because$

Hence, DO = 4 cm

#### Question 4:

In the given figure, if ∠ADE = ∠B, show that ∆ADE ∼ ∆ABC. If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE.

Given:
$\angle$ADE = $\angle$ABC and
Let DE be x cm
Therefore, by AA similarity theorem, $△ADE~△ABC$

Hence, DE = 2.8 cm

#### Question 5:

ABC ∼ ∆ PQR and the perimeters of ∆ ABC and ∆ PQR are 36 cm and 24 cm respectively. If PQ = 10 cm, find AB.

We know that the ratio of perimeters of two similar triangles is same as the ratio of their corresponding sides.
Therefore,

Hence, AB = 15 cm

#### Question 6:

In the given figure, ∠CAB = 90° and AD BC. Show that ∆ BDA ∼ ∆ BAC. If AC = 75 cm, AB = 1 m and BC = 1.25 m, find AD.

$⇒\frac{AD}{AC}=\frac{AB}{BC}$

#### Question 7:

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.

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

Hence, BC = 8.1 cm

#### Question 8:

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

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

CD  = $\frac{8×8}{4}=16$ cm

#### Question 9:

P and Q are points on the sides AB and AC, respectively, of ∆ ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm, show that BC = 3PQ.

We have:

This completes the proof.

#### Question 10:

ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersects AE at F, prove that AF × FB = EF × FD.

We have:

$\because$ DA $\parallel$ BC
$\therefore$       (Alternate angles)
$△DAF~△BEF$          (AA similarity theorem)

or,
This completes the proof.

#### Question 11:

In the given figure, DBBC, DEAB and ACBC.
Prove that $\frac{BE}{DE}=\frac{AC}{BC}.$

In , we have:

This completes the proof.

#### Question 12:

A vertical stick of length 7.5 m casts a shadow 5 m long on the ground and at the same time a tower casts a shadow 24 m long. Find the height of the tower.

Let AB be the vertical stick and BC be its shadow.
Given:
AB = 7.5 m, BC = 5 m

Let PQ be the tower and QR be its shadow.
Given:
QR = 24 m
Let the length of PQ be x m.

In , we have:

Therefore, by AA similarity theorem, we get:
$△ABC~△PQR$
$⇒\frac{AB}{BC}=\frac{PQ}{QR}$

Therefore, PQ = 36 m

Hence, the height of the tower is 36 m.

#### Question 13:

In an isosceles ∆ ABC, the base AB is produced both ways to P and Q, such that AP × BQ = AC2.
Prove that ∆ ACP ∼ ∆ BCQ.

Disclaimer: It should be $△APC~△BCQ$ instead of ∆ACP ∼ ∆BCQ
It is given that $△$ABC is an isosceles triangle.
Therefore,
CA = CB

Also,

Thus, by SAS similarity theorem, we get:
$△APC~△BCQ$
This completes the proof.

#### Question 14:

In the given figure,
Prove that ∆ ACB ∼ ∆ DCE.

We have:

#### Question 1:

ABC ∼ ∆ DEF and their areas are respectively 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

It is given that $△ABC~△DEF$.
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.

Hence, BC = 11.2 cm

#### Question 2:

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.

It is given that $△ABC~△PQR$.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.

Hence, QR = 6 cm

#### Question 3:

ABC ∼ ∆PQR and ar(∆ABC) = 4 ar(∆PQR). If BC = 12 cm, find QR.

Hence, QR = 6 cm

#### Question 4:

The areas of two similar triangles are 169 cm2 and 121 cm2 respectively. If the longest side of the larger triangle is 26 cm. find the longest side of the smaller triangle.

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Let the longest side of smaller triangle be x cm.

Hence, the longest side of the smaller triangle is 22 cm.

#### Question 5:

ABC ∼ ∆DEF and their areas are respectively 100 cm2 and 49 cm2. If the altitude of ∆ABC is 5 cm, find the corresponding altitude of ∆DEF.

It is given that ∆ABC ∼ ∆DEF.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Let the altitude of
ABC be AP, drawn from A to BC to meet BC at P and the altitude of ∆DEF be DQ, drawn from D to meet EF at Q.

Then,

Hence, the altitude of DEF is 3.5 cm

#### Question 6:

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively​.

It is given that $△ABC~△DEF$.
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 their areas is 4 : 9

#### Question 7:

The areas of two similar triangles are 81 cm2 and 49 cm2, respectively. If the altitude of one triangle is 6.3 cm, find the corresponding altitude of the other triangle.

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.

Hence, the altitude of the other triangle is 4.9 cm.

#### Question 8:

The areas of two similar triangles are 100 cm2 and 64 cm2 respectively. If a median of the similar triangle is 5.6 cm, find the corresponding median of the other.

Let the two triangles be ABC and PQR with medians AM and PN, respectively.

Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of their corresponding medians.

Hence, the median of the larger triangle is 7 cm.

#### Question 9:

In the given figure, 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 cm, prove that are of ∆APQ is $\frac{1}{16}$ of the area of ∆ABC.

We have:

Also,
By SAS similarity , we can conclude that ∆APQ$~$∆ABC.

Hence proved.

#### Question 10:

In the given figure, DEBC. If DE = 3 cm, BC = 6 cm and ar(∆ADE) = 15 cm2, find the area of ∆ABC.

It is given that DE ∥ BC

By AA similarity , we can conclude that $△ADE~△ABC$.

Hence, area of triangle ABC is 60 cm2.​

#### Question 11:

ABC is right-angled at A and AD ⊥ BC. If BC = 13 cm and AC = 5 cm, find the ratio of the areas of ∆ABC and ∆ADC.

In , we have:

By AA similarity, we can conclude that $△BAC~△ADC$.
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.

Hence, the ratio of areas of both the triangles is 169 : 25

#### Question 12:

In the given figure, DEBC and DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED.

It is given that DE $\parallel$ BC.

Applying AA similarity theorem, we can conclude that $△ADE~△ABC$.

#### Question 13:

In ∆ABC, D and E are the midpoints of AB and AC, respectively. Find the ratio of the areas of ∆ADE and ∆ABC.

It is given that D and E are midpoints of AB and AC.
Applying midpoint theorem, we can conclude that DE $\parallel$ BC.
Hence, by B.P.T., we get:

Also, .

Applying SAS similarity theorem, we can conclude that $△ADE~△ABC$.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.

#### Question 1:

The sides of certain triangles are given below. Determine which of them are right triangles.
(i) 9 cm, 16 cm, 18 cm
(ii) 7 cm, 24 cm, 25 cm
(iii) 1.4 cm, 4.8 cm, 5 cm
(iv) 1.6 cm, 3.8 cm, 4 cm
(v)

For the given triangle to be right-angled, the sum of square of the two sides must be equal to the square of the third side.
Here, let the three sides of the triangle be a, b and c.
(i)
a = 9 cm, b = 16 cm and c = 18 cm
Then,

Thus, the given triangle is not right-angled.

(ii)
a = 7 cm, b = 24 cm and c = 25 cm
Then,

Thus, the given triangle is a right-angled.

(iii)
a = 1.4 cm, b = 4.8 cm and c = 5 cm
Then,

Thus, the given triangle is right-angled.

(iv)
a = 1.6 cm, b = 3.8cm and c = 4 cm
Then,

Thus, the given triangle is not right-angled.

(v)
p = (a $-$ 1) cm,  q = 2$\sqrt{a}$ cm and r = (a + 1) cm
Then,

Thus, the given triangle is right-angled.

#### Question 2:

A man goes 80 m due east and then 150 m due north. How far is he from the starting point?

Let the man starts from point A and goes 80 m due east to B.
Then, from B, he goes 150 m due north to C.

We need to find AC.
In right-angled triangle ABC, we have:

Hence, the man is 170 m away from the starting point.

#### Question 3:

A man goes 10 m due south and then 24 due west. How far is he from the starting point?

Let the man starts from point D and goes 10 m due south and stops at E. He then goes 24 m due west at F.
In right $△$DEF, we have:
DE = 10 m, EF = 24 m

Hence, the man is 26 m away from the starting point.

#### Question 4:

A 13 m long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.

Let AB and AC be the ladder and height of the building.
It is given that:
AB = 13 m and AC = 12 m
We need to find the distance of the foot of the ladder from the building, i.e, BC.
In right-angled triangle ABC, we have:

Hence, the distance of the foot of the ladder from the building is 5 m

#### Question 5:

A ladder is placed in such a way that its foot is at a distance of 15 m from a wall and its top reaches a window 20 m above the ground. Find the length of the ladder.

Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, respectively.
We have:
AB = 20 m and BC = 15 m
Applying Pythagoras theorem in right-angled triangle ABC, we get:

Hence, the length of the ladder is 25 m.

#### Question 6:

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

Let the two poles be DE and AB and the distance between their bases be BE.
We have:
DE = 9 m, AB = 14 m and BE = 12 m
Draw a line parallel to BE from D, meeting AB at C.
Then, DC = 12 m  and AC = 5 m
We need to find AD, the distance between their tops.

Applying Pythagoras theorem in right-angled triangle ACD, we have: ​

Hence, the distance between the tops of the two poles is 13 m.

#### Question 7:

In the given figure, O is a point inside a ∆PQR such that ∠POR = 90°, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that ∆PQR is right-angled.

Applying Pythagoras theorem in right-angled triangle POR, we have:

In ∆ PQR,

Therefore, by applying Pythagoras theorem, we can say that ∆PQR is right-angled at P.

#### Question 8:

ABC is an isosceles triangle, right-angled at C.Prove that AB2 = 2AC2.

∆ABC is an isosceles triangle with AC = BC.
Also, ∆ABC is a right-angled triangle.
Therefore, applying Pythagoras theorem, we get:

This completes the proof.

#### Question 9:

ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ∆ABC is a right triangle

Given:
AC = BC

It is evident that the sum of squares of the two sides of triangle ABC is equal to square of the third side.

Therefore, by applying Pythagoras theorem, we conclude that  ∆ABC is right-angled at C.

#### Question 10:

ABC is an isosceles triangle with AB = AC = 13 cm. The length of the altitude from A on BC is 5 cm. Find BC.

It is given that $△ABC$ is an isosceles triangle.
Also, AB = AC = 13 cm
Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.
are right-angled triangles.
Applying Pythagoras theorem, we have:

Hence,
BC = 2(BD) = 2  = 24 cm

#### Question 11:

Find the length of altitude AD of an isosceles ∆ABC in which AB = AC = 2a units and BC = a units.

In isosceles ∆ ABC, we have:
AB = AC = 2a units and BC = a units
Let AD be the altitude drawn from A that meets BC at D.
Then, D is the midpoint of BC.
BD = BC =
Applying Pythagoras theorem in right-angled ∆ABD, we have:

#### Question 12:

ABC is an equilateral triangle of a side 2a units. Find each of its altitudes.

Let AD, BE and CF be the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, respectively.
Then, D, E and F are the midpoints of BC, AC and AB, respectively.

In right-angled ∆ABD, we have:
AB = 2a and BD = a
Applying Pythagoras theorem, we get:

Similarly,
BE = $a\sqrt{3}$ units and CF = $a\sqrt{3}$ units

#### Question 13:

Find the height of an equilateral triangle of side 12 cm.

Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. Then, D will be the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:

Hence, the height of the given triangle is 6$\sqrt{3}$ cm.

#### Question 14:

Find the length of a diagonal of a rectangle whose adjacent sides are 30 cm and 16 cm.

Let ABCD be the rectangle with diagonals AC and BD meeting at O.
According to the question:
AB = CD = 30 cm and BC = AD = 16 cm

Applying Pythagoras theorem in right-angled triangle ABC, we get:

Diagonals of a rectangle are equal.
Therefore, AC = BD = 34 cm

#### Question 15:

Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.

Let ABCD be the rhombus with diagonals (AC = 24 cm and BD = 10 cm) meeting at O.
We know that the diagonals of a rhombus bisect each other at right angles.
Applying Pythagoras theorem in right-angled triangle AOB, we get:

Hence, the length of each side of the rhombus is 13 cm.

#### Question 16:

In ∆ABC, D is the midpoint of BC and AE ⊥ BC. If AC > AB, show that
$A{B}^{2}=A{D}^{2}-BC·DE+\frac{1}{4}B{C}^{2}.$

In right-angled triangle AEB, applying Pythagoras theorem, we have:
...(i)
In right-angled triangle AED, applying Pythagoras theorem, we have:

...(ii)

This completes the proof.

#### Question 17:

In the given figure, D is the midpoint of side BC and AEBC. 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) $\left({b}^{2}+{c}^{2}\right)=2{p}^{2}+\frac{1}{2}{a}^{2}$
(iv) $\left({b}^{2}-{c}^{2}\right)=2ax$

(i)
In right-angled triangle AEC, applying Pythagoras theorem, we have:

(ii)
In right-angled triangle AEB, applying Pythagoras theorem, we have:

(iii)

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

(iv)
Subtracting (ii) from (i), we get:

#### Question 18:

In ∆ABC, AB = AC. Side BC is produced to D. Prove that

Draw AE$\perp$BC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:

Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.
So, BE = CE
and DE+ CE = DE + BE = BD

#### Question 19:

ABC is an isosceles triangle, right-angled at B. similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ∆ABE and ∆ACD.

Applying Pythagoras theorem in right-angled triangle ABC, we get:

...(i)

$\because$ $△\mathrm{ACD}~△\mathrm{ABE}$

#### Question 1:

A man goes 24 m due west and them 10 m due north. How far is he from the starting point?
(a) 34 m
(b) 17 m
(c) 26 m
(d) 28 m

(c) 26 m

Suppose, the man starts from point A and goes 24 m due west to point B. From here, he goes 10 m due north and stops at C.
In right triangle ABC, we have:
AB = 24 m, BC = 10 m
Applying Pythagoras theorem, we get:

#### Question 2:

A man goes 12 due south and then 35 m due west. How far is he from the starting point?
(a) 47 m
(b) 23.5 m
(c) 23 m
(d) 37 m

(d) 37 m

Suppose, the man starts from point A and goes 12 m due south to point B. From here, he goes 35 m due west and stops at C.
In right triangle ABC, we have:
AB = 12 m, BC = 35 m
Applying Pythagoras theorem, we get:

#### Question 3:

Two poles of height 13 m and 7 m respectively stand vertically on a plane ground at a distance of 8 m from each other. The distance between their tops is
(a) 9 m
(b) 10 m
(c) 11 m
(d) 12 m

(b) 10 m

Let AB and DE be the two poles.
According to the question:
AB = 13 m
DE = 7 m
Distance between their bottoms = BE = 8 m
Draw a perpendicular DC to AB from D, meeting AB at C. We get:
DC = 8m, AC = 6 m
Applying Pythagoras theorem in right-angled triangle ACD, we have:

#### Question 4:

A vertical stick 1.8 m long casts a shadow 45 cm long on the ground. At the same time, what is the length of the shadow of a pole 6 m high?
(a) 2.4 m
(b) 1.35 m
(c) 1.5 m
(d) 13.5 m

(c) 1.5 m

Let AB and AC be the vertical stick and its shadow, respectively.
According to the question:
AB = 1.8 m
AC = 45 cm = 0.45 m
Again, let DE and DF be the pole and its shadow, respectively.
According to the question:
DE = 6 m
DF = ?
Now, in right-angled triangles ABC and DEF, we have:

#### Question 5:

A vertical pole 6 m long casts a shadow of length 3.6 m on the ground. What is the height of a tower which casts a shadow of length 18 m at the same time?
(a) 10.8 m
(b) 28.8 m
(c) 32.4 m
(d) 30 m

(d) 30 m

Let AB and AC be the vertical pole and its shadow, respectively.
According to the question:
AB = 6 m
AC = 3.6 m
Again, let DE and DF be the tower and its shadow.
According to the question:
DF = 18 m
​DE = ?
Now, in right-angled triangles ABC and DEF, we have:

#### Question 6:

A ladder 25 m long just reaches the top of a building 24 m high from the ground. What is the distance of the foot of the ladder from the building?
(a) 7 m
(b) 14 m
(c) 21 m
(d) 24.5 m

(a) 7 m

Let the ladder BC reaches the building at C.
Let the height of building where the ladder reaches be AC.
According to the question:
BC = 25 m
AC = 24 m
In right-angled triangle CAB, we apply Pythagoras theorem to find the value of AB.

#### Question 7:

A ladder 15 m long reaches a window that is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. The width of the street is
(a) 27 m
(b) 21 m
(c) 24 m
(d) 18 m

(b) 21 m

Let the ladder be at point C on the ground.
At first, the ladder is placed towards the left side of the street and it reaches the window AB, which is 9 m high from the ground.
Secondly, the ladder is placed towards the right side of the street and it reaches the window DE, which is 12 m high from the ground.

Applying Pythagoras theorem in right-angled triangles ABC and DEC, we get:

#### Question 8:

In the given figure, O is a point inside a ∆PQR such that ∠POR = 90°, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and ∠QPR = 90°, then QR = ?

(a) 28 cm
(b) 25 cm
(c) 26 cm
(d) 32 cm

(c) 26 cm

Applying Pythagoras theorem in right-angled triangle POR, we get:

PR = 10 cm

Now, applying Pythagoras theorem in right-angled triangle QPR, we get:

#### Question 9:

The hypotenuse of a right triangle is 25 cm. The other two sides are such that one is 5 cm longer. The lengths of these sides are
(a) 10 cm, 15 cm
(b) 15 cm, 20 cm
(c) 12 cm, 17 cm
(d)13 cm, 18 cm

(b) 15 cm, 20 cm

It is given that the length of hypotenuse is 25 cm.
Let the other two sides be x cm and (x$-$5) cm.
Applying Pythagoras theorem, we get:

Now,
x $-$ 5 = 20 $-$ 5 = 15 cm

#### Question 10:

The height of an equilateral triangle having each side 12 cm, is
(a)
(b)
(c)
(d)

(b)

Let ABC be the equilateral triangle with AD as its altitude from A.

In right-angled triangle ABD, we have:

#### Question 11:

ABC is an isosceles triangle with AB = AC = 13 cm and the length of altitude from A on BC is 5 cm. Then, BC = ?
(a) 12 cm
(b) 16 cm
(c) 18 cm
(d) 24 cm

(d) 24 cm

In triangle ABC, let the altitude from A on BC meets BC at D.
We have:
AD = 5 cm, AB = 13 cm and D is the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:

Therefore, BC = 2BD = 24 cm

#### Question 12:

The measures of three angles of a triangle are in the ratio 1 : 2 : 3. Then, the triangle is
(a) right-angled
(b) equilateral
(c) isosceles
(d) obtuse-angled

(a) right-angled

Let the angles be x$°$, 2x$°$ and 3x$°$.
In a triangle, the sum of its angles is equal to 180$°$.
Therefore,
x + 2x + 3x = 180
or 6x = 180
or x = 30

Hence, the angles are 30$°$, 60$°$ and 90$°$.

#### Question 13:

For a ∆ABC, which is true?
(a) ABAC = BC
(b) (ABAC) > BC
(c) (AB AC) < BC
(d) None of these

(c) (AB − AC) < BC

In a triangle, the difference of lengths of its two sides is always less than the third side.
Hence, (AB − AC) < BC is true.

#### Question 14:

In a ∆ABC, it is given that AB = 6 cm, AC = 8 cm and AD is the bisector of ∠A. Then, BD : DC =?

(a) 3 : 4
(b) 9 : 16
(c) 4 : 3
(d) $\sqrt{3}:2$

(a) 3 : 4

In ∆ ABD and ∆ACD, we have:

Now,

#### Question 15:

In a ∆ABC, it is given that AD is the internal bisector of ∠A. If BD = 4 cm, DC = 5 cm and AB = 6 cm, then AC = ?

(a) 4.5 cm
(b) 8 cm
(c) 9 cm
(d) 7.5 cm

(d) 7.5 cm
It is given that AD bisects angle A.
Therefore, applying angle bisector theorem, we get:

Hence, AC = 7.5 cm

#### Question 16:

In a ∆ABC, it is given that AD is the internal bisector of ∠A. If AB = 10 cm, AC = 14 cm and BC = 6 cm, then CD = ?

(a) 4.8 cm
(b) 3.5 cm
(c) 7 cm
(d) 10.5 cm

(b) 3.5 cm

It is given that AD bisects angle A.
Therefore, applying angle bisector theorem, we get:

Hence, CD = 3.5 cm

#### Question 17:

In a triangle, the perpendicular from the vertex to the base bisects the base. The triangle is
Figure

(a) right-angle
(b) isosceles
(c) scalene
(d) obtuse-angled

(b) isosceles

In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.

#### Question 18:

In an equilateral triangle ABC, if AD ⊥ BC, then which of the following is true?

Applying Pythagoras theorem in right-angled triangles ABD and ADC, we get:

#### Question 19:

In a rhombus of side 10 cm, one of the diagonals is 12 cm long. The length of the second diagonal is
(a) 20 cm
(b) 18 cm
(c) 16 cm
(d) 22 cm

(c) 16 cm

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
Also, diagonals of a rhombus bisect each other at right angles.
If AC =12 cm, AO = 6 cm
Applying Pythagoras theorem in right-angled triangle AOB. we get:

Hence, the length of the second diagonal BD is 16 cm.

#### Question 20:

The lengths of the diagonals of a rhombus are 24 cm and 10 cm. The length of each side of the rhombus is
(a) 12 cm
(b) 13 cm
(c) 14 cm
(d) 17 cm

(b) 13 cm

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We have:
AC = 24 cm and BD = 10 cm
We know that diagonals of a rhombus bisect each other at right angles.
Therefore applying Pythagoras theorem in right-angled triangle AOB, we get:

Hence, the length of each side of the rhombus is 13 cm.

#### Question 21:

If the diagonals of a quadrilateral divide each other proportionally, then it is a
(a) parallelogram
(b) trapezium
(c) rectangle
(d) square

(b) trapezium
Diagonals of a trapezium divide each other proportionally.

#### Question 22:

In the given figure, ABCD is a trapezium whose diagonals AC and BD intersect at O such that OA = (3x −1) cm, OB = (2x + 1) cm, OC = (5x − 3) cm and OD = (6x − 5) cm. Then, x = ?

(a) 2
(b) 3
(c) 2.5
(d) 4

(a) 2

We know that the diagonals of a trapezium are proportional.
Therefore,

#### Question 23:

The line segments joining the midpoints of the adjacent sides of a quadrilateral form
(a) a parallelogram
(b) a rectangle
(c) a square
(d) a rhombus

(a) a parallelogram

The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.

#### Question 24:

If the bisector of an angle of a triangle bisects the opposite side, then the triangle is
(a) scalene
(b) equilateral
(c) isosceles
(d) right-angled

(c) isosceles

Let AD be the angle bisector of angle A in triangle ABC.
Applying angle bisector theorem, we get:
$\frac{AB}{AC}=\frac{BD}{DC}$

It is given that AD bisects BC.
Therefore, BD = DC

Therefore, the triangle is isosceles.

#### Question 25:

In ∆ABC, it is given that

(a) 30°
(b) 40°
(c) 45°
(d) 50°

(a) 30$°$
We have:

$\frac{AB}{AC}=\frac{BD}{DC}$

Applying angle bisector theorem, we can conclude that AD bisects $\angle$A.

#### Question 26:

In ∆ABC, DE ∥ BC so that AD = 2.4 cm, AE = 3.2 cm and EC = 4.8 cm. Then, AB = ?

(a) 3.6 cm
(b) 6 cm
(c) 6.4 cm
(d) 7.2 cm

(b) 6 cm

It is given that DE$\parallel$BC.

Applying basic proportionality theorem, we have:

Therefore, AB = AD + BD = 2.4 + 3.6 = 6 cm

#### Question 27:

In ∆ABCDE is drawn parallel to BC, cutting AB and AC at D and E, respectively, such that AB = 7.2 cm, AC = 6.4 cm and AD = 4.5 cm. Find AE.

(a) 5.4 cm
(b) 4 cm
(c) 3.6 cm
(d) 3.2 cm

(b) 4 cm

It is given that DE$\parallel$BC.

Applying basic proportionality theorem, we get:

#### Question 28:

In ∆ABC, DEBC so that AD = (7x − 4) cm, AE = (5x − 2) cm, DB = (3x + 4) cm and EC = 3x cm. Then, we have:

(a) x = 3
(b) x = 5
(c) x = 4
(d) x = 2.5

(c) x = 4

It is given that $DE\parallel BC$.
Applying Thales' theorem, we get:

#### Question 29:

In
If AC = 5.6 cm, then AE = ?

(a) 4.2 cm
(b) 3.1 cm
(c) 2.8 cm
(d) 2.1 cm

(d) 2.1 cm

It is given that DE$\parallel BC$.
Applying Thales' theorem, we get:

#### Question 30:

ABC ∼ ∆DEF and the perimeters of ∆ABC and ∆DEF are 30 cm and 18 cm respectively. If BC = 9 cm, then EF = ?
(a) 6.3 cm
(b) 5.4 cm
(c) 7.2 cm
(d) 4.5 cm

(b) 5.4 cm

ABC ∼ ∆DEF

Therefore,

#### Question 31:

ABC ∼ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25 cm, what is the perimeter of ∆ABC?
(a) 35 cm
(b) 28 cm
(c) 42 cm
(d) 40 cm

(a) 35 cm

$\because$ ∆ABC ∼ ∆DEF

#### Question 32:

In ∆ABC, it is given that AB = 9 cm, BC = 6 cm and CA = 7.5 cm. Also, in ∆DEF , EF = 8 cm and ∆DEF ∼ ∆ABC. Then, perimeter of ∆DEF is
(a) 22.5 cm
(b) 25 cm
(c) 27 cm
(d) 30 cm

(d) 30 cm

Perimeter of $△$ABC = AB + BC + CA = 9 + 6 + 7.5 = 22.5 cm

$\because$ ∆DEF ∼ ∆ABC

#### Question 33:

It is given that ∆ABC ∼ ∆DEF. If ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm, then which of the following is true?
(a) DE = 12 cm, ∠F = 50°
(b) DE = 12 cm, ∠F = 100°
(c) EF = 12 cm, ∠D = 100°
(d) EF = 12 cm, ∠F = 30°

(b) DE = 12 cm, $\angle$F = 100°

Disclaimer: In the question, it should be ∆ABC ∼ ∆DFE  instead of  ∆ABC ∼ ∆DEF.

In triangle ABC,

$\because$ ∆ABC ∼ ∆DFE

#### Question 34:

In the given figure ∠BAC = 90° and AD ⊥ BC. Then,

(a) BC CD = BC2
(b) AB AC = BC2

#### Question 35:

In ∆ABC and ∆DEF, it is given that $\frac{AB}{DE}=\frac{BC}{FD},$ then
(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠F = ∠F

(c) $\angle$B = $\angle$D

#### Question 36:

In ∆DEF and ∆PQR, it is given that ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?
(a) $\frac{EF}{PR}=\frac{DF}{PQ}$
(b) $\frac{DE}{PQ}=\frac{EF}{RP}$
(c) $\frac{DE}{QR}=\frac{DF}{PQ}$
(d) $\frac{EF}{RP}=\frac{DE}{QR}$

(b)

In ∆DEF and ∆PQR, we have:

#### Question 37:

If ∆ABC ∼ ∆EDF and ∆ABC is not similar to ∆DEF, then which of the following is not true?
(a) BCEF = ACFD
(b) ABEF = ACDE
(c) BCDE = ABEF
(d) BCDE = ABFD

(c) BCDE = ABEF

ABC ∼ ∆EDF
Therefore,

#### Question 38:

In ∆ABC and ∆DEF, it is given that ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) similar as well as congruent

(b) similar but not congruent

In ∆ABC and ∆DEF, we have:

#### Question 39:

If in ∆ABC and ∆PQR, we have: $\frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ}$, then
(a) ∆PQR ∼ ∆CAB
(b) ∆PQR ∼ ∆ABC
(c) ∆CBA ∼ ∆PQR
(d) ∆BCA ∼ ∆PQR

(a) ∆PQR ∼ ∆CAB

In ∆ABC and ∆PQR, we have:

#### Question 40:

In the given figure, two lines segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°, then ∠PBA = ?

(a) 50°
(b) 30°
(c) 60°
(d) 100°

(d) 100°

#### Question 41:

It is given that ∆ABC ∼ ∆DEF and the corresponding sides of these triangles are in the ratio 3 : 5. Then, ar(∆ABC) : ar(∆DEF) = ?
(a) 3 : 5
(b) 5 : 3
(c) 9 : 25
(d) 25 : 9

(c) 9 : 25
It is given that ∆ABC ∼ ∆DEF.
Therefore,

#### Question 42:

It is given that ∆ ABC ∼ ∆PQR and
(a) $\frac{2}{3}$
(b) $\frac{3}{2}$
(c) $\frac{4}{9}$
(d) $\frac{9}{4}$

(d) 9 : 4
It is given that ∆ ABC ∼ ∆PQR and .
Therefore,

#### Question 43:

In an equilateral ∆ABC, D is the midpoint of AB and E is the midpoint of AC. Then, ar(∆ABC) : ar(∆ADE) = ?

(a) 2 : 1
(b) 4 : 1
(c) 1 : 2
(d) 1 : 4

(b) 4:1
In ∆ABC, D is the midpoint of AB and E is the midpoint of AC.
Therefore, by midpoint theorem, DE$\parallel$BC.
Also, by Basic Proportionality Theorem,
$\frac{AD}{DB}=\frac{AE}{EC}$

#### Question 44:

In ∆ABC and ∆DEF, we have: $\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{5}{7},$ then
ar(∆ABC) : ar(∆DEF) = ?

(a) 5 : 7
(b) 25 : 49
(c) 49 : 25
(d) 125 : 343

(b) 25 : 49

#### Question 45:

ABC ∼ ∆DEF such that ar(∆ABC) = 36 cm2 and ar(∆DEF) = 49 cm2.
Then, the ratio of their corresponding sides is
(a) 36 : 49
(b) 6 : 7
(c) 7 : 6
(d) $\sqrt{6}:\sqrt{7}$

(b) 6:7

$\because$ ∆ABC ∼ ∆DEF

#### Question 46:

Two isosceles triangles have their corresponding angles equal and their areas are in the ratio 25 : 36. The ratio of their corresponding heights is
(a) 25 : 36
(b) 36 : 25
(c) 5 : 6
(d) 6 : 5

(c) 5:6
Let x and y be the corresponding heights of the two triangles.
It is given that the corresponding angles of the triangles are equal.
Therefore, the triangles are similar. (By AA criterion)
Hence,

#### Question 47:

The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is
(a) congruent to the original triangle
(b) similar to the original triangle
(c) an isosceles triangle
(d) an equilateral triangle

(b) similar to the original triangle

The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is similar to the original triangle.

#### Question 48:

If $∆ABC~∆QRP,\frac{\mathrm{ar}\left(∆ABC\right)}{\mathrm{ar}\left(∆PQR\right)}=\frac{9}{4},$ AB = 18 cm and BC = 15 cm, then PR = ?
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d)

(b) 10 cm

#### Question 49:

In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and ∠AOC = 45°. Then, ∆OAC and ∆ODB are

(a) equilateral and similar
(b) equilateral but not similar
(c) isosceles and similar
(d) isosceles but not similar

(c) isosceles and similar

In ∆AOC and ∆ODB, we have:

#### Question 50:

In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2, then ∠C = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

(d) 90°

Given:
AC = BC

Applying Pythagoras theorem, we conclude that ∆ABC is right angled at C.
or,

#### Question 51:

In ∆ABC, if AB = 16 cm, BC = 12 cm and AC = 20 cm, then ∆ABC is
(a) acute-angled
(b) right-angled
(c) obtuse-angled

(b) right-angled

We have:

Hence, ∆ABC is a right-angled triangle.

#### Question 52:

Which of the following is a true statement?
(a) Two similar triangles are always congruent.
(b) Two figures are similar if they have the same shape and size.
(c) Two triangles are similar if their corresponding sides are proportional.
(d) Two polygons are similar if their corresponding sides are proportional.

(c) Two triangles are similar if their corresponding sides are proportional.
According to the statement:

#### Question 53:

Which of the following is false statement?
(a) If the areas of two similar triangles are equal, then the triangles are congruent.
(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides.
(c) The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding medians.
(d) The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides.

Because the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

#### Question 54:

Match the following columns:

 Column I Column II (a) In a given ∆ABC, DE ∥ BC and $\frac{AD}{DB}=\frac{3}{5}.$ If AC = 5.6 cm, then AE = ...... cm. (p) 6 (b) If ∆ABC ∼ ∆DEF such that 2AB = 3DE and BC = 6 cm, then EF = ...... cm. (q) 4 (c) If ∆ABC ∼ ∆PQR such that ar(∆ABC) : ar(∆PQR) = 9 : 16 and BC = 4.5 cm, then QR = ...... cm. (r) 3 (d) In the given figure, AB ∥ CD and OA = (2x + 4) cm, OB = (9x − 21) cm, OC = (2x − 1) cm and OD = 3 cm. Then x = ? Figure (s) 2.1

(a) - (s)
Let AE be x.
Therefore, EC = 5.6 $-$ x
It is given that DE $\parallel$ BC.
Therefore, by B.P.T., we get:

(b) - (q)

(c) - (p)

(d) - (r)

#### Question 55:

Match the following columns:

 Column I Column II (a) A man goes 10 m due east and then 20 m due north. His distance from the starting point is ...... m. (p) $25\sqrt{3}$ (b) In an equilateral triangle with each side 10 cm, the altitude is ...... cm. (q) $5\sqrt{3}$ (c) The area of an equilateral triangle having each side 10 cm is ...... cm2. (r) $10\sqrt{5}$ (d) The length of diagonal of a rectangle having length 8 m and breadth 6 m is ...... m . (s) 10

(a) - (r)
Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.
Then, in right-angled triangle ABC, we have:

Hence, the man is 10$\sqrt{3}$ m away from the starting point.

(b) - (q)
Let the triangle be ABC with altitude AD.
In right-angled triangle ABD, we have:

(c) - (p)
Area of an equilateral triangle with side a =

(d) - (s)
Let the rectangle be ABCD with diagonals AC and BD.
In right-angled triangle ABC, we have:

#### Question 56:

Assertion (A)
In ∆ABC, if DE ∥ BC intersects AB in D and AC in E, then $\frac{AD}{AB}=\frac{AE}{AC}.$
Reason (R)
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then these sides are divided in the same ratio.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

The Reason (R) is clearly true by Thales' theorem.
It is given that DE$\parallel$BC.
Applying Thales' theorem, we have:

#### Question 57:

Assertion (A)
ABC ∼ ∆DEF such that BC = 4.5 cm and EF = 3.5 cm and ar(∆ABC) = 81 cm2,  then ar(∆DEF) = 49 cm2.
Reason (R)
The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding medians.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

The Reason(R) is clearly true.
$\because$ABC ∼ ∆DEF

Hence, Assertion (A) is true but Reason (R) is not the correct explanation of (A).

#### Question 58:

Assertion (A)
ABC ∼ ∆DEF, such that ar(∆ABC) = 100 cm2 and ar(∆DEF) = 144 cm2. If AB = 24 cm, then DE = 36 cm.
Reason (R)
If ∆ABC ∼ ∆DEF, then $\frac{\mathrm{ar}\left(∆ABC\right)}{\mathrm{ar}\left(∆DEF\right)}=\frac{A{B}^{2}}{D{E}^{2}}=\frac{B{C}^{2}}{E{F}^{2}}=\frac{A{C}^{2}}{D{F}^{2}}.$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(d) Assertion (A) is false and Reason (R) is true.

Reason (R) is clearly true.

Hence, Assertion (A) is false.

#### Question 59:

Assertion (A)
In an isosceles ∆ABC, if ∠C = 90°, then AB2 = 3AC2.
Reason (R)
In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2, then ∠C = 90°.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(d) Assertion (A) is false and Reason (R) is true.

In triangle ABC, let BC be equal to AC.
We have:

Thus,

Assertion (A) is false.
This proves that Reason (R) is true.

Hence, Assertion (A) is false and Reason (R) is true.

#### Question 60:

Look at the statements given below:
I. ∆ABC ∼ ∆DEF and the altitudes of these triangles are in the ratio 1 : 2, then ar(∆ABC) : ar(∆DEF) = 1 : 4.
II. In ∆ABC, DE ∥ BC and AD : DB = 1 : 2, then $\frac{\mathrm{ar}\left(∆ADE\right)}{\mathrm{ar}\left(∆ABC\right)}=\frac{1}{4}.$
III. In a ∆ABC, P and Q are points on AB and AC respectively such that AB = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, then BC = 3PQ.

Which is true?
(a) I only
(b) II only
(c) I and II
(d) I and III

(d) I and III  are true.

I .
$\because$ ∆ABC ∼ ∆DEF

II.
We have:
AD : DB = 1 : 2

III.
We have:
AP = 3 cm
PB = 6 cm
AQ =  5 cm
QC =  10 cm
BC =  3PQ

Thus, I and III  are true, II is false.

#### Question 1:

ABC ∼ ∆DEF and their perimeters are 32 cm and 24 cm respectively. If AB = 10 cm, then DE =?
(a) 8 cm
(b) 7.5 cm
(c) 15 cm
(d) $5\sqrt{3}\mathrm{cm}$

(b) 7.5 cm

$\because$ ∆ABC ∼ ∆DEF

#### Question 2:

In the given figure, DE ∥ BC. If DE = 5 cm, BC = 8 cm and AD = 3.5 cm, then AB = ?

(a) 5.6 cm
(b) 4.8 cm
(c) 5.2 cm
(d) 6.4 cm

(a) 5.6 cm
$\because$ DE ∥ BC

#### Question 3:

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

(a) 12 m
(b) 13 m
(c) 14 m
(d) 15 m

(b) 13 m

Let the poles be AB and CD.
It is given that:
AB = 6 m and CD = 11 m
Let AC be 12 m.
Draw a perpendicular from B on CD, meeting CD at E.
Then,
BE = 12 m
We have to find BD.
Applying Pythagoras theorem in right-angled triangle BED, we have: ​

#### Question 4:

The areas of two similar triangles are 25 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 3.5 cm, then the corresponding altitude of the other triangle is
(a) 5.6 cm
(b) 6.3 cm
(c) 4.2 cm
(d) 7 cm

(c)
We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let h be the altitude of the other triangle.
Therefore,

#### Question 5:

If ∆ABC ∼ ∆DEF such that 2 AB = DE and BC = 6 cm, find EF.

$\because$ABC ∼ ∆DEF

#### Question 6:

In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

$\because$ DE BC

#### Question 7:

A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Let the ladder be AB and BC be the height of the window from the ground.

We have:
AB = 10 m and BC = 8 m
Applying Pythagoras theorem in right-angled triangle ACB, we have:

Hence, the foot of the ladder is 6 m away from the base of the wall.

#### Question 8:

Find the length of the altitude of an equilateral triangle of side 2a cm.

Let the triangle be ABC with AD as its altitude. Then, D is the midpoint of BC.
In right-angled triangle ABD, we have:

Hence, the length of the altitude of an equilateral triangle of side 2a cm is $\sqrt{3}a$ cm.

#### Question 9:

ABC ∼ ∆DEF such that ar(∆ABC) = 64 cm2 and ar(∆DEF) = 169 cm2. If BC = 4 cm, find EF.

$\because$ ∆ABC ∼ ∆DEF

#### Question 10:

In an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm2. Find ar(∆COD).

In ∆AOB and ∆COD, we have:

#### Question 11:

The corresponding sides of two similar triangles are in the ratio 2 : 3. If the area of the smaller triangle is 48 cm2, find the area of the larger triangle.

It is given that the triangles are similar.
Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides.

#### Question 12:

In the given figure, LM  CB and LN  CD.
Prove that $\frac{AM}{AB}=\frac{AN}{AD}.$

Therefore, applying Thales' theorem, we have:

This completes the proof.

#### Question 13:

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

Let the triangle be ABC with AD as the bisector of $\angle A$ which meets BC at D.
We have to prove:

Draw CE $\parallel$ DA, meeting BA produced at E.
CE $\parallel$DA
Therefore,

This completes the proof.

#### Question 14:

In an equilateral triangle with side a, prove that area = $\frac{\sqrt{3}}{4}{a}^{2}.$

Let ABC be the equilateral triangle with each side equal to a.
Let AD be the altitude from A, meeting BC at D.
Therefore, D is the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we have:

Therefore,

This completes the proof.

#### Question 15:

Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We know that the diagonals of a rhombus bisect each other at right angles.
$\therefore$ If AC = 24 cm and BD =10 cm, AO = 12 cm and BO = 5 cm
Applying Pythagoras theorem in right-angled triangle AOB, we get:

Hence, the length of each side of the given rhombus is 13 cm.

#### Question 16:

Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.

Let the two triangles be ABC and PQR.
We have:
$△ABC~△PQR$,
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p

We have to prove:

$△ABC~△PQR$; therefore, their corresponding sides will be proportional.

This completes the proof.

#### Question 17:

In the given figure, ∆ABC and ∆DBC have the same base BC. If AD and BC intersect at O, prove that $\frac{\mathrm{ar}\left(∆ABC\right)}{\mathrm{ar}\left(∆DBC\right)}=\frac{AO}{DO}.$

This completes the proof.

#### Question 18:

In the given figure, XYAC and XY divides ∆ABC into two regions, equal in area. Show that $\frac{AX}{AB}=\frac{\left(2-\sqrt{2}\right)}{2}.$

#### Question 19:

In the given figure, ∆ABC is an obtuse triangle, obtuse-angles at B. If AD CB, Prove that AC2 = AB2 + BC2 + 2BC ⋅ BD.

Applying Pythagoras theorem in right-angled triangle ADC, we get:

Applying Pythagoras theorem in right-angled triangle ADB, we get:

This completes the proof.

#### Question 20:

In the given figure, PA, QB and RC are perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, show that $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}.$