Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 4 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among Class 10 students for Math Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### 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:

M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that

$\left(\mathrm{i}\right)\frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Given: ABCD is a parallelogram
To prove:
$\left(\mathrm{i}\right)\frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Proof: In △DMC and △NMB
$\angle$DMC =$\angle$NMB      (Vertically opposite angle)
$\angle$DCM =$\angle$NBM       (Alternate angles)
By AAA- similarity
△DMC ~ △NMB
$\therefore \frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}}$

Adding 1 to both sides, we get

#### Question 6:

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 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, M and N are points on AB and AC respectively such that BM = CN. If ∠B = ∠C then show that MN∥BC.

In △ABC, ∠B = ∠C
∴AB = AC (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB − BM = AC − BM
⇒AB − BM = AC − CN           (∵BM = CN)
⇒AM = AN
∴∠AMN = ∠ANM (Angles opposite to equal sides are equal)
Now, In △ABC,
∠A + ∠B + ∠C = 180o                     .....(1)
(Angle Sum Property of triangle)
Again In In △AMN,
∠A +∠AMN + ∠ANM = 180o          ......(2)
(Angle Sum Property of triangle)
From (1) and (2), we get
∠B + ∠C = ∠AMN + ∠ANM
⇒2∠B  = 2∠AMN
⇒∠B  = ∠AMN
Since, ∠B  and ∠AMN are corresponding angles.
∴ MN∥BC

#### Question 9:

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 10:

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 11:

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 12:

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 13:

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 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:

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 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 8:

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 9:

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 10:

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 11:

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 12:

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

In , we have:

This completes the proof.

#### Question 13:

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 14:

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 15:

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

We have:

#### Question 16:

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

In △ABC, P and Q are mid points of AB and AC respectively.
So, PQ|| BC, and $\mathrm{PQ}=\frac{1}{2}\mathrm{BC}$                                              .....(1)

Similarly, In △ADC, $\mathrm{QR}=\frac{1}{2}\mathrm{AD}=\frac{1}{2}\mathrm{BC}$                          .....(2)
Now, In △BCD, $\mathrm{SR}=\frac{1}{2}\mathrm{BC}$                                               .....(3)

Similarly, In △ABD, $\mathrm{PS}=\frac{1}{2}\mathrm{AD}=\frac{1}{2}\mathrm{BC}$                           .....(4)
Using (1), (2), (3) and (4)
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.

#### Question 17:

In a circle, two chords AB and CD intersect at a point inside the circle. Prove that
$\left(\mathrm{a}\right)△\mathrm{PAC}~△\mathrm{PDB}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\mathrm{PA}.\mathrm{PB}=\mathrm{PC}.\mathrm{PD}$

Given: AB and CD are two chords
To Prove:
$\left(\mathrm{a}\right)△\mathrm{PAC}~△\mathrm{PDB}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\mathrm{PA}.\mathrm{PB}=\mathrm{PC}.\mathrm{PD}$

Proof: In
$\angle \mathrm{APC}=\angle \mathrm{DPB}$  (Vertically Opposite angles)
$\angle \mathrm{CAP}=\angle \mathrm{BDP}$  (Angles in the same segment are equal)
By AA similarity-criterion $△\mathrm{PAC}~△\mathrm{PDB}$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are porportional.
$\therefore \frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{PB}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{PA}.\mathrm{PB}=\mathrm{PC}.\mathrm{PD}\phantom{\rule{0ex}{0ex}}$

#### Question 18:

Two chords AB and CD of a circle intersect at a point outside the circle. Prove that
$\left(\mathrm{a}\right)△\mathrm{PAC}~△\mathrm{PDB}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\mathrm{PA}.\mathrm{PB}=\mathrm{PC}.\mathrm{PD}$

Given: AB and CD are two chords
To Prove:
$\left(\mathrm{a}\right)△\mathrm{PAC}~△\mathrm{PDB}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\mathrm{PA}.\mathrm{PB}=\mathrm{PC}.\mathrm{PD}$

Proof:
$\angle \mathrm{ABD}+\angle \mathrm{ACD}={180}^{\circ }$           .....(1)
(Opposite angles of a cyclic quadrilateral are supplementary)
$\angle \mathrm{PCA}+\angle \mathrm{ACD}={180}^{\circ }$           ....(2)
(Linear Pair Angles)
Using (1) and (2), we get
$\angle \mathrm{ABD}=\angle \mathrm{PCA}$
$\angle \mathrm{A}=\angle \mathrm{A}$          (Common)
By AA similarity-criterion $△\mathrm{PAC}~△\mathrm{PDB}$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$\therefore \frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{PB}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{PA}.\mathrm{PB}=\mathrm{PC}.\mathrm{PD}\phantom{\rule{0ex}{0ex}}$

#### Question 19:

In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD⊥AC. If  DP⊥AB and DQ⊥BC then
prove that
(a) DQ2 = DP.QC
(b) DP2 = DQ.AP

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse  then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.
(a) Now using the same property in In △BDC, we get

△CQD ∼ △DQB
$\frac{\mathrm{CQ}}{\mathrm{DQ}}=\frac{\mathrm{DQ}}{\mathrm{QB}}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{DQ}}^{2}=\mathrm{QB}.\mathrm{CQ}$

Now. since all the angles in quadrilateral BQDP are right angles.
Hence, BQDP is a rectangle.
So, QB = DP and DQ = PB
$\therefore {\mathrm{DQ}}^{2}=\mathrm{DP}.\mathrm{CQ}$

(b)

Similarly, △APD ∼ △DPB

#### 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:

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

Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A.
Now, In right triangle ABC
By using Pythagoras theorem, we have

Hence, the stake should be driven $6\sqrt{7}$m far from the base of the pole.

#### 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 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 9:

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 10:

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 11:

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 12:

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 13:

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 14:

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 15:

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 16:

In the given figure, ∠ACB = 900 and CD ⊥ AB. Prove that

$\frac{{\mathrm{BC}}^{2}}{{\mathrm{AC}}^{2}}=\frac{\mathrm{BD}}{\mathrm{AD}}$

Given: ∠ACB = 900 and CD ⊥ AB
To Prove: $\frac{{\mathrm{BC}}^{2}}{{\mathrm{AC}}^{2}}=\frac{\mathrm{BD}}{\mathrm{AD}}$
Proof:
In
$\angle \mathrm{ACB}=\angle \mathrm{CDB}={90}^{\circ }$  (Given)
$\angle \mathrm{ABC}=\angle \mathrm{CBD}$  (Common)
By AA similarity-criterion $△\mathrm{ACB}~△\mathrm{CDB}$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.

In
$\angle \mathrm{ACB}=\angle \mathrm{ADC}={90}^{\circ }$  (Given)
$\angle \mathrm{CAB}=\angle \mathrm{DAC}$  (Common)
By AA similarity-criterion $△\mathrm{ACB}~△\mathrm{ADC}$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.

Dividing (2) by (1), we get
$\frac{{\mathrm{BC}}^{2}}{{\mathrm{AC}}^{2}}=\frac{\mathrm{BD}}{\mathrm{AD}}$

#### 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 20:

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

Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in $1\frac{1}{2}$ hours =
Distance covered by plane B in $1\frac{1}{2}$ hours =
Now, In right triangle ABC
By using Pythagoras theorem, we have

Hence, the distance between two planes after $1\frac{1}{2}$ hours is $300\sqrt{61}$ m.

#### Question 21:

In △ABC, AD is a median and AL ⊥ BC.
Prove that

(a)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2                 .....(1)
Again, In right triangle ACL
Using Pythagoras theorem, we have

(b)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2                 .....(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have

(c) Adding (2) and (4), we get

#### Question 22:

Naman is doing fly-fishing in a stream. The tip of his fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away from him and 2.4 m from the point directly under the tip of the road. Assuming that the string(from the top of his road to the fly) is taut, how much string does he have out (see the adjoining figure)? If he pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from him after seconds.

Naman pulls in the string at the rate of 5 cm per second.
Hence, after 12 seconds the length of the string he will pulled is given by
12 × 5 = 60 cm or 0.6 m
Now, In △BMC
By using Pythagoras theorem, we have
BC2 = CM2 + MB2
= (2.4)2 + (1.8)2
= 9
∴ BC = 3 m
Now, BC' = BC − 0.6
= 3 − 0.6
= 2.4 m
Now, In △BC'M
By using Pythagoras theorem, we have
C'M2 = BC'2 − MB2
= (2.4)2 − (1.8)2
= 2.52
∴ C'M = 1.6 m
The horizontal distance of the fly from him after 12 seconds is given by
C'A = C'M + MA
= 1.6 + 1.2
= 2.8 m

#### Question 1:

State the two properties which are necessary for given two triangles to be similar.

The two triangles are similar if and only if

1. The corresponding sides are in proportion.
2. The corresponding angles are equal.

#### Question 2:

State the basic proportionality theorem.

If a line is drawn parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio.

#### Question 3:

State the converse of Thale's 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.

#### Question 4:

State the mid point theorem.

The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side.

#### Question 5:

State the AAA-similarity criterion.

If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and hence the triangles are similar.

#### Question 6:

State the AA-similarity criterion.

If two angles of a triangle are correspondingly equal to the two angles of another triangle, then the two triangles are similar.

#### Question 7:

State the SSS-similarity criterion.

If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

#### Question 8:

State the SAS-similarity criterion.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.

#### Question 9:

State the Pythagoras' theorem.

The square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here, the hypotenuse is the longest side and it's always opposite the right angle.

#### Question 10:

State the converse of Pythagoras' theorem.

If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle

#### Question 11:

If D, E and F are respectively the midpoints of sides AB, BC and CA of △ABC then what is the ratio of the areas of △DEF and △ABC?

By using mid theorem i.e., the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

Since, the opposite sides of the quadrilateral are parallel and equal.
Hence, BDFE is a parallelogram
Similarly, DFCE is a parallelogram.
Now, In △ABC and △EFD
∠ABC = ∠EFD         (Opposite angles of a parallelogram)
∠BCA = ∠EDF       (Opposite angles of a parallelogram)
By AA similarity criterion, △ABC ∼ △EFD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$\therefore \frac{\mathrm{area}\left(△\mathrm{DEF}\right)}{\mathrm{area}\left(△\mathrm{ABC}\right)}={\left(\frac{\mathrm{DF}}{\mathrm{BC}}\right)}^{2}={\left(\frac{\mathrm{DF}}{2\mathrm{DF}}\right)}^{2}=\frac{1}{4}$
Hence, the ratio of the areas of △DEF and △ABC is 1 : 4.

#### Question 12:

Two triangles ABC and PQR are such that AB = 3 cm, AC = 6 cm, ∠A = 700 PR = 9 cm, ∠P = 700 and PQ = 4.5 cm. Show that
△ABC ∼ △PQR and state similarity theorem.

Now, In △ABC and △PQR
∠A = ∠P = 700          (Given)

By SAS similarity criterion, △ABC ∼ △PQR

#### Question 13:

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

When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal.
Here, △ABC ∼ △DEF

#### Question 14:

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

∠ADE = ∠ABC         (Corresponding angles in DE∥BC)
∠AED = ∠ACB         (Corresponding angles in DE∥BC)
By AA similarity criterion, △ADE ∼ △ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
$\therefore \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$
$⇒\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{AE}+\mathrm{EC}}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{x+3x+4}=\frac{x+3}{x+3+3x+19}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{4x+4}=\frac{x+3}{4x+22}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{x}{2x+2}=\frac{x+3}{2x+11}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+11x=2{x}^{2}+2x+6x+6\phantom{\rule{0ex}{0ex}}⇒3x=6\phantom{\rule{0ex}{0ex}}⇒x=2$
Hence, the value of x is 2.

#### Question 15:

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 is 6 m.

Let AB be a ladder and B is the window at 8 m above the ground C.
Now, In right triangle ABC
By using Pythagoras theorem, we have

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

#### Question 16:

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

We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB = BC = CA = 2a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
∴DC = a
By using Pythagoras theorem, we have
${\mathrm{AC}}^{2}={\mathrm{CD}}^{2}+{\mathrm{DA}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(2a\right)}^{2}={a}^{2}+{\mathrm{DA}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{DA}}^{2}=4{a}^{2}-{a}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{DA}}^{2}=3{a}^{2}$

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

#### Question 17:

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

We have △ABC ∼ △DEF
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$\therefore \frac{\mathrm{area}\left(△\mathrm{ABC}\right)}{\mathrm{area}\left(△\mathrm{DEF}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{64}{169}={\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{8}{13}\right)}^{2}={\left(\frac{4}{\mathrm{EF}}\right)}^{2}$

#### Question 18:

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

In △AOB and COD
∠ABO = ∠CDO         (Alternte angles in AB∥CD)
∠AOB = ∠COD         (Vertically opposite angles)
By AA similarity criterion, △AOB ∼ △COD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.

#### Question 19:

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

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.

#### Question 20:

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

We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides..
Suppose ABC is an equilateral triangle having AB = BC = CA = a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
$\therefore \mathrm{DC}=\frac{1}{2}a$
By using Pythagoras theorem, we have
${\mathrm{AC}}^{2}={\mathrm{CD}}^{2}+{\mathrm{DA}}^{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}={\left(\frac{1}{2}a\right)}^{2}+{\mathrm{DA}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{DA}}^{2}={a}^{2}-\frac{1}{4}{a}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{DA}}^{2}=\frac{3}{4}{a}^{2}$

#### Question 21:

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

Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 12 cm and BO = 5 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + BO2
= 122 + 52
= 144 + 25
= 169
∴AB2 = 169
⇒ AB = 13 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 13 cm

#### Question 22:

Two traingles DEF and GHK are such that ∠D = 48and ∠H = 57 . If △DEF ∼ △GHK, then find the measure of  ∠F.

If two traingle are similar then the corresponding angles of the two tringles are equal.
Here, △DEF ∼ △GHK
∴∠E = ∠H = 57
Now, In △DEF
∠D + ∠E + ∠F = 180      (Angle sum property of triangle)
⇒ ∠F = 180  −  48  −  57  =  75

#### Question 23:

In the given figure, MN∥BC and AM : MB = 1 : 2.
Find$\frac{\mathrm{area}\left(△\mathrm{AMN}\right)}{\mathrm{area}\left(△\mathrm{ABC}\right)}$

We have
AM : MB = 1 : 2
$⇒\frac{\mathrm{MB}}{\mathrm{AM}}=\frac{2}{1}$
Adding 1 to both sides, we get
$⇒\frac{\mathrm{MB}}{\mathrm{AM}}+1=\frac{2}{1}+1\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{MB}+\mathrm{AM}}{\mathrm{AM}}=\frac{2+1}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AB}}{\mathrm{AM}}=\frac{3}{1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Now, In △AMN and △ABC
∠AMN = ∠ABC         (Corresponding angles in MN∥BC)
∠ANM = ∠ACB         (Corresponding angles in MN∥BC)
By AA similarity criterion, △AMN ∼ △ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$\therefore \frac{\mathrm{area}\left(△\mathrm{AMN}\right)}{\mathrm{area}\left(△\mathrm{ABC}\right)}={\left(\frac{\mathrm{AM}}{\mathrm{AB}}\right)}^{2}={\left(\frac{1}{3}\right)}^{2}=\frac{1}{9}$

#### Question 24:

In a triangle BMP and CNR it is given that PB = 5 cm, MP = 6 cm, BM = 9 cm and NR = 9 cm. If △BMP ∼ △CNR, then find the perimeter of the △CNR

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Here, △BMP ∼ △CNR

Perimeter of △CNR = CN + NR + CR = 13.5 + 9 + 7.5 = 30 cm

#### Question 25:

Each of the equal sides of an isosceles triangle is 25 cm. Find the length of its altitude if the base is 14 cm.

We know that the altitude drawn from the vertex opposite to the non equal side bisects the non equal side.
Suppose ABC is an isosceles triangle having equal sides AB and BC.
So, the altitude drawn from the vertex will bisect the opposite side.
Now, In right triangle ABD
By using Pythagoras theorem, we have

#### Question 26:

A man goes 12 m due south and then 35 m due west. How far is he from the starting point?

In right triangle SOW
By using Pythagoras theorem, we have

Hence, the man is 37 m away from the starting point.

#### Question 27:

If the length of the sides BC, CA and AB of a △ABC are a, b and c respectively and AD is the bisector of ∠A then find the length of BD and DC.

Let DC = x
∴ BD = ax
By using angle bisector theore in △ABC, we have
$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}\phantom{\rule{0ex}{0ex}}⇒\frac{c}{b}=\frac{a-x}{x}\phantom{\rule{0ex}{0ex}}⇒cx=ab-bx\phantom{\rule{0ex}{0ex}}⇒x\left(b+c\right)=ab\phantom{\rule{0ex}{0ex}}⇒x=\frac{ab}{\left(b+c\right)}$
Now,
$a-x=a-\frac{ab}{b+c}\phantom{\rule{0ex}{0ex}}=\frac{ab+ac-ab}{b+c}\phantom{\rule{0ex}{0ex}}=\frac{ac}{\left(a+b\right)}$

#### Question 28:

In the given figure, ∠AMN = ∠MBC = 760 If p, q and r are the lengths of AM, MB and BC respectively, then express the length of MN  in terms of p, q and r.

In △AMN and △ABC
∠AMN = ∠ABC = 760     (Given)
∠A = ∠A         (Common)
By AA similarity criterion, △AMN ∼ △ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
$\therefore \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{MN}}{\mathrm{BC}}$
$⇒\frac{\mathrm{AM}}{\mathrm{AM}+\mathrm{MB}}=\frac{\mathrm{MN}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{a+b}=\frac{\mathrm{MN}}{c}\phantom{\rule{0ex}{0ex}}⇒\mathrm{MN}=\frac{ac}{a+b}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 29:

The length of the diagonals of a rhombus are 40 cm and 42 cm. Find the length of each sides of the rhombus

Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 20 cm and BO = 21 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + OB2
= 202 + 212
= 400 + 441
= 841
∴AB2 = 841
⇒ AB = 29 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 29 cm

#### Question 30:

For each of the following statements state whether true(T) or false(F).
(i) Two circles with different radii are similar.
(ii) Any two rectangles are similar.
(iii) If two traingles are similar, the their corresponding angles are equal and their corresponding sides are equal.
(iv) The length of the line segment joining the mid points of any two sides of a trinagle is equal to the half the length of the third side.
(v) In △ABC, AB = 6 cm, ∠A = 450 and AC = 8 cm and in △DEF, DF = 9 cm, ∠D = 450 and DE = 12 cm, then △ABC ∼ △DEF.
(vi) The polygon formed by joining the mid points of the sides of a quadrilateral is a rhombus.
(vii) The ratio of areas of two similar triangles is equal to the ratio of their corresponding angle-bisector segments.
(viii) The ratio of the perimeters of two similar triangles is same as the ratio of their corresponding medians.
(ix) If O is any point inside a rectangle ABCD then OA2 + OC2 = OB2 + OD2
(x) The sum of the squares on the sides of a rhombus is equal to the sum of the squares on its diagonals.

(ii) True
Two circles of any radii are similar to each other.

(i) False
Two rectangles are similar if their corresponding sides are proportional.

(iii) Falase
If two traingles are similar, the their corresponding angles are equal and their corresponding sides are proportional.

(iv) True
Suppose ABC is a triangle and M, N are

Construction: DE is expanded to F sich that EF = DE
To Prove = $\mathrm{DE}=\frac{1}{2}\mathrm{BC}$
AE = EC        (E is the mid point of AC)
DE = EF        (By construction)
AED = CEF      (Vertically Opposite angle)
By SAS criterion, △ADE ≅ △CEF
⇒ BD = CF
Since, ∠ADE and ∠EFC are alternate angle
Hence, AD ∥ CF and BD ∥ CF
When two sides of a quadrilateral are parallel, then it is a parallelogram
∵DF = BC and BD ∥ CF
∴BDFC is a parallelogram
Hence, DF = BC
⇒ DE + EF = BC
$⇒\mathrm{DE}=\frac{1}{2}\mathrm{BC}$

(v) False
In △ABC, AB = 6 cm, ∠A = 45 and AC = 8 cm and in △DEF, DF = 9 cm, ∠D = 45 and DE = 12 cm, then △ABC ∼ △DEF.

In △ABC and △DEF
∠A = ∠D = 45
$\frac{\mathrm{AB}}{\mathrm{AC}}\ne \frac{\mathrm{DE}}{\mathrm{DF}}$
So △ABC is not similar to △DEF

(vi) False
The polygon formed by joining the mid points of the sides of a quadrilateral is a parallelogram.

(vii) True

Given: △ABC ∼ △DEF
To Prove = $\frac{\mathrm{Ar}\left(△\mathrm{ABC}\right)}{\mathrm{Ar}\left(△\mathrm{DEF}\right)}={\left(\frac{\mathrm{AP}}{\mathrm{DQ}}\right)}^{2}$
Proof: In △ABP and △DEQ
∠BAP = ∠EDQ        (As ∠A = ∠D, so their Half is also equal)
∠B = ∠E                   (△ABC ∼ △DEF)
By AA criterion, △ABP ∼ △DEQ
$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AP}}{\mathrm{DQ}}$           .....(1)
Since, △ABC ∼ △DEF

(viii) True

Given: △ABC ∼ △DEF
To Prove = $\frac{\mathrm{Perimeter}\left(△\mathrm{ABC}\right)}{\mathrm{Perimeter}\left(△\mathrm{DEF}\right)}=\frac{\mathrm{AP}}{\mathrm{DQ}}$
Proof: In △ABP and △DEQ
∠B = ∠E                  (∵△ABC ∼ △DEF)
∵△ABC ∼ △DEF
$\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{2\mathrm{BP}}{2\mathrm{EQ}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BP}}{\mathrm{EQ}}$

By SAS criterion, △ABP ∼ △DEQ
$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AP}}{\mathrm{DQ}}$           .....(1)
Since, △ABC ∼ △DEF

(ix) True

Suppose ABCD is a rectangle with O is any point inside it.
Construction: Join OA, OB, OC, OD and draw two parallel lines SQ ∥ AB ∥ DC and PR ∥ BC ∥ AD
To prove: OA2 + OC2 = OB2 + OD2
Proof:
OA2 + OC2 = (AS2 + OS2) + (OQ2 + QC2)             [Using Pythagoras theorem in right triangle AOP and COQ]
= (BQ2 + OS2) + (OQ2 + DS2
= (BQ2 + OQ2) + (OS2 + DS2)                                 [Using Pythagoras theorem in right triangle BOQ and DOS]
= OB2 + OD2
Hence, LHS = RHS

(x) True

Suppose ABCD is a rhombus having AC and BD its diagonals.
Since, the diagonals of a rhombus perpendicular bisect each other.
Hence, AOC is a right angle triangle
In right triangle AOC
By using Pythagoras theorem, we have

#### 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:

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 3:

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 4:

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 5:

The shadow of a 5-m-long stick is 2 m long. At the same time the length of the shadow of a 12.5 m high tree (in m) is
(a) 3.0
(b) 3.5
(c) 4.5
(d) 5.0

Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.
Suppsose DA and BA are the shadows of DE and BC respectively.
∠A = ∠A             (Common)
By AA-similarity criterion

If two triangles are similar, then the the ratio of their corresponding sides are equal.

Hence, the correct answer is option (d).

#### 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:

In the given figure, O is the point inside a △MNP such that ∠MOP = 900 , OM = 16 cm and OP = 12 cm. If MN = 21 cm and ∠NMP = 900 then
NP = ?
(a) 25 cm
(b) 29 cm
(c) 33 cm
(d) 35 cm

Now, In right triangle MOP
By using Pythagoras theorem, we have

Now, In right triangle MPN
By using Pythagoras theorem, we have

Hence, the correct answer is option (b).

#### Question 8:

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 9:

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 10:

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 11:

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 12:

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 13:

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, the CD = ?
(a) 4.8 cm
(b) 3.5 cm
(c) 7 cm
(d) 10.5 cm

By using angle bisector theore in △ABC, we have
$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}\phantom{\rule{0ex}{0ex}}⇒\frac{10}{14}=\frac{6-x}{x}\phantom{\rule{0ex}{0ex}}⇒10x=84-14x\phantom{\rule{0ex}{0ex}}⇒24x=84\phantom{\rule{0ex}{0ex}}⇒x=3.5$
Hence, the correct answer is option (b).

#### Question 14:

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 15:

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 16:

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 17:

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 18:

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 19:

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 20:

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 21:

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 22:

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 23:

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 24:

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 25:

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 26:

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 27:

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 28:

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 29:

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 30:

ABC and BDE are two equilateral triangles such that D is the mid point of BC. Ratio of the areas of triangles ABC and BDE is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1

Given: ABC and BDE are two equilateral triangles
Since, D is the mid point of BC and BDE is also an equilateral traingle.
Hence, E is also the mid point of AB.
Now, D and E are the mid points of BC and AB.
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it.

Now, In △ABC and △EBD
∠BED = ∠BAC             (Corresponding angles)
∠B = ∠B             (Common)
By AA-similarity criterion
△ABC ∼ △EBD

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$\therefore \frac{\mathrm{area}\left(△\mathrm{ABC}\right)}{\mathrm{area}\left(△\mathrm{DBE}\right)}={\left(\frac{\mathrm{AC}}{\mathrm{ED}}\right)}^{2}={\left(\frac{2\mathrm{ED}}{\mathrm{ED}}\right)}^{2}=\frac{4}{1}$
Hence, the correct answer is option (d).

#### Question 31:

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 32:

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

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

#### Question 33:

In △ABC, AB = $6\sqrt{3}$ cm, AC = 12 cm and BC = 6 cm. Then ∠B is
(a) 45o
(b) 60o
(c) 90o
(d) 120o

Since, the square of the longest side is equal to the sum of the square of two sides, so △ABC is a right angled triangle.
∴The angle opposite to AC i.e. ∠B = 90o
Hence, the correct answer is option (c)

#### Question 34:

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 35:

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 36:

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 37:

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 38:

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 39:

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 40:

Corresponding sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(a) 2 : 3
(b) 4 : 9
(c) 9 : 4
(d) 16 : 81

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.

Hence, the correct answer is option (d).

#### Question 41:

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 42:

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 43:

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 44:

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 45:

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 46:

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 47:

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 48:

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 49:

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 50:

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 51:

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 52:

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 53:

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 54:

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 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 $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}.$