ML Aggarwal Solution Class 10 Chapter 13 Similarity Exercise 13.3

 Exercise 13.3

Question 1

Given that ∆s ABC and PQR are similar.

Find:

(i) The ratio of the area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3.

(ii) the ratio of their corresponding sides if area of ∆ABC : area of ∆PQR = 25 : 36.

Sol :

(i) ∴ ∆ABC ~ ∆PQR

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$

(By theorem 15.1)

But BC = QR =1 : 3

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{(1)^{2}}{(3)^{2}}=\frac{1}{9}$

Hence area of $\Delta \mathrm{ABC}:$ area of $\Delta \mathrm{PQR}$

=1: 9

(ii) $\because \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$

(By theorem 15.1)

But area of ΔABC=area of ΔPQR

=25 : 36

$\therefore \frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}=\frac{25}{36} \Rightarrow\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^{2}=\left(\frac{5}{6}\right)^{2}$

$\Rightarrow  \frac{B C}{Q R}=\frac{5}{6}$

$\Rightarrow \mathrm{BC}: \mathrm{QR}=5: 6$

Question 2

∆ABC ~ DEF. If area of ∆ABC = 9 sq. cm., area of ∆DEF =16 sq. cm and BC = 2.1 cm., find the length of EF.

Sol :

Let EF = x

Given that

∆ABC ~ ∆DEF,

$\therefore \frac{\text { area } \Delta \mathrm{ABC}}{\text { area } \Delta \mathrm{DEF}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$

$\Rightarrow \frac{9}{16}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$

$\Rightarrow \frac{(2 \cdot 1)^{2}}{x^{2}}=\frac{9}{16}$

$\Rightarrow \frac{2 \cdot 1}{x}=\frac{3}{4}$

(Taking square root)

$\Rightarrow 3 x=4 \times 2 \cdot 1$

$\Rightarrow x=\frac{4 \times 2 \cdot 1}{3}$

∴x=2.8

Hence EF=2.8 cm

Question 3

∆ABC ~ ∆DEF. If BC = 3 cm, EF = 4 cm and area of ∆ABC = 54 sq. cm. Determine the area of ∆DEF.

Sol :

∆ABC ~ ∆DEF

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DEF}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$

$\Rightarrow \frac{54}{\text { area of } \Delta \mathrm{DEF}}=\frac{(3)^{2}}{(4)^{2}}$

$\Rightarrow \frac{54}{\text { area of } \Delta \mathrm{DEF}}=\frac{9}{16}$

$\Rightarrow$ area of $\Delta \mathrm{DEF}=\frac{54 \times 16}{9}=6 \times 16$

=96 cm

Question 4

The area of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.

Sol :

Let ABC ~ ∆DEF, AL and DM are their altitudes

then area of ∆ABC = 36 cm²

area of ∆DEF = 25 cm² and AL = 2.4 cm.

Let DM = x

Now ∆ABC ~ ∆DEF

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AL}^{2}}{\mathrm{DM}^{2}}$

$\Rightarrow \quad \frac{36}{25}=\frac{(2 \cdot 4)^{2}}{x^{2}}$

$\Rightarrow \quad 36 x^{2}=(2 \cdot 4)^{2} \times 25$

$\Rightarrow \quad x^{2}=\frac{(2 \cdot 4)^{2} \times 25}{36}=\frac{576 \times 25}{100 \times 36}=\frac{16}{4}$

$=4=(2)^{2}$

$\therefore x=2 \mathrm{~cm}$

Hence altitude of the other triangle

=2 cm

Question 5

(a) In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm², find the area of ∆QOA. (2006)

(b) In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm.

(i) Prove that ∆OAB ~ ∆OCD.

(ii) Find CD and OB.

(iii) Find the ratio of areas of ∆OAB and ∆OCD.

Sol :

In ∆AOQ and ∆BOP, we have

∠ OAQ = ∠ OBP [Each = 90°]

∠ AOQ= ∠BOP

[Vertically opposite angles]

∆AOQ ~ ∆BOP [A.A. similarity]

$\Rightarrow \frac{\text { Area of } \Delta \mathrm{AOQ}}{\text { Area of } \Delta \mathrm{BOP}}=\frac{\mathrm{OQ}^{2}}{\mathrm{PO}^{2}}$

$[\because$ Area's of similar triangles are proportional to the squares of their corresponding sides]

$\frac{\text { Area of } \Delta \mathrm{AOQ}}{120}=\frac{(9)^{2}}{(6)^{2}}$

$\Rightarrow \frac{\text { Area of } \Delta \mathrm{AOQ}}{120}=\frac{81}{36}$

$\Rightarrow$ Area of $\Delta \mathrm{AOQ}=\frac{81 \times 120}{36}=9 \times 30 \mathrm{~cm}^{2}$

$=270 \mathrm{~cm}^{2}$

(b) Given : In the figure AB || CD

AO=10 cm, OC=5 cm

 AB=6.5 cm and OD=2.8 cm

To prove : (i) ΔOAB∼ΔOCD

(ii) Find CD and OB

(iii) Find the ratio of areas of ΔOAB and ΔOCD

Proof : In the ΔOAB and ΔOCD

(i) ∴∠AOB=∠COD

(vertically opposite angles)

$\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (Alternate angles)

$\angle \mathrm{OBA}=\angle \mathrm{ODC}$ (Alternate angles)

∴ΔOAB〜ΔOCD (AAA axiom)

(ii) ∴$\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}=\frac{\mathrm{AB}}{\mathrm{CD}}$

$\Rightarrow \frac{10}{5}=\frac{\mathrm{OB}}{2.8}=\frac{6.5}{\mathrm{CD}}$

(a) $\frac{\mathrm{OB}}{2.8}=\frac{10}{5}$

$\Rightarrow \mathrm{OB}=\frac{10}{5} \times 2.8=5.6 \mathrm{~cm}$

$\therefore \mathrm{OB}=5.6 \mathrm{~cm}$

(b) $\frac{6.5}{\mathrm{CD}}=\frac{10}{5}$

$\Rightarrow C D=\frac{6.5 \times 5}{10}$

$\mathrm{CD}=\frac{32.5}{10}=3.25 \mathrm{~cm}$

(iii) $\because \Delta \mathrm{OAB} \sim \Delta \mathrm{OCD}$ (Proved)

$\therefore \frac{\operatorname{ar}(\Delta \mathrm{OAB})}{\operatorname{ar}(\Delta \mathrm{OCD})}$

$=\frac{A B^{2}}{C D^{2}}=\frac{(6.5)^{2}}{(3.25)^{2}}$

$=\frac{6.5 \times 6.5}{3.25 \times 3.25}$

$=\frac{2 \times 2}{1}=\frac{4}{1}$

$\operatorname{ar}(\Delta \mathrm{OAB}): \operatorname{ar}(\Delta \mathrm{OCD})=4: 1$

Question 6

(a) In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm, find the area of ∆ABC.

(b) In the figure (iii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of ∆ADE and trapezium DBCE.

Sol :

(a) In the figure,

DE || BC

∠D = ∠B and ∠E = ∠C

(Corresponding angles)

Now in $\Delta \mathrm{ADE}$ and $\Delta \mathrm{ABC}$

$\angle \mathrm{D}=\angle \mathrm{B}, \angle \mathrm{E}=\angle \mathrm{C}$ (proved)

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

$\therefore \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$ (AAA postulate)

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}$

$=\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}$

$\Rightarrow \frac{28}{\text { area of } \Delta \mathrm{ABC}}$

$=\frac{(6)^{2}}{(9)^{2}}=\frac{36}{81}$

$\Rightarrow$ area of $\Delta \mathrm{ABC}=\frac{28 \times 81}{36}=63$

$\therefore$ Area of $\Delta \mathrm{ABC}=63 \mathrm{~cm}^{2}$

(b) In the figure , DE || BC

$\therefore \angle \mathrm{D}=\angle \mathrm{B}$ and $\angle \mathrm{E}=\angle \mathrm{C}$

(corresponding angles)

Now in ΔADE and ΔABC

∠D=∠B, ∠E=∠C (proved)

∴∠A=∠A (common)

∴ΔADE〜ΔABC (AAA postulate)

But $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{1}{2}$

$\Rightarrow \frac{D B}{A D}=\frac{2}{1}$

Adding 1 both sides

$\frac{\mathrm{DB}}{\mathrm{AD}}+1=\frac{2}{1}+1$

$\Rightarrow \frac{\mathrm{AD}+\mathrm{DB}}{\mathrm{AD}}=\frac{2+1}{1}$

$\Rightarrow \frac{A B}{A D}=\frac{3}{1}$

$\Rightarrow \frac{A D}{A B}=\frac{1}{3}$

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

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{\mathrm{AD}^{2}}{\mathrm{AB}^{2}}=\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$

$\Rightarrow$ area of $\Delta \mathrm{ABC}=9$ area of $\Delta \mathrm{ADE}$

$\Rightarrow$ area of trapezium DBCE

$\Rightarrow$ area of $\Delta \mathrm{ABC}-$ area of $\Delta \mathrm{ADE}$

$=9$ area of $\Delta \mathrm{ADE}-$ area of $\Delta \mathrm{ADE}$

$=8$ area of $\Delta \mathrm{ADE}$

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of trapezium DBCE }}=\frac{1}{8}$

$\therefore$ area of $\Delta \mathrm{ADE}$ : area of trepezium DBCE

=1 : 8

Question 7

In the given figure, DE || BC.

(i) Prove that ∆ADE and ∆ABC are similar.

(ii) Given that $A D=\frac{1}{2} B D$, calculate DE if BC=4.5 cm

(iii) If area of ∆ABC = 18cm², find the area of trapezium DBCE

Sol :

(i) Given : In ∆ABC, DE || BC.

To prove : ∆ADE ~ ∆ABC

Proof: In ∆ADE and ∆ABC,

∠A = ∠A (common)

∠ADE = ∠ABC (corresponding angles)

∴ ∆ADE ~ ∆ABC. (AA axiom)

(ii) ∴ ∆ADE ~ ∆ABC

$\therefore \quad \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}$

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

$\Rightarrow \frac{\frac{1}{2} B D}{\frac{1}{2} B D+B D}=\frac{D E}{4.5}$

(∵BC=4.5)

$\Rightarrow  \frac{\frac{1}{2} B D}{\frac{3}{2} B D}=\frac{D E}{4 \cdot 5}$

$\Rightarrow  \frac{1}{2} \times \frac{2}{3}=\frac{\mathrm{DE}}{4 \cdot 5}$

$\Rightarrow \frac{1}{3}=\frac{\mathrm{DE}}{4 \cdot 5}$

$\therefore \mathrm{DE}=\frac{4 \cdot 5}{3}=1 \cdot 5 \mathrm{~cm}$ 

(iii) Area of $\Delta \mathrm{ABC}=18 \mathrm{~cm}^{2}$

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{\mathrm{DE}^{2}}{\mathrm{BC}^{2}}$

(Area of similar triangles are proportional to the square of their corresponding sides)

$\frac{\text { area of } \Delta \mathrm{ADE}}{18}=\left(\frac{\mathrm{DE}}{\mathrm{BC}}\right)^{2}$

$\Rightarrow \quad \frac{\text { area of } \Delta \mathrm{ADE}}{18}=\left(\frac{\mathrm{AD}}{\mathrm{AB}}\right)^{2}$

$\Rightarrow \quad \frac{\text { area of } \Delta \mathrm{ADE}}{18}=\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$

[proved in (ii)]

$\Rightarrow$ area of $\Delta \mathrm{ADE}=18 \times \frac{1}{9}=2$

$\therefore$ area of tripezium DBCE

=area of ΔABC-area of ΔADE

=18-2=16$\mathrm{cm}^{2}$

Question 8

In the given figure, AB and DE are perpendicular to BC.

(i) Prove that ∆ABC ~ ∆DEC

(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm, calculate CD.

(iii) Find the ratio of the area of ∆ABC : area of ∆DEC

Sol :

(i) To prove : ∆ABC ~ ∆DEC

In ∆ABC and ∆DEC

$\angle \mathrm{ABC}=\angle \mathrm{DEC}=90^{\circ}$ 

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

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

(ii) $\frac{A C}{C D}=\frac{A B}{D E}$

(Corresponding sides of similar triangles are proportional)

$\frac{15}{\mathrm{CD}}=\frac{6}{4}$

$\therefore \mathrm{CD}=\frac{15 \times 4}{6}$

(iii) $\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DEC}}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}$

$=\frac{6^{2}}{4^{2}}=\frac{36}{16}=\frac{9}{4}=9: 4$

Question 9

In the adjoining figure, ABC is a triangle.

$\mathrm{DE}$ is parallel to $\mathrm{BC}$ and $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{2}$

(i) Determine the ratios $\frac{\mathrm{AD}}{\mathrm{AB}}, \frac{\mathrm{DE}}{\mathrm{BC}}$

(ii) Prove that ∆DEF is similar to ∆CBF.

Hence, find $\frac{E F}{F B}$

(iii) What is the ratio of the areas of ∆DEF and ∆CBF ? (2007)

Sol :

(i) $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{2}$ (given)

$\therefore \frac{\mathrm{DB}}{\mathrm{AD}}=\frac{2}{3}$

or $\frac{\mathrm{DB}}{\mathrm{AD}}+1=\frac{2}{3}+1$

or $\frac{\mathrm{DB}+\mathrm{AD}}{\mathrm{AD}}=\frac{2+3}{3}$

or $\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{5}{3}$

$\Rightarrow \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{3}{5}$

In ΔADE and ΔABC

∠ADE=∠B (Corresponding ∠s)

∠AED=∠C (Corresponding ∠s)

∴By AA similarity

ΔADE〜ΔABC

$\therefore \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{DE}}{\mathrm{BC}}$

$\Rightarrow \frac{\mathrm{DE}}{\mathrm{BC}}=\frac{3}{5}$

(ii) In ΔDEF and ΔCBF

∠1=∠2 (Alternate ∠s)

∠3=∠4 (Alternate ∠s)

∠5=∠6 (Vertically oppo. ∠s)

∴ΔDEF〜ΔCBF

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

(iii) As the ratio of the area of two similar triangles is equal to the ratio of the square of any two corresponding sides.

$\therefore \frac{\text { Area of } \Delta \mathrm{DFE}}{\text { Area of } \Delta \mathrm{BFC}}=\frac{\mathrm{DE}^{2}}{\mathrm{BC}^{2}}$

$=\left(\frac{\mathrm{DE}}{\mathrm{BC}}\right)^{2}=\left(\frac{3}{5}\right)^{2}=\frac{9}{25}$

Question 10

In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:

(i) area ∆APO : area ∆ABC.

(ii) area ∆APO : area ∆CQO. (2008)

Sol :

In the figure,

PQ || BC and PO is produced to Q such that CQ || BA

and AP : PB = 2 : 3.

Figure to be added

(i) Now in ΔAPO and ΔABC

∠A=∠A (Common)

∠APO=∠ABC (Corresponding angles)

∴ΔAPO~ΔABC (AA axiom)

∵Areas of the similar triangle are proportional to the square of their corresponding sides

$\therefore \frac{\operatorname{area}(\Delta \mathrm{APO})}{\operatorname{area}(\Delta \mathrm{ABC})}=\frac{\mathrm{AP}^{2}}{\mathrm{AB}^{2}}=\frac{\mathrm{AP}^{2}}{(\mathrm{AP}+\mathrm{PB})^{2}}$

$=\frac{(2)^{2}}{(2+3)^{2}}=\frac{4}{(5)^{2}}=\frac{4}{25}$

area $(\triangle \mathrm{APO}):$ area $(\Delta \mathrm{ABC})=4: 25$

(ii) In $\Delta \mathrm{APO}$ and $\Delta \mathrm{CQO}$

$\angle \mathrm{AOP}=\angle \mathrm{COQ}$

(vertically opposite angles)

$\angle \mathrm{APQ}=\angle \mathrm{OQC}$ (Alternate angles)

∴ΔAPQ～ΔCQO

$\therefore \frac{\operatorname{area}(\Delta \mathrm{APO})}{\operatorname{area}(\Delta C Q O)}=\frac{A P^{2}}{C Q^{2}}=\frac{A P^{2}}{P B^{2}}$

$(\because P B=C Q)$

$=\frac{(2)^{2}}{(3)^{2}}=\frac{4}{9}$

area (ΔAPO) : area (ΔCQO)

=4 : 9

Question 11

(a) In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of ∆AOB and ∆COD.

(b) In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find

(i) AB

(ii) BC

(iii) area of ∆ADM : area of ∆ANB.

(c) In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find

(i) EF : AD

(ii) area of ∆BEF : area of ∆ABD

(iii) area of ∆ABD : area of trap. AFED

(iv) area of ∆FEO : area of ∆OBC.

Sol :

(a) In trapezium ABCD, AB || DC.

∠OAB = ∠OCD [alternate angles]

∠OBA = ∠ODC

∆AOB ~ ∆COD

$\therefore \frac{\text { area of } \Delta \mathrm{AOB}}{\text { area of } \Delta \mathrm{COD}}=\frac{\mathrm{AB}^{2}}{\mathrm{CD}^{2}}$

$=\frac{(2 \mathrm{CD})^{2}}{\mathrm{CD}^{2}}$ $(\because A B=2 C D)$

$=\frac{4 \mathrm{CD}^{2}}{\mathrm{CD}^{2}}=\frac{4}{1}$

∴area of ΔAOB : area of ΔCOD= 4 : 1

(b) In ||gm ABCD, AM⊥DC and AN⊥CB 

Now area of ||gm ABCD= DC×AM or

BC×AN

∴DC×AM=BC×AN=area of ||gm

⇒DC×6=BC×10=45

(i) ∴$\mathrm{DC}=\frac{45}{6}=\frac{15}{2}=7 \cdot 5 \mathrm{~cm}$

∴AB=7.5 cm (∵AB=DC)

(ii) Now in ΔADM and ΔABN

∠D=∠B (opposite angles of a ||gm)

∠M=∠N (each 90°)

$\therefore \Delta \mathrm{ADM} \sim \Delta \mathrm{ABN}$

$\therefore \frac{\text { area of } \Delta \mathrm{ADM}}{\text { area of } \Delta \mathrm{ABN}}=\frac{\mathrm{AD}^{2}}{\mathrm{AB}^{2}}$

$=\frac{B C^{2}}{A B^{2}}=\frac{(4 \cdot 5)^{2}}{(7 \cdot 5)^{2}}$

$=\frac{20 \cdot 25}{56 \cdot 25}=\frac{2025}{5625}$

$=\frac{81}{225}=\frac{9}{25}$

$\therefore$ area of $\Delta$ ADM : area of $\Delta$ ABN

= 9 : 25

(c) In ||gm ABCD, E is a point on AB , CE intersects the diagonal BD at O

EF || BC and AE : EB=2 : 3

In $\Delta \mathrm{ABD}, \mathrm{EF} \| \mathrm{BC}$ or $\mathrm{AD}$

(i) $\therefore \frac{\mathrm{AB}}{\mathrm{BE}}=\frac{\mathrm{AD}}{\mathrm{EF}}$

$\Rightarrow \frac{\mathrm{EF}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{AB}}$

But $\frac{A E}{E B}=\frac{2}{3}$

$\Rightarrow \frac{A E}{E B}+1=\frac{2}{3}+1$

$\Rightarrow \frac{A E+E B}{E B}=\frac{2+3}{3}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{EB}}=\frac{5}{3}$

$\Rightarrow \frac{\mathrm{BE}}{\mathrm{AB}}=\frac{3}{5}$

⇒EF : AD= 3 : 5

(ii) ∵ΔBEF~ΔABD

$\frac{\text { area of } \Delta \mathrm{BEF}}{\text { area of } \Delta \mathrm{ABD}}=\frac{(\mathrm{EF})^{2}}{(\mathrm{AD})^{2}}$

$=\frac{(3)^{2}}{(5)^{2}}=\frac{9}{25}$

$\therefore$ area of $\Delta \mathrm{BEF}:$ area of $\Delta \mathrm{ABD}=9: 25$

(iii) $\frac{\text { area of } \Delta \mathrm{ABD}}{\text { area of } \Delta \mathrm{BEF}}=\frac{25}{9}$ (from (ii))

25 area of $\Delta \mathrm{BEF}=9$ area of $\Delta \mathrm{ABD}$

$\Rightarrow 25$ (area of $\Delta \mathrm{ABD}$ - area of trap AEFD)

$=9$ area of $\Delta \mathrm{ABD}$

$\Rightarrow 25$ area of $\Delta \mathrm{ABD}-25$ area of AFED

$=9$ area of $\Delta \mathrm{ABD}$

$\Rightarrow 25$ area of trap AEFD $=25$ area of $\Delta \mathrm{ABD}-9$ area of $\Delta \mathrm{ABD}$

$\Rightarrow 25$ area of $A E F D=16$ area of $\Delta A B D$

$\Rightarrow \frac{\text { area of } \Delta \mathrm{ABD}}{\text { area of trap } \mathrm{AEFD}}=\frac{25}{16}$

$\Rightarrow$ area of $\Delta \mathrm{ABD}:$ area of trap $\mathrm{AEFD}$

= 25 : 16

(iv) In $\Delta$ FEO and $\Delta$ OBC

$\angle \mathrm{EOF}=\angle \mathrm{BOC}$

(Vertically opposite angles)

$\angle \mathrm{F}=\angle \mathrm{OBC}$ (Alternate angles)

$\therefore \Delta \mathrm{FEO} \sim \Delta \mathrm{OBC}$

$\therefore \frac{\text { area of FEO }}{\text { area of } \Delta \mathrm{OBC}}=\frac{\mathrm{EF}^{2}}{\mathrm{BC}^{2}}$

$=\frac{E F^{2}}{A D^{2}}=\frac{9}{25}$ [From (i)]

$\therefore$ area of $\Delta \mathrm{FEO}:$ area of $\Delta \mathrm{OBC}=9: 25$

Question 12

In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q. If area of ∆CPQ = 20 cm², find

(i) area of ∆BPQ.

(ii) area ∆CDP.

(iii) area of || gm ABCD.

Sol :

In the figure, ABCD is a parallelogram.

P is a point on BC such that BP : PC = 1 : 2

and DP is produced to meet ABC produced at Q.

Area ∆CPQ = 20 cm²

Draw QN $\perp$ CB (Produced)

$\frac{\operatorname{area} \Delta \mathrm{BPQ}}{\text { area } \Delta \mathrm{CPQ}}=\frac{\frac{1}{2} \mathrm{BP} \times \mathrm{QN}}{\frac{1}{2} \mathrm{PC} \times \mathrm{QN}}$

$=\frac{\mathrm{BP}}{\mathrm{PC}}=\frac{1}{2}$ (given)

(i) ∴Area $\Delta \mathrm{BPQ}=\frac{1}{2}$ area $\Delta \mathrm{CPQ}=\frac{1}{2} \times 20$ $\mathrm{cm}^{2}=10 \mathrm{~cm}^{2}$

(ii) Now in $\Delta$ CDP and $\Delta$ BQP,

$\angle \mathrm{CPD}=\angle \mathrm{QPB}$ (Vertically opposite angles)

$\angle \mathrm{PDC}=\angle \mathrm{POB}$ (Alternate angles)

$\therefore \Delta \mathrm{CDP} \sim \Delta \mathrm{BQP}$ (AA axiom)

$\therefore \frac{\text { area } \Delta \mathrm{CDP}}{\text { area } \Delta \mathrm{BQP}}=\frac{\mathrm{PC}^{2}}{\mathrm{BP}^{2}}$

⇒$\frac{\operatorname{areaCDP}}{\operatorname{areaBQP}}=\frac{(2)^{2}}{(1)^{2}}$

$=\frac{4}{1}$

$\therefore$ area $\Delta \mathrm{CDP}=4 \times$ area $\Delta \mathrm{BQP}$

$=4 \times 10=40 \mathrm{~cm}^{2}$

(iii) Now area of ||gm ABCD=2 Area ΔDCQ

{ΔDCQ and ||gm ABCD are one the same base and between the same parallels}

=2 area (ΔDCP+ΔCPQ)

=2(40+20)$\mathrm{cm}^{2}$

$=2 \times 60 \mathrm{~cm}^{2}=120 \mathrm{~cm}^{2}$

Question 13

(a) In the figure (i) given below, DE || BC and the ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.

(b) In the figure (ii) given below, AB || DC and AB = 2 DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find (i) ED (ii) BE (iii) area of ∆EDC : area of trapezium ABCD.

Sol :

(a) In ∆ABC, DE || BC

Now in ∆ABC and ∆ADE

∠A = ∠A (common)

∠D = ∠B and ∠E = ∠C

(Corresponding angles)

∆ADE ~ ∆ABC

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}$...(i)

But $\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of trap } \mathrm{DBCE}}=\frac{4}{5}$

$\Rightarrow \frac{\text { area of trap } \mathrm{DBCE}}{\text { area of } \Delta \mathrm{ADE}}=\frac{5}{4}$

$\Rightarrow \frac{\text { area of trap } \mathrm{DBCE}}{\text { area of } \Delta \mathrm{ADE}}+1$

$=\frac{5}{4}+1$

(Adding 1 both sides)

$\Rightarrow \frac{\text { area of trap } \mathrm{DBCE}+\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ADE}}$

$=\frac{5+4}{4}=\frac{9}{4}$

$\Rightarrow \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{ADE}}=\frac{9}{4}$

$\Rightarrow \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{4}{9}$

Now from (i) $\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}=\frac{4}{9}$

$\Rightarrow \frac{(2)^{2}}{(3)^{2}}$

$\Rightarrow \frac{\mathrm{DE}}{\mathrm{BC}}=\frac{2}{3}$

$\Rightarrow \mathrm{DE}: \mathrm{BC}=2: 3$

(b) In the figure , DC||AB, AB=2 DC

AD=3 cm, BC=4 cm

In ΔEAB, DC||AB

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

$=\frac{A B}{D C}=\frac{2 D C}{D C}=\frac{2}{1}$

(i) ∴EA=2, DA=2×3=6

ED=EA-DA=6-3=3cm

(ii) $\frac{E B}{C B}=\frac{2}{1}$

$\Rightarrow E B=2 C B=2 \times 4=8 \mathrm{~cm}$

$\therefore B E=8 \mathrm{~cm}$

(iii) $\therefore \operatorname{In} \Delta \mathrm{EAB}, \mathrm{DC} \| \mathrm{AB}$

$\therefore \Delta \mathrm{EDC} \sim \Delta \mathrm{EAB}$

$\therefore \frac{\text { area of } \Delta \mathrm{EDC}}{\text { area of } \Delta \mathrm{ABE}}=\frac{\mathrm{DC}^{2}}{\mathrm{AB}^{2}}$

$=\frac{\mathrm{DC}^{2}}{(2 \mathrm{DC})^{2}}$

$=\frac{D C^{2}}{4 D C^{2}}=\frac{1}{4}$

$=\frac{1}{4}$

$\therefore$ area of $\mathrm{ABE}=4$ area of $\Delta \mathrm{EDC}$

$\Rightarrow$ area of $\Delta \mathrm{EDC}+$ area of trap $\mathrm{ABCD}=4$ area of $\Delta$ EDC

area of $\operatorname{trap} A B C D=4$ area of $\Delta E D C-$ area of $\Delta$ EDC

$\Rightarrow$ area of $\operatorname{trap} \mathrm{ABCD}=3$ area of $\Delta \mathrm{EDC}$

$\Rightarrow \frac{\text { area of } \Delta \mathrm{EDC}}{\text { area of trap } \mathrm{ABCD}}=\frac{1}{3}$

$\Rightarrow$ area of $\Delta E D C:$ area of trap. $A B C D=1: 3$

Question 14

(a) In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BB = 12 cm., find

(i) BP

(ii) the ratio of areas of ∆APB and ∆DPC.

(b) In the figure given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm.

(i) Prove that ∆ACD is similar to ∆BCA

(ii) Find BC and CD

(iii) Find the area of ∆ACD : area of ∆ABC.

Sol :

(a) In trapezium ABCD, DC || AB

AB=9 cm, DC=6 cm, BD=12 cm

(i) In $\Delta \mathrm{APB}$ and $\Delta \mathrm{CPD}$

$\angle \mathrm{APB}=\angle \mathrm{CPD}$ 

(Vertically opposite angles)

$\angle \mathrm{PAB}=\angle \mathrm{PCD}$ (alternate angles)

$\therefore \Delta \mathrm{APB} \sim \Delta \mathrm{CPD}$ (alternate angles)

$\therefore \Delta \mathrm{APB} \sim \Delta \mathrm{CPD}$ (AA postulate)

$\therefore \frac{\mathrm{BP}}{\mathrm{PD}}=\frac{\mathrm{AB}}{\mathrm{CD}}$

$\Rightarrow \frac{\mathrm{BP}}{12-\mathrm{BP}}=\frac{9}{6}$

$\Rightarrow  6 \mathrm{BP}=108-9 \mathrm{BP}$

$\Rightarrow  6 \mathrm{BP}+9 \mathrm{BP}=108$

$\Rightarrow 15 \mathrm{BP}=108$

$\Rightarrow \mathrm{BP}=\frac{108}{15}=7 \cdot 2 \mathrm{~cm}$

(ii) Again 

∵ΔAPB～ΔCPD

$\therefore \frac{\text { area of } \Delta \mathrm{APB}}{\text { area of } \Delta \mathrm{CPD}}=\frac{\mathrm{AB}^{2}}{\mathrm{CD}^{2}}$

$=\frac{(9)^{2}}{(6)^{2}}=\frac{81}{36}=\frac{9}{4}$

(b) In ΔACD and ΔBCA

∠C=∠C (common)

∠ABC=∠CAD (Given)

∴ΔACD≅ΔBCA (by AA axiom)

$\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{CD}}{\mathrm{CA}}=\frac{\mathrm{AD}}{\mathrm{AB}}$

$\Rightarrow \frac{4}{\mathrm{BC}}=\frac{\mathrm{CD}}{4}=\frac{5}{8}$

$\therefore \frac{4}{\mathrm{BC}}=\frac{5}{8}$

$\mathrm{BC}=\frac{4 \times 8}{5}=\frac{32}{5}=6.4 \mathrm{~cm}$

$\therefore \frac{C D}{4}=\frac{5}{8}$

$\Rightarrow \mathrm{CD}=\frac{5}{8} \times 4=\frac{5}{2}=2.5 \mathrm{~cm}$

(iii) $\because \Delta \mathrm{ACD} \cong \Delta \mathrm{BCA}$

$\therefore \frac{\operatorname{area}(\Delta \mathrm{ACB})}{\operatorname{area}(\Delta \mathrm{BCA})}=\frac{\mathrm{AC}^{2}}{\mathrm{AB}^{2}}$

$=\frac{(4)^{2}}{(8)^{2}}=\frac{16}{64}=\frac{1}{4}$

area $(\Delta \mathrm{ACD}):$ area $(\Delta \mathrm{ABC})=1: 4$

Question 15

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:

(i) ∆ADE ~ ∆ACB.

(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.

(iii) Find, area of ∆ADE : area of quadrilateral BCED.

Sol :

(i) Consider DADE and DACB

∠A = ∠A (Common)

m∠B = m∠E = 90°

Thus by angle-angle similarity, triangles,

∆ACB ~ ∆ADE

(ii) Consider ∆ADE and ∆ACB

Since they are similar triangles,

the sides are proportional Thus, we have,

$\frac{\mathrm{AE}}{\mathrm{AB}}=\frac{\mathrm{AD}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}$..(i)

Consider $\Delta \mathrm{ABC}$

By applying Pythagoras Theorem, we have, $\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}$

$\Rightarrow \mathrm{AB}^{2}+5^{2}=13^{2} \Rightarrow \mathrm{AB}^{2}+25=169$

$\Rightarrow \mathrm{AB}^{2}=169-25=144$

$\Rightarrow \mathrm{AB}=12 \mathrm{~cm}$

From equation (1), we have

$\frac{4}{12}=\frac{\mathrm{AD}}{13}=\frac{\mathrm{DE}}{5}$

$\Rightarrow \frac{1}{3}=\frac{\mathrm{AD}}{13}$

$\Rightarrow \mathrm{AD}=\frac{13}{3} \mathrm{~cm}=4.33 \mathrm{~cm}$

Also, $\frac{4}{12}=\frac{\mathrm{DE}}{5}$

$\Rightarrow \mathrm{DE}=\frac{20}{12}=\frac{5}{3} \mathrm{~cm}=1.67 \mathrm{~cm}$

(iii) We need to find the area of ΔADE and quadrilateral BCED

Area of $\Delta \mathrm{ADE}=\frac{1}{2} \times \mathrm{AE} \times \mathrm{DE}$

$=\frac{1}{2} \times 4 \times \frac{5}{3}$

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

Area of quadrilateral BCED=Area of ΔABC-Area of ΔADE

$=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}-\frac{10}{3}$

Area of quad. $\mathrm{BCED}=$ Area of $\Delta \mathrm{ABC}-$ Area of $\Delta \mathrm{ADE}$

$=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}-\frac{10}{3}$

$=\frac{1}{2} \times 5 \times 12-\frac{10}{3}$

$=30-\frac{10}{3}$

$=\frac{90-10}{3}=\frac{80}{3} \mathrm{~cm}^{2}$

Thus ratio of areas of $\Delta \mathrm{ADE}$ to quadrilateral

$\mathrm{BCED}=\frac{\frac{10}{3}}{\frac{80}{3}}=\frac{1}{8}$

Question 16

Two isosceles triangles have equal vertical angles and their areas are in the ratio 7 : 16. Find the ratio of their corresponding height.

Sol :

In two isosceles ∆s ABC and DEF

$\angle \mathrm{A}=\angle \mathrm{D}$ (given)

$\therefore \angle \mathrm{B}+\angle \mathrm{C}=\angle \mathrm{E}+\angle \mathrm{F}$

But $\angle \mathrm{B}=\angle \mathrm{C}$ and $\angle \mathrm{E}=\angle \mathrm{F}$

(opposite angles of equal sides)

$\therefore \angle B=\angle E$ and $\angle C=\angle F$

$\therefore \Delta A B C \sim \Delta D E F$

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AL}^{2}}{\mathrm{DM}^{2}}$

(Cor. of theorem 13.1)

$\Rightarrow \frac{\mathrm{AL}^{2}}{\mathrm{DM}^{2}}=\frac{7}{16}$

$\Rightarrow \frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\sqrt{7}}{4}$

Hence $\mathrm{AL}: \mathrm{DM}=\sqrt{7}: 4$

Question 17

On a map drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements :

AB = 3 cm, BC = 4 cm and ∠ABC = 90°. Calculate

(i) the actual length of AB in km.

(ii) the area of the plot in sq. km:

Sol :

Scale factor k = 1 : 250000 = $\frac{1}{250000}$

Length on map,

AB = 3 cm, BC = 4 cm

$\therefore$ Length of $A B$ of Actual plot $=\frac{1}{k}$ (Length of AB on the map)

=250000 (3 cm)

$=\frac{250000 \times 3}{100 \times 1000} \mathrm{~km}=\frac{15}{2}=7 \cdot 5 \mathrm{~km}$

(ii) Area of plot on the map $=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}$

$=\frac{1}{2} \times 3 \times 4=6 \mathrm{~cm}^{2}$

$\therefore$ Area of actual plot $=\frac{1}{k^{2}} \times 6 \mathrm{~cm}^{2}$

$=(250000)^{2} \times 6 \mathrm{~cm}^{2}$

$=\frac{250000 \times 250000 \times 6}{100000 \times 100000} \mathrm{~km}^{2}$

$=\frac{25}{4} \times 6=\frac{75}{2}=37 \cdot 5 \mathrm{~km}^{2}$

Question 18

On a map drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD has the following measurements AB = 12 cm and BG = 16 cm.

Calculate:

(i) the distance of a diagonal of the plot in km.

(ii) the area of the plot in sq. km.

Sol :

Scale factor $(\mathrm{k})=\frac{1}{25000}$

Measurements of plot ABCD on the map are

AB = 12 cm and BC = 16 cm.

Diagonal $A C=\sqrt{A B^{2}+B C^{2}}$

$=\sqrt{(12)^{2}+(16)^{2}}=\sqrt{144+256}$

$=\sqrt{400}=20 \mathrm{~cm}$

and area=AB×BC

$=12 \mathrm{~cm} \times 16 \mathrm{~cm}=192 \mathrm{~cm}^{2}$

(i) Now actual length of $\mathrm{AC}=\frac{1}{k}$

(Length of AC on map)

$=25000 \times 20 \mathrm{~cm}=\frac{25000 \times 20}{100 \times 1000} \mathrm{~km}$

=5 km

(ii) Area of plot $=\left(\frac{1}{k}\right)^{2}$ (Area of plot on map)

Question 19

The model of a building is constructed with the scale factor 1 : 30.

(i) If the height of the model is 80 cm, find the actual height of the building in metres.

(ii) If the actual volume of a tank at the top of the building is 27 m³, find the volume of the tank on the top of the model. (2009)

Sol :

(i) $\frac{\text { Height of model }}{\text { Height of actual building }}=\frac{1}{30}$

$\frac{80}{\mathrm{H}}=\frac{1}{30}$

$\Rightarrow \mathrm{H}=2400 \mathrm{~cm}=24 \mathrm{~m}$

(ii) $\frac{\text { Volume of model }}{\text { Volume of tank }}=\left(\frac{1}{30}\right)^{3}$

$\frac{\mathrm{V}}{27}=\frac{1}{27000}$

$\Rightarrow \mathrm{V}=\frac{1}{1000} \mathrm{~m}^{3}=1000 \mathrm{~cm}^{3}$

Question 20

A model of a ship is made to a scale of 1 : 200.

(i) If the length of the model is 4 m, find the length of the ship.

(ii) If the area of the deck of the ship is 160000 m², find the area of the deck of the model.

(iii) If the volume of the model is 200 litres, find the volume of the ship in m³.

(100 litres = m³)

Sol :

Scale = 1 : 200

(i) Length of a model of ship = 4 m

$\therefore$ Length of the ship $=\frac{4 \times 200}{1}=800 \mathrm{~m}$

(ii) Area of deck of ship $=160000 \mathrm{~m}^{2}$

$\therefore$ Area of deck of the model

$=160000 \times\left(\frac{1}{200}\right)^{2} \mathrm{~m}^{2}$

$=160000 \times \frac{1}{40000}=4 \mathrm{~m}^{2}$

(iii) Volume of the model of the ship =200 l

Volume of ship $=200 \times\left(\frac{200}{1}\right)^{3} l$

$=200 \times 8000000 l$

$=\frac{200 \times 8000000}{1000} \mathrm{~m}^{3}=1600000 \mathrm{~m}^{3}$