ML Aggarwal Solution Class 10 Chapter 13 Similarity Exercise 13.1

 Exercise 13.1

Question 1

State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm):

Sol :

Given

(i) In ∆ABC and PQR

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{3.2}{4}=\frac{4}{5}$

$\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{3.6}{4.5}=\frac{4}{5}$

$\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{3}{5.4}=\frac{5}{9}$

$\because$ All the sides are not proportional.

$\therefore$ The triangles are not similar.

(ii) In ΔDEF and ΔLMN

$\angle \mathrm{E}=\angle \mathrm{N}=40^{\circ}$

$\frac{\mathrm{DE}}{\mathrm{LN}}=\frac{4}{2}$

$=\frac{2}{1}$ and $\frac{\mathrm{EF}}{\mathrm{MN}}=\frac{4.8}{2.4}=\frac{2}{1}$

$\therefore \Delta \mathrm{DEF} \sim \Delta \mathrm{LMN}$  (SAS axiom)

Question 2

It is given that ∆DEF ~ ∆RPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P ? Why?

Sol :

∆DEF ~ ∆RPQ

∠D = ∠R and ∠F = ∠Q not ∠P

No, ∠F ≠ ∠P

Question 3

If in two right triangles, one of the acute angle of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles are similar? Why?

Sol :

In two right triangles,

one of the acute angles of the one triangle is

equal to an acute angle of the other triangle.

The triangles are similar. (AAA axiom)

Question 4

In the given figure, BD and CE intersect each other at the point P. Is ∆PBC ~ ∆PDE? Give reasons for your answer.

Sol :

In the given figure, two line segments intersect each other at P.

In ∆BCP and ∆DEP

∠BPC = ∠DPE

$\frac{5}{10}=\frac{6}{12}\left(\right.$ each $\left.=\frac{1}{2}\right)$

$\therefore \Delta B C D \sim \Delta D E P$  (SAS axiom)

Question 5

It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm.

Find the lengths of the remaining sides of the triangles.

Sol :

∆ABC ~ ∆EDF

AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm

$\because \Delta \mathrm{ABC} \sim \Delta \mathrm{EDF}$

$\therefore \frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{AC}}{\mathrm{EF}}=\frac{\mathrm{BC}}{\mathrm{DF}}$

$\Rightarrow \frac{5}{12}=\frac{7}{\mathrm{EF}}=\frac{\mathrm{BC}}{15}$

$\Rightarrow \frac{7}{\mathrm{EF}}=\frac{5}{12}$

$\Rightarrow \mathrm{EF}=\frac{7 \times 12}{5}=\frac{84}{5}=16.8 \mathrm{~cm}$

and $\frac{5}{12}=\frac{B C}{15}$

$\Rightarrow \mathrm{BC}=\frac{5 \times 15}{12}$

$=\frac{25}{4}=6.25 \mathrm{~cm}$

Question 6

(a) If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, then find the perimeter of ∆ABC.

(b) If ∆ABC ~ ∆PQR, Perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then find the length of AC.

Sol :

(a) ∆ABC ~ ∆DEF

AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm

Perimeter of ΔABC

$\because \Delta \mathrm{ABC} \sim \Delta \mathrm{DEF}$

$\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{EF}}$

$\Rightarrow \frac{4}{6}=\frac{A C}{12}=\frac{B C}{9}$

$\frac{\Lambda C}{12}=\frac{4}{6}$ $\Rightarrow AC=\frac{12 \times 4}{6}=8 \mathrm{~cm}$

and $\frac{B C}{E F}=\frac{4}{6} \Rightarrow \frac{B C}{9}=\frac{4}{6}$

$\Rightarrow \mathrm{BC}=\frac{9 \times 4}{6}=6 \mathrm{~cm}$

$\therefore$ Perimeter of $\Delta \mathrm{ABC}=\mathrm{AB}+\mathrm{BC}+\mathrm{AC}$

=4+6+8=18 cm

(b) 

$\triangle A B C-\Delta P Q R$

Given Perimeter of $\Delta \mathrm{ABC}=32 \mathrm{~cm}$ and Perimeter of $\Delta \mathrm{PQR}=48 \mathrm{~cm}$

Side $P R=6 \mathrm{~cm}$

$\because \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$

$\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}}$

$\Rightarrow \frac{32}{48}=\frac{A C}{6}$

$\Rightarrow A C=\frac{32 \times 6}{48}=4 \mathrm{~cm}$

Hence, the length of AC=4 cm

Question 7

Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4 cm, 7 cm and 8 cm.

Sol :

Let ∆ABG ~ ∆DEF in which shortest side of

∆ABC is BC = 6 cm.

In ∆DEF, DE = 8 cm, EF = 4 cm and DF = 7 cm

$\because \Delta \mathrm{ABC} \sim \Delta \mathrm{DEF}$

$\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$

$\Rightarrow \frac{\mathrm{AB}}{8}=\frac{6}{4}=\frac{\mathrm{AC}}{7}$

Now, $\frac{\mathrm{AB}}{8}=\frac{6}{4}$

$\Rightarrow \mathrm{AB}=\frac{8 \times 6}{4}=12 \mathrm{~cm}$

$\frac{A C}{7}=\frac{6}{4}$

$\Rightarrow A C=\frac{7 \times 6}{4}=\frac{21}{2}=10.5 \mathrm{~cm}$

Question 8

(a) In the figure given below, AB || DE, AC = , 3 cm, CE = 7.5 cm and BD = 14 cm. Calculate CB and DC.

(b) In the figure (2) given below, CA || BD, the lines AB and CD meet at G.

(i) Prove that ∆ACO ~ ∆BDO.

(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.

Sol :

(a) In the given figure,

AB||DE, AC = 3 cm, CE = 7.5 cm, BD = 14 cm

To calculate CB and DC

In ΔABC and ΔCDE

$\angle \mathrm{ACB}=\angle \mathrm{DCE}$

(Vertically opposite angles)

$\angle \mathrm{BAC}=\angle \mathrm{CED}$ (Alternate agnles)

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{CDE}$ (AA axiom)

$\therefore \frac{\mathrm{AC}}{\mathrm{CE}}=\frac{\mathrm{BC}}{\mathrm{CD}}$

$\Rightarrow \frac{3}{7.5}=\frac{\mathrm{BC}}{\mathrm{CD}}$

$\Rightarrow \mathrm{BC} \times 7.5=3 \mathrm{CD}$ $\left\{\begin{aligned} \because \mathrm{BD} &=14 \mathrm{~cm} \\ \mathrm{CB} &=14-\mathrm{DC} \end{aligned}\right\}$

Let $\mathrm{BC}=x,$ then $\mathrm{CD}=(14-x) \mathrm{cm}$

$x \times 7.5=3 \times(14-x)$

$7.5 x=42-3 x$

$7.5 x+3 x=42$

$\Rightarrow 10.5 x=42$

$x=\frac{42}{10.5}=4$

$\therefore B C=4 \mathrm{~cm}$ and $D C=14-4=10 \mathrm{~cm}$

(b) In the given figure, CA||BD

AB and CD intersect at O

To prove:

(i) ΔACOΔBDO

(ii) BD=2.4cm ,OD=4cm

OB=3.2cm, AC=3.6cm then calculate

OA and OC

Proof :

(i) In ΔACO and ΔBDO

∠AOC=∠BOD (vertically opposite angles)

∠A=∠B

∴ΔACO~ΔBDO (AA axiom)

BD=2.4cm, OD=4cm, OB=3.2cm

AC=3.6cm

∵ΔACO~ΔBOD

∴$\frac{A O}{O B}=\frac{C O}{O D}=\frac{A C}{B D}$

$\Rightarrow \frac{\mathrm{AO}}{3.2}=\frac{\mathrm{CO}}{4}=\frac{3.6}{2.4}$

$\Rightarrow \frac{\mathrm{AO}}{3.2}=\frac{3.6}{2.4}$

$\Rightarrow \mathrm{AO}=\frac{3.6 \times 3.2}{2.4}=4.8 \mathrm{~cm}$

$\frac{\mathrm{CO}}{4}=\frac{3.6}{2.4}$

$\Rightarrow \mathrm{CO}=\frac{3.6 \times 4}{2.4}=6 \mathrm{~cm}$

Question 9

(a) In the figure

(i) given below, ∠P = ∠RTS.

Prove that ∆RPQ ~ ∆RTS.

(b) In the figure (ii) given below,

∠ADC = ∠BAC. Prove that CA² = DC x BC.

Sol :

(a) In the given figure, ∠P = ∠RTS

To prove : ∆RPQ ~ ∆RTS

Proof : In ∆RPQ and ∆RTS

∠R = ∠R (common)

∠P = ∠RTS (given)

∆RPQ ~ ∆RTS (AA axiom)

(b) In the given figure, ∠ADC = ∠BAC

To prove : CA² = DC x BC

Proof : In ∆ABC and ∆ADC

∠C = ∠C (common)

∠BAC = ∠ADC (given)

∴∆ABC ~ ∆ADC (AA axiom)

$\therefore \frac{\mathrm{CA}}{\mathrm{DC}}=\frac{\mathrm{BC}}{\mathrm{CA}}$

(Corresponding sides are proportional)

$\therefore \mathrm{CA} \times \mathrm{CA}=\mathrm{DC} \times \mathrm{BC}$

$\Rightarrow \mathrm{CA}^{2}=\mathrm{DC} \times \mathrm{BC}$

Question 10

(a) In the figure (1) given below, AP = 2PB and CP = 2PD.

(i) Prove that ∆ACP is similar to ∆BDP and AC || BD.

(ii) If AC = 4.5 cm, calculate the length of BD.

(b) In the figure (2) given below,

∠ADE = ∠ACB.

(i) Prove that ∆s ABC and AED are similar.

(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.

(c) In the figure (3) given below, ∠PQR = ∠PRS. Prove that triangles PQR and PRS are similar. If PR = 8 cm, PS = 4 cm, calculate PQ.

Sol :

In the given figure,

AP = 2PB, CP = 2PD

To prove:

(i) $\Delta \mathrm{ACP} \sim \Delta \mathrm{BDP}$ and $\mathrm{AC} \| \mathrm{BD}$

(ii) If AC=4.5 cm, find length of BD

Proof :

(i) $\mathrm{AP}=2 \mathrm{~PB} \Rightarrow \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{2}{1}$

and $\mathrm{CP}=2 \mathrm{PD} \Rightarrow \frac{\mathrm{CP}}{\mathrm{PD}}=\frac{2}{1}$

and $\angle \mathrm{APC}=\angle \mathrm{BPD}$

(Vertically opposite angles)

$\therefore \Delta \mathrm{ACP} \sim \Delta \mathrm{BDP}$ (SAS axiom)

$\therefore \angle \mathrm{CAP}=\angle \mathrm{PBD}$

But these are alternate angles

$\therefore \mathrm{AC} \| \mathrm{BD}$

(ii) $\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{2}{1}$

$\therefore A C=2 B D \Rightarrow 2 B D=4.5 \mathrm{~cm}$

$\therefore B D=\frac{4.5}{2}=2.25 \mathrm{~cm}$

(b) In the given figure, $\angle \mathrm{ADE}=\angle \mathrm{ACB}$

To prove :

(i) $\triangle \mathrm{ABC} \sim \Delta \mathrm{AED}$

(ii) If AE=3 cm, BD=1 cm and AB=6cm find AC

Proof : 

In ΔABC and ΔAED

∠A=∠A (common)

∠ACB=∠ADE (given)

∴$\triangle \mathrm{ABC} \sim \Delta \mathrm{AED}$  (AA axiom)

∴$\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{AC}}{\mathrm{AD}}$

$\frac{B C}{D E}=\frac{6}{3}=\frac{A C}{5}$

$\left\{\begin{aligned} \because & \mathrm{AD}=\mathrm{AB}-\mathrm{BD} \\ &=6-1=5 \mathrm{~cm} \end{aligned}\right\}$

$\frac{6}{3}=\frac{\mathrm{AC}}{5}$

$\Rightarrow \mathrm{AC}=\frac{6 \times 5}{3}=10 \mathrm{~cm}$

(c) In the given figure, ∠PQR=∠PRS

To prove:

(i) ΔPQR～ΔPRS

(ii) If $\mathrm{PR}=8 \mathrm{~cm}, \mathrm{PS}=4 \mathrm{~cm},$ find $\mathrm{PQ}$

Proof : In ΔPQR and ΔPRS

∠P=∠P (common)

∠PQR=∠PRS (given)

∴$\Delta \mathrm{PQR} \sim \Delta \mathrm{PRS}$ (AA axiom)

∴$\frac{P Q}{P R}=\frac{P R}{P S}=\frac{Q R}{S R}$

(Sides are proportional)

$\frac{P Q}{8}=\frac{8}{4}$

$\Rightarrow P Q=\frac{8 \times 8}{4}=16 \mathrm{~cm}$

∴PQ=16cm

Question 11

In the given figure, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC. Prpve that BM x NP = CN x MP.

Sol :

In the given figure, ABC in which AB = AC.

P is a point on BC such that PM ⊥ AB and PN ⊥ AC

To prove : BM x NP = CN x MP

Proof: $\operatorname{In} \Delta \mathrm{ABC}, \mathrm{AB}=\mathrm{AC}$

$\therefore \angle \mathrm{B}=\angle \mathrm{C}$ (Angles opposite to equal sides)

Now in ΔBMP and ΔCNP 

$\angle \mathrm{M}=\angle \mathrm{N}$ (each 90°)

$\therefore \angle \mathrm{B}=\angle \mathrm{C}$ (proved)

$\therefore \Delta \mathrm{BMP} \sim \Delta \mathrm{CNP}$ (AA axiom)

$\therefore \frac{\mathrm{BM}}{\mathrm{CN}}=\frac{\mathrm{MP}}{\mathrm{NP}}$  (sides are proportional)

$\therefore \mathrm{BM} \times \mathrm{NP}=\mathrm{CN} \times \mathrm{MP}$

(By cross multiplying)

Hence proved

Question 12

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

Sol :

Given : ∆ABC ~ ∆PQR .

To prove : Ratio in their perimeters k

the same as the ratio in their corresponding sides.

Proof : $\because \triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$

$\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{RP}}$

$=\frac{A B+B C+C A}{P Q+Q R+R P}$

$=\frac{\text { Perimeter of } \Delta \mathrm{ABC}}{\text { Perimeter of } \Delta \mathrm{PQR}}$

Question 13

In the given figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that $\frac{A O}{O O}=\frac{B O}{O D}$

Using the above result, find the value(s) of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.

Sol :

ABCD is a trapezium in which AB || DC

Diagonals AC and BD intersect each other at O.

Figure to be added

To prove :

(i) $\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{OD}}$

(ii) If OA=3 x-19,OB=x-4, OC=x-3, OD=4

Find the value of x

Proof :

(i) In ΔAOB and ΔCOD

∠AOB=∠COD

(vertically opposite angles)

∠OAB=∠OCD

∴$\Delta \mathrm{AOB} \sim \Delta \mathrm{COD}$

∴$\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$

(ii) ∵ OA=3x-19, OB=x-4, OC=x-3,OD=4

then $\frac{3 x-19}{x-3}=\frac{x-4}{4}$

By cross multiplication

$(x-3)(x-4)=4(3 x-19)$

$x^{2}-4 x-3 x+12=12 x-76$

$\Rightarrow x^{2}-7 x-12 x+12+76=0$

$\Rightarrow x^{2}-19 x+88=0$

$\Rightarrow x^{2}-8 x-11 x+88=0$

$\Rightarrow x(x-8)-11(x-8)=0$

$(x-8)(x-11)=0$

Either x-8=0, then x=8

or x-11=0, then x=11

∴x=8, 11

Question 14

In ∆ABC, ∠A is acute. BD and CE are perpendicular on AC and AB respectively. Prove that AB x AE = AC x AD.

Sol :

In ∆ABC, ∠A is acute

BD and CE are perpendiculars on AC and AB respectively

To prove : $A B \times A E=A C \times A D$

Proof : $\operatorname{In} \Delta \mathrm{ADB}$ and $\Delta \mathrm{AEC}$

$\angle A=\angle A$  (common)

$\angle \mathrm{ADB}=\angle \mathrm{AEC}$ (each 90°)

∴$\Delta \mathrm{ADB} \sim \Delta \mathrm{AEC}$ (AA axiom)

∴$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}$

⇒AB×AE=AC×AD

(By cross multiplication)

Question 15

In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC. Prove that $\frac{B E}{D E}=\frac{A C}{B C}$

Sol :

In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC

To prove : 

$\frac{B E}{D E}=\frac{A C}{B C}$

Proof: In ∆ABC and ∆DEB

$\angle \mathrm{C}=90^{\circ}$

$\therefore \angle \mathrm{A}+\angle \mathrm{ABC}=90^{\circ}$...(i)

and $\operatorname{In} \Delta \mathrm{DEB}$

$\angle \mathrm{DBE}+\angle \mathrm{ABC}=90^{\circ}$..(ii)

From (i)

∠A=∠DBE

Now in ΔABC and ΔDBE

∠C=∠E (each 90°)

∴ΔABC～ΔDBE (AA axiom)

$\therefore \frac{\mathrm{AC}}{\mathrm{BE}}=\frac{\mathrm{BC}}{\mathrm{DE}} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{BE}}{\mathrm{DE}}$

Question 16

(a) In the figure (1) given below, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that ∆ABE ~ ∆CFB.

(b) In the figure (2) given below, PQRS is a parallelogram; PQ = 16 cm, QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.

(i) Prove that triangle RLQ is similar to triangle PLN. Hence, find PN.

(ii) Name a triangle similar to triangle RLM. Evaluate RM.

Sol :

(a) In the given figure, ABCD is a ||gm

E is a point on AD and produced

and BE intersects CD at F.

To prove : ∆ABE ~ ∆CFB

Proof : In ∆ABE and ∆CFB

∠A = ∠C (opposite angles of a ||gm)

∠ABE = ∠BFC (alternate angles)

∆ABE ~ ∆CFB (AA axiom)

(b) In the given figure, PQRS is a ||gm PQ = 16 cm,

QR = 10 cm

L is a point on PR such that

RL : LP=2 : 3

QL is produced to meet RS at M and PS produced at N

To prove :

(i) $\Delta \mathrm{RLQ} \sim \Delta \mathrm{PLN}$ and find PN

(ii) Name triangle similar to ΔRLM and evaluate RM

Proof : In ΔRLQ and ΔRLN

∠RLQ=∠NLP (vertically opposite angles)

∠RQL=∠LNP (alternate angles)

∴ΔRLQ～ΔPLN

∴$\frac{\mathrm{QR}}{\mathrm{PN}}=\frac{\mathrm{RL}}{\mathrm{LP}}=\frac{2}{3}$

$\Rightarrow \frac{10}{\mathrm{PN}}=\frac{2}{3}$

$\Rightarrow \mathrm{PN}=\frac{10 \times 3}{2}=15 \mathrm{~cm}$

(ii) In $\Delta \mathrm{RLM}$ and $\Delta \mathrm{QLP}$

$\angle \mathrm{RLM}=\angle \mathrm{QLP}$ (vertically opposite angles)

$\angle \mathrm{LRM}=\angle \mathrm{LPQ}$ (alternate angles)

$\therefore \Delta R L M \sim \Delta Q L P$

$\therefore \frac{R M}{P Q}=\frac{R L}{L P}=\frac{2}{3}$

$\frac{\mathrm{RM}}{16}=\frac{2}{3}$

$\Rightarrow \mathrm{RM}=\frac{16 \times 2}{3}=\frac{32}{3}$

$=10 \frac{2}{3} \mathrm{~cm}$

Question 17

The altitude BN and CM of ∆ABC meet at H. Prove that

(i) CN . HM = BM . HN .

(ii) $\frac{H O}{H B}=\sqrt{\frac{O M . H N}{B M . H M}}$

(iii) $\triangle M H N \sim \Delta B H C$.

Sol :

In the given figure, BN ⊥ AC and CM ⊥ AB of ∆ABC

which intersect each other at H.

To prove:

(i) CN.HM = BM.HN

(ii) $\frac{H C}{H B}=\sqrt{\frac{C N \cdot H N}{B M \cdot H M}}$

(iii) $\Delta M H N \sim \Delta B H C$

Construction: Join MN

Proof :

(i) In $\Delta \mathrm{BHM}$ and $\Delta \mathrm{CHN}$

$\angle \mathrm{BHM}=\angle \mathrm{CHN}$

(vertically opposite angles)

$\angle M=\angle N$ (each 90°)

$\therefore \Delta \mathrm{BHM} \sim \Delta \mathrm{CHN}$ (AA axiom)

$\therefore \frac{\mathrm{HM}}{\mathrm{HN}}=\frac{\mathrm{BM}}{\mathrm{CN}}=\frac{\mathrm{HB}}{\mathrm{HC}}$

By cross multiplication,

CN.HM=BM.HN

Now, $\frac{\mathrm{HC}}{\mathrm{HB}}=\sqrt{\frac{\mathrm{HN} \times \mathrm{CN}}{\mathrm{HM} \times \mathrm{BM}}}$

$=\sqrt{\frac{\mathrm{CN} \times \mathrm{HN}}{\mathrm{BM} \times \mathrm{HM}}}$

$\because M$ and N divide AB and AC in the same ratio

∴MN||BC

(ii) Now in $\Delta \mathrm{MHN}$ and $\Delta \mathrm{BHC}$

$\angle \mathrm{MHN}=\angle \mathrm{BHC}$

(vertically opposite angles)

$\angle \mathrm{MNH}=\angle \mathrm{HBC}$ (alternate angles)

$\therefore \Delta \mathrm{MHN} \sim \Delta \mathrm{BHC}$ (AA axiom)

Question 18

In the given figure, CM and RN are respectively the medians of ∆ABC and ∆PQR. If ∆ABC ~ ∆PQR, prove that:

(i) ∆AMC ~ ∆PNR

(ii) $\frac{C M}{R N}=\frac{A B}{P Q}$

(iii) ∆CMB ~ ∆RNQ.

Sol :

In the given figure, CM and RN are medians of ∆ABC and ∆PQR

respectively and ∆ABC ~ ∆PQR

To prove:

(i) ∆AMC ~ ∆PNR

(ii) $\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{AB}}{\mathrm{PQ}}$

(iii) $\Delta \mathrm{CMB} \sim \Delta \mathrm{RNQ}$

Proof $: \because \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$

$\angle \mathrm{A}=\angle \mathrm{P}, \angle \mathrm{B}=\angle \mathrm{Q}$ and $\angle \mathrm{C}=\angle \mathrm{R}$ and corresponding sides are also proportional

i.e., $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{RP}}$

Now in $\Delta \mathrm{AMC}$ and $\Delta \mathrm{PNR}$,

$\angle \mathrm{A}=\angle \mathrm{P}$

$\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AM}}{\mathrm{PN}}$  $\left\{\because \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\frac{1}{2} \mathrm{AB}}{\frac{1}{2} \mathrm{PQ}}=\frac{\mathrm{AM}}{\mathrm{PN}}\right\}$

$\therefore \Delta \mathrm{AMC} \sim \Delta \mathrm{PNR}$ (SAS axiom)

(ii) $\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{AM}}{\mathrm{PN}}=\frac{2 \mathrm{AM}}{2 \mathrm{PN}}=\frac{\mathrm{AB}}{\mathrm{PQ}}$

(iii) Now in $\Delta \mathrm{CMB}$ and $\Delta \mathrm{RNQ}$

$\angle B=\angle Q$

$\frac{B C}{Q P}=\frac{B M}{Q N}$ $\left\{\because \frac{\mathrm{BM}}{\mathrm{QN}}=\frac{\frac{1}{2} \mathrm{AB}}{\frac{1}{2} \mathrm{PQ}}=\frac{\mathrm{AB}}{\mathrm{PQ}}\right\}$

$\therefore \Delta \mathrm{CMB} \sim \Delta \mathrm{RNQ}$ (SAS axiom)

Question 19

In the given figure, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. Prove that

(i) EF = FC

(ii) AG : GD = 2 : 1

Sol :

In the given figure,

AD and BE are the medians of ∆ABC

intersecting each other at G

DF || BE is drawn

To prove :

Proof :

(i) D is the mid-point of BC and DF||BE

∴F is the mid-point of CE

$\Rightarrow \mathrm{EF}=\mathrm{FC}=\frac{1}{2} \mathrm{EC}$

$\Rightarrow \mathrm{EF}=\frac{1}{2} \mathrm{AE}$

Now in $\Delta \mathrm{AGE}$ and $\Delta \mathrm{ADF}$

$\because \mathrm{GE}$ or $\mathrm{BG} \| \mathrm{DF}$

$\therefore \Delta \mathrm{AGE} \sim \Delta \mathrm{ADF}$

$\Rightarrow \frac{A G}{G D}=\frac{A E}{E F}=\frac{1}{\frac{1}{2}}=1 \times \frac{2}{1}=\frac{2}{1}$

$\therefore A G: G D=2: 1$

Question 20

(a) In the figure given below, AB, EF and CD are parallel lines. Given that AB =15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate

(i) EF

(ii) AC.

(b) In the figure given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm, BE = x and AE = y. Find the values of x and y.

Sol :

In the given figure,

AB || EF || CD

AB = 15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm

Calculate :

(i) EF and (ii) AC

Proof: In $\Delta \mathrm{EFG}$ and $\Delta \mathrm{CGD}$

$\angle \mathrm{EGF}=\angle \mathrm{CGD}$

(Vertically opposite angles)

$\angle \mathrm{FEG}=\angle \mathrm{GCD}$ (alternate angles)

$\therefore \Delta \mathrm{EFG} \sim \Delta \mathrm{CGD}$ (AA axiom)

$\therefore \frac{E G}{G C}=\frac{E F}{C D}$

$\Rightarrow \frac{5}{10}=\frac{\mathrm{EF}}{18}$

$\Rightarrow \mathrm{EF}=\frac{5 \times 18}{10}=9 \mathrm{~cm}$

$\therefore \mathrm{EF}=9 \mathrm{~cm}$

(ii) In $\Delta \mathrm{ABC}$ and $\Delta \mathrm{EFC}$

EF||AB

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{EFC}$

$\therefore \frac{\mathrm{AC}}{\mathrm{EC}}=\frac{\mathrm{AB}}{\mathrm{EF}}$

$\Rightarrow \frac{\mathrm{AC}}{5+10}=\frac{15}{9}$

$\Rightarrow \frac{\mathrm{AC}}{15}=\frac{15}{9}$

$\Rightarrow \mathrm{AC}=\frac{15 \times 15}{9}=25 \mathrm{~cm}$

$\therefore \mathrm{AC}=25 \mathrm{~cm}$

(b) In the given figure, AF||BE||CD

AF=7.5 cm, CD=4.5 cm, ED=3 cm

BE=x and AE=y

To find the value of x and y

In ΔAEF and ΔCED

$\angle \mathrm{AEF}=\angle \mathrm{CED}$ (vertically opposite angles) 

$\angle \mathrm{F}=\angle \mathrm{C}$ (alternate angles) 

$\therefore \Delta \mathrm{AEF} \sim \Delta \mathrm{CED}$ (AA axiom)

$\therefore \frac{\mathrm{AF}}{\mathrm{CD}}=\frac{\mathrm{AE}}{\mathrm{ED}} \Rightarrow \frac{7.5}{4.5}=\frac{y}{3}$

$\Rightarrow y=\frac{7.5 \times 3}{4.5}=5.0 \mathrm{~cm}$

Similarly in $\triangle \mathrm{ACD}, \mathrm{BE} \| \mathrm{CD}$

$\therefore \Delta \mathrm{ABE} \sim \Delta \mathrm{ACD}$

$\frac{\mathrm{EB}}{\mathrm{CD}}=\frac{\mathrm{AE}}{\mathrm{AD}}$

$\therefore \frac{x}{\mathrm{CD}}=\frac{y}{y+3}$

$\Rightarrow \frac{x}{4.5}=\frac{5}{5+3}=\frac{5}{8}$

$x=\frac{4.5 \times 5}{8}=\frac{22.5}{8}=\frac{225}{10 \times 8}$

$=\frac{45}{16}=2 \frac{13}{16} \mathrm{~cm}$

Question 21

In the given figure, ∠A = 90° and AD ⊥ BC If BD = 2 cm and CD = 8 cm, find AD.

Sol :

In ∆ABC, we have ∠A = 90°

Also, AD ⊥ BC

Now,

In ∆ABC, we have,

∠BAC = 90°

⇒ ∠BAD + ∠DAC = 90°…..(i)

In $\Delta \mathrm{ADC},$ we have,

$\angle \mathrm{ADC}=90^{\circ}$

So, $\angle \mathrm{DCA}+\angle \mathrm{DAC}=90^{\circ}$...(ii)

From (i) and (ii), we have

$\angle \mathrm{BAD}+\angle \mathrm{DAC}=\angle \mathrm{DCA}+\angle \mathrm{DAC}$

$\Rightarrow \angle \mathrm{BAD}=\angle \mathrm{DCA}$ ...(iii)

In $\triangle B D A$ and $\Delta A D C,$

$\angle B D A=\angle A D C$ (90° each)

$\angle B A D=\angle D C A$ (proved above)

So, $\Delta B D A \sim \Delta A D C$ (AA similarity)

$\Rightarrow \frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{AD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$

(Corresponding sides of similar Δ's are proportional)

$\Rightarrow \frac{B D}{A D}=\frac{A D}{D C}$

$\Rightarrow A D^{2}=B D \times C D$

$\Rightarrow A D^{2}=2 \times 8=16$

$\Rightarrow A D=4 \mathrm{~cm}$

Question 22

A 15 metres high tower casts a shadow of 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

Sol :

Height of a tower AB = 15 m

and its shadow BC = 24 m

At the same time and position

Let the height of a telephone pole DE = x m

and its shadow EF = 16 m

$\because$ The time is same

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{DEF}$

$\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{CD}}{\mathrm{EF}}$

$\Rightarrow \frac{15}{x}=\frac{24}{16}$

$\Rightarrow x=\frac{15 \times 16}{24}=10$

$\therefore$ Height of pole $=10 \mathrm{~m}$

Question 23

A street light bulb is fixed on a pole 6 m above the level of street. If a woman of height casts a shadow of 3 m, find how far she is away from the base of the pole?

Sol :

Height of height pole(AB) = 6m

and height of a woman (DE) = 1.5 m

Here shadow EF = 3 m

$\because$ Pole and woman are standing in the same line 

$\therefore A D \| D F$

$\therefore \Delta \mathrm{AFB} \sim \Delta \mathrm{DFE}$

$\therefore \frac{\mathrm{FB}}{\mathrm{FF}}=\frac{\mathrm{AB}}{\mathrm{DF}}$

$\Rightarrow \frac{3+x}{3}=\frac{6}{15}=\frac{4}{1}$

$\Rightarrow 3+x=12$

$\Rightarrow x=12-3=9 \mathrm{~m}$

$\therefore$ Woman is 9 m away from the pole.