# RS Aggarwal Solutions Class 10 Ex 4A

Question 1: In each of the given pairs of triangles, find which pair of triangles are similar. State the similarity criterion and write the similarity relation in symbolic form.

(i)

Solution:

In $\Delta$ABC and $\Delta$PQR

$\angle$A = $\angle$Q = 50o

$\angle$B = $\angle$P = 60o

$\angle$C = $\angle$R = 70o

Therefore, $\Delta$ABC ~ $\Delta$QPR (by AAA similarity)

Question 2: In the given figure, $\Delta$ODC ~ $\Delta$OBA, $\angle$BOC = 115o and $\angle$CDO = 70o.

Find:

(i) $\angle$DOC

(ii) $\angle$DCO

(iii) $\angle$OAB

(iv) $\angle$OBA

Solution:

$\Delta$ODC ~ $\Delta$OBA

$\angle$BOC = 115o and $\angle$CDO = 70o

(i) $\angle$DOC = (180o$\angle$BOC) = (180o â€“ 115o) = 65o

(ii) $\angle$OCD = 180o$\angle$CDO – $\angle$DOC

Therefore, $\angle$OCD = 180o â€“ (70o + 65o) = 45o

(iii) Now, $\Delta$ODC ~ $\Delta$OBA

$\angle$AOB = $\angle$COD = 65o [Vertically opposite angles]

$\angle$OAB = $\angle$OCD = 45o

$\angle$OBA = $\angle$ODC = 70o [Alternate angles]

So, $\angle$OAB = 45o

(iv) $\angle$OBA = 70o

Question 3: In the given figure, $\Delta$OAB ~ $\Delta$OCD. If AB = 8 cm, BO = 6.4 cm, OC = 3.5 cm and CD = 5 cm, find (i) OA and (ii) DO.

Solution:

Given: $\Delta$OAB ~ $\Delta$OCD

AB = 8 cm, BO = 6.4 cm, OC = 3.5 cm and CD = 5 cm

$\Rightarrow$ $\frac{OA}{OC}=\frac{AB}{CD}=\frac{BO}{DO}$

$\Rightarrow$ $\frac{OA}{3.5}=\frac{8}{5}=\frac{6.4}{DO}$

$\Rightarrow$ $\frac{OA}{3.5}=\frac{8}{5}$ and $\frac{6.4}{DO}=\frac{8}{5}=$

OA = $\frac{3.5\times 8}{5}=5.6$

And DO = $\frac{6.4\times 5}{8}=4$

OA = 5.6 cm and DO = 4 cm

Question 4: In the given figure, if $\angle$ADE = $\angle$B, show that $\Delta$ADE ~ $\Delta$ABC. If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE.

Solution:

Given: $\angle$ADE = $\angle$B,

AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm

Proof:

In $\Delta$ADE ~ $\Delta$ABC,

$\angle$A = $\angle$A (Common)

$\angle$ADE = $\angle$B (Given)

Therefore, $\Delta$ADE ~ $\Delta$ABC (AA Criterion)

$\Rightarrow$ $\frac{AD}{AB}=\frac{DE}{BC}$

$\Rightarrow$ $\frac{3.8}{(3.6+2.1)}=\frac{DE}{4.2}$

$\Rightarrow$ $\frac{3.8}{5.7}=\frac{DE}{4.2}$

DE = $\frac{3.8\times 4.2}{5.7}=2.8$

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.

Solution:

Given: $\Delta$ABC ~ $\Delta$PQR in such a way that perimeter of respective $\Delta$ABC = 32 cm and $\Delta$PQR = 24 cm and PQ = 12 cm. Then, we have to find AB.

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

Therefore, $\frac{Perimeter of \Delta ABC}{Perimeter of \Delta PQR}=\frac{AB}{PQ}$

$\Rightarrow$ $\frac{32}{24}=\frac{AB}{12}$

$\Rightarrow$ AB = $\frac{32\times 12}{24}= 17$

Therefore, AB = 17 cm

Question 6: In the given figure, $\angle$CAB = 90o and AD $\perp$ BC. Show that $\Delta$BDA ~ $\Delta$BAC. If AC = 75 cm, AB = 1 cm and BC = 1.25 m, find AD.

Solution:

Given: AB = 100 cm, BC = 125 cm, AC = 75 cm

Proof:

In $\Delta$BAC and $\Delta$BDA

$\angle$BAC = $\angle$BDA = 90o

$\angle$B = $\angle$B (Common)

$\Delta$BAC ~ $\Delta$BDA (by AA similarities)

$\Rightarrow$ $\frac{BA}{BC}=\frac{AD}{AC}$

$\Rightarrow$ $\frac{100}{125}=\frac{AD}{75}$

$\Rightarrow$ AD = $\frac{100\times 75}{125}= 60$

Question 7: In the given figure, $\angle$ABC = 90o and BD $\perp$ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

Solution:

Given that AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm

In $\Delta$CBA and $\Delta$CDB

$\angle$CBA = $\angle$CDB = 90o

$\angle$C = $\angle$C (Common)

Therefore, $\Delta$CBA ~ $\Delta$CDB (by AA similarities)

$\Rightarrow$ $\frac{BC}{CD}=\frac{BA}{BD}$

$\Rightarrow$ $\frac{BC}{5.4}=\frac{5.7}{3.8}$

$\Rightarrow$ BC = $\frac{5.7\times 5.4}{3.8}= 8.1$

Hence, BC = 8.1 cm

Question 8: In the given figure, $\angle$ABC = 90o and BD $\perp$ AC. If BD = 8 cm, AD = 4 cm, find CD.

Solution:

Given that BD = 8 cm and AD = 4 cm

In $\Delta$DBA and $\Delta$DCB, we have

$\angle$BDA = $\angle$CDB = 90o

$\angle$DBA = $\angle$DCB (each = 90o â€“ $\angle$A)

Therefore, $\Delta$DBA ~ $\Delta$DCB (by AA similarities)

Therefore, $\frac{BC}{CD}=\frac{AD}{BD}$

$\Rightarrow$ $CD=\frac{BD^{2}}{AD}$

$\Rightarrow$ $CD=\frac{(8)^{2}}{4}=\frac{64}{4}=16cm$

Hence, CD = 16 cm

Question 9: P and Q are points on the sides AB and AC respectively of a $\Delta$ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm, show that BC = 3PQ.

Solution:

Given: P is a point on AB.

Then, AB = AP + PB = (2 + 4) cm = 6 cm

Also Q is a point on AC.

Then, AC = AQ + QC = (3 + 6) cm = 9 cm

Therefore, $\frac{AP}{AB}=\frac{2}{6}=\frac{1}{3}$

and $\frac{AQ}{AC}=\frac{3}{9}=\frac{1}{3}$

Therefore, $\frac{AP}{AB}=\frac{AQ}{AC}$

Thus, in $\Delta$APQ and $\Delta$ABC

$\angle$A = $\angle$A (Common)

and $\frac{AP}{AB}=\frac{AQ}{AC}$

Therefore, $\Delta$APQ ~ $\Delta$ABC (by SAS similarity)

$\Rightarrow$ $\frac{AP}{AB}=\frac{PQ}{BC}=\frac{AQ}{AC}$

Therefore, $\frac{PQ}{BC}=\frac{AQ}{AC}$

$\Rightarrow$ $\frac{PQ}{BC}=\frac{3}{9}=\frac{1}{3}$

Therefore, BC = 3PQ [Hence proved]

Question 10: ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersects AE at F, prove that AF x FB = EF x FD.

Solution:

Given: ABCD is a parallelogram and E is a point on BC.

Diagonal DB intersects AE at F.

To prove: AF x FB = EF x FD

Proof: In $\Delta$AFD and $\Delta$EFB

$\angle$AFD = $\angle$EFB (Vertically opposite angles)

$\angle$DAF = $\angle$BEF (Alternate angles)

Therefore, $\Delta$AFD $\approx$ $\Delta$EFB [By AAA similarity]

Therefore, $\frac{AF}{EF}=\frac{FD}{FB}$

AF . FB = EF . FD

Hence proved

Question 11: In the given figure, DB $\perp$ BC, DE $\perp$ AB and AC $\perp$ BC.

Prove that $\frac{BE}{DE}=\frac{AC}{BC}$

Solution:

In the given figure: DB $\perp$ BC, DE $\perp$ AB and AC $\perp$ BC and AB is the transversal

Therefore, $\angle$DBE = $\angle$BAC [Alternate angles]

In $\Delta$BDE and $\Delta$ABC

$\angle$DEB = $\angle$ACB = 90o

$\angle$DBE = $\angle$BAC

$\Delta$DBE ~ $\Delta$ABC [By AA similarity]

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

Hence proved

Question 12: A vertical pole 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.

Solution:

Let AB be the vertical stick and let AC be its shadow.

Then, AB = 7.5 m and AC = 5 m

Let DE be the vertical tower and let DF be its shadow

Then, DF = 24 m, Let DE = x meters

Now, in $\Delta$BAC and $\Delta$EDF,

$\Delta$BAC ~ $\Delta$EDF by SAS criterion

$\Rightarrow$ $\frac{AB}{DE}=\frac{AC}{DF}$

$\Rightarrow$ $\frac{7.5}{x}=\frac{5}{24}$

$\Rightarrow$ x = $\frac{7.5\times 24}{5}=36m$

Therefore, height of the vertical tower is 36 m.

Question 13: In an isosceles $\Delta$ABC, the base AB has produced both ways in P and Q such that AP x BQ = AC2.

Prove that $\Delta$ACP ~ $\Delta$BCQ.

Solution:

In $\Delta$ACP and $\Delta$BCQ

CA = CB

$\angle$CAB = $\angle$CBA

$\Rightarrow$ 180o â€“ $\angle$CAB = 180o â€“ $\angle$CBA

$\Rightarrow$ $\angle$CAP = $\angle$CBQ

Now, AP x BQ = AC2

$\Rightarrow$ $\frac{AP}{AC}=\frac{AC}{BQ}$

$\Rightarrow$ $\frac{AP}{AC}=\frac{BC}{BQ}$

Thus, $\angle$CAP = $\angle$CBQ and $\frac{AP}{AC}=\frac{BC}{BQ}$

Therefore, $\Delta$ACP ~ $\Delta$BCQ

Question 14: In the given figure, $\angle$1 = $\angle$2 and $\frac{AC}{BD}=\frac{CB}{CE}$. Prove that $\Delta$ACB ~ $\Delta$DCE.

Solution:

$\angle$1 = $\angle$2 (Given)

$\frac{AC}{BD}=\frac{CB}{CE}$ $\Rightarrow$ $\frac{AC}{CB}=\frac{BD}{CE}$ (Given)

Also, $\angle$2 = $\angle$1

Thus, $\frac{AC}{CB}=\frac{BD}{CE}$ and $\angle$2 = $\angle$1

Therefore, by SAS similarity criterion $\Delta$ACB ~ $\Delta$DCE

Question 15: ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the midpoints of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

Solution:

Given: ABCD is a quadrilateral in which AD = BC. P, Q, R, S are the midpoints of AB, AC, CD and BD.

To prove: PQRS is a rhombus

Proof: In $\Delta$ABC

Since P and Q are mid points of AB and AC

Therefore, PQ || BC and PQ = Â½ AC = Â½ DA (Mid-point theorem)

Similarly,

In $\Delta$CDA,

Since R and Q are mid points of CD and AC

Therefore, RQ || DA and RQ = Â½ DA

In $\Delta$BDA,

Since S and P are mid points of BD and AB

Therefore, SP || DA and SP = Â½ DA

In $\Delta$CDB,

Since S and R are mid points of BD and CD

Therefore, SR || BC and SR = Â½ BC = Â½ DA

Therefore, SP || RQ and PQ || SR and PQ = RQ = SP = SR

Hence, PQRS is a rhombus.

## EXERCISE – 4C

Question 1: $\Delta$ABC ~ $\Delta$DEF and their areas are respectively 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Solution:

Given: $\Delta$ABC ~ $\Delta$DEF,

Area of $\Delta$ABC = 64 cm2 and area of $\Delta$DEF = 121 cm2

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

Therefore, $\frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{BC^{2}}{EF^{2}}$

$\Rightarrow$ $\frac{64}{121}=\frac{x^{2}}{(15.4)^{2}}$ where BC = x

x2 = $\frac{64}{121}\times (15.4)^{2}$

x = $\sqrt{\frac{64}{121}\times 15.4\times 15.4}=\left ( \frac{8}{11}\times 15.4 \right )=11.2cm$

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.

Solution:

Given: $\Delta$ABC ~ $\Delta$PQR,

Let, the area of $\Delta$ABC = 9 cm2. Then area of $\Delta$PQR = 16 cm2.

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

Therefore, $\frac{area of \Delta ABC}{area of \Delta PQR}=\frac{9}{16}=\frac{BC^{2}}{QR^{2}}$

$\Rightarrow$ $\frac{9}{16}=\frac{(4.5)^{2}}{QR^{2}}$

$\Rightarrow$ QR2 = (4.5)2 x $\frac{16}{9}$

$\Rightarrow$ QR = $\sqrt{(4.5)^{2}\times \frac{16}{9}}$

$\Rightarrow$ QR = 4.5 x $\frac{4}{3}$ = 6 cm

Hence, QR = 6 cm

Question 3: $\Delta$ABC ~ $\Delta$PQR and ar($\Delta$ABC) = 4ar($\Delta$PQR). If BC = 12 cm, find QR.

Solution:

$\Delta$ABC ~ $\Delta$PQR,

area of $\Delta$ABC = 4.area of $\Delta$PQR.

Let area of $\Delta$PQR = x. Then the area of $\Delta$ABC = 4x.

We know that the ratios of the areas of two similar triangle is equal to the ratio of the square of their corresponding sides.

Therefore, $\frac{area of \Delta ABC}{area of \Delta PQR}=\frac{BC^{2}}{QR^{2}}$

$\frac{4x}{x}=\frac{(12)^{2}}{QR^{2}}\Rightarrow 4=\frac{(12)^{2}}{y^{2}}$

$\Rightarrow$ 4y2 = 144

$\Rightarrow$ y = 6

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.

Solution:

Given: $\Delta$ACB ~ $\Delta$DEF such that ar($\Delta$ABC) = 169 cm2 and ar($\Delta$DEF) = 121 cm2

We know that the ratio of the area of similar triangles is equal to the ratio of the square of their corresponding sides.

Therefore, $\frac{area of \Delta ABC}{area of \Delta DEF}=\frac{BC^{2}}{EF^{2}}$

$\Rightarrow$ $\frac{169}{121}=\frac{(26)^{2}}{EF^{2}}$

EF2 = (26)2 x $\frac{121}{169}$

EF = $\sqrt{(26)^{2}\times \frac{121}{169}}=26\times \frac{11}{13}=22cm$

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

Question 5: $\Delta$ABC ~ $\Delta$DEF and their areas are respectively 100 cm2 and 49 cm2. If the altitude of $\Delta$ABC is 5 cm, find the corresponding altitude of $\Delta$DEF.

Solution:

Given: $\Delta$ACB ~ $\Delta$DEF

ar($\Delta$ABC) = 100 cm2 and ar($\Delta$DEF) = 49 cm2

Let AL and DM be the corresponding altitude of ABC and DEF respectively such that AL = 5 cm and let DM = x cm

We know that the ratio of the area of two similar triangles is equal to the ratio of the square of corresponding altitudes.

Therefore, $\frac{area of \Delta ABC}{area of \Delta DEF}=\frac{AL^{2}}{DM^{2}}$

$\Rightarrow$ $\frac{100}{49}=\frac{(5)^{2}}{x^{2}}$

$\Rightarrow$ x2 = (5)2 x $\frac{49}{100}$

$\Rightarrow$ x = $\sqrt{(5)^{2}\times \frac{49}{100}}=5\times \frac{7}{10}=3.5cm$

Hence, DM = 3.5 cm

Therefore, the required altitude is 3.5 cm

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

Solution:

Given: $\Delta$ACB ~ $\Delta$DEF

Let AL and DM be the corresponding altitudes of $\Delta$ABC and $\Delta$DEF respectively such that AL = 6 cm and DM = 9 cm.

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

$\frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{AL^{2}}{DM^{2}}$

$\frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{6^{2}}{9^{2}}=\frac{36}{81}=\frac{4}{9}=4:9$

Hence, ratio of their areas = 4 : 9

#### Practise This Question

Choose the correct option and justify your choice.
(iii)sin2A = 2sinA is true when A=