Similarity — Exercise 13.2

In the figure (i) given below, if DE ∥ BC, AD = 3 cm, BD = 4 cm and BC = 5 cm, find (i) AE : EC (ii) DE.

Answer

(i) Considering △ABC and △ADE,

∠ A = ∠ A (Common angle)
∠ ADE = ∠ABC (Alternate angle)

So, by AA rule of similarity △ABC ~ △ADE.

$\therefore \dfrac{AE}{AC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{AE}{AE + EC} = \dfrac{AD}{AD + BD} \\[1em] \Rightarrow \dfrac{AE}{AE + EC} = \dfrac{3}{3 + 4} \\[1em] \Rightarrow 7AE = 3(AE + EC) \\[1em] \Rightarrow 7AE = 3AE + 3EC \\[1em] \Rightarrow 7AE - 3AE = 3EC \\[1em] \Rightarrow 4AE = 3EC \\[1em] \Rightarrow \dfrac{AE}{EC} = \dfrac{3}{4}.$

Hence, AE : EC = 3 : 4.

(ii) Since, △ABC ~ △ADE

$\therefore \dfrac{DE}{BC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{DE}{BC} = \dfrac{AD}{AD + BD} \\[1em] \Rightarrow \dfrac{DE}{5} = \dfrac{3}{3 + 4} \\[1em] \Rightarrow DE = \dfrac{15}{7} \\[1em] \Rightarrow DE = 2\dfrac{1}{7}. \\[1em]$

Hence, DE = $2\dfrac{1}{7}.$

In the figure (ii) given below, PQ ∥ AC, AP = 4 cm, PB = 6 cm and BC = 8 cm, find CQ and BQ.

Answer

Considering △ABC and △PBQ,

∠B = ∠B (Common angle)
∠BPQ = ∠BAC (Corresponding angle are equal)

So, by AA rule of similarity △ABC ~ △PBQ.

$\therefore \dfrac{BQ}{BC} = \dfrac{BP}{AB} \\[1em] \Rightarrow \dfrac{BQ}{8} = \dfrac{6}{AP + PB} \\[1em] \Rightarrow \dfrac{BQ}{8} = \dfrac{6}{4 + 6} \\[1em] \Rightarrow BQ = \dfrac{48}{10} \\[1em] \Rightarrow BQ = 4.8$

CQ = BC - BQ = 8 - 4.8 = 3.2 .

Hence, BQ = 4.8 cm and CQ = 3.2 cm.

In the figure (iii) given below, if XY ∥ QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, find PY and XY.

Answer

Considering △PQR and △PXY,

∠P = ∠P (Common angle)
∠PXY = ∠PQR (Corresponding angles)

So, by AA rule of similarity △PQR ~ △PXY. Since ratio of corresponding sides is same,

$\therefore \dfrac{PX}{QX} = \dfrac{PY}{YR} \\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{PY}{4.5} \\[1em] \Rightarrow PY = \dfrac{4.5}{3} \\[1em] \Rightarrow PY = 1.5 \\[1em]$

As triangles are similar so ratio of corresponding sides is same,

$\therefore \dfrac{XY}{QR} = \dfrac{PX}{PQ} \\[1em] \Rightarrow \dfrac{XY}{9} = \dfrac{1}{1 + 3} \\[1em] \Rightarrow XY = \dfrac{9}{4} \\[1em] \Rightarrow XY = 2.25$

Hence, the value of PY = 1.5 cm and XY = 2.25 cm.

In the adjoining figure, DE ∥ BC.

(i) If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, find the value of x.

(ii) If DB = x - 3, AB = 2x, EC = x - 2 and AC = 2x + 3, find the value of x.

Answer

(i) Considering △ABC and △ADE,

∠A = ∠A (Common angle)
∠ADE = ∠ABC (Corresponding angles)

So, by AA rule of similarity △ABC ~ △ADE. Since ratio of corresponding sides is same,

$\therefore \dfrac{AD}{AB} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC} \\[1em] \Rightarrow \dfrac{x}{x + x - 2} = \dfrac{x + 2}{x + 2 + x - 1} \\[1em] \Rightarrow \dfrac{x}{2x - 2} = \dfrac{x + 2}{2x + 1} \\[1em] \Rightarrow x(2x + 1) = (x + 2)(2x - 2) \\[1em] \Rightarrow 2x^2 + x = 2x^2 - 2x + 4x - 4 \\[1em] \Rightarrow 2x^2 - 2x^2 + x + 2x - 4x = 4 \\[1em] \Rightarrow -x = -4 \\[1em] \Rightarrow x = 4.$

Hence, the value of x = 4.

(ii) Since, triangles are similar so, ratio of corresponding sides is same,

$\therefore \dfrac{AD}{AB} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{AB - DB}{AB} = \dfrac{AC - EC}{AC} \\[1em] \Rightarrow \dfrac{2x - (x - 3)}{2x} = \dfrac{2x + 3 - (x - 2)}{2x + 3} \\[1em] \Rightarrow \dfrac{x + 3}{2x} = \dfrac{x + 5}{2x + 3} \\[1em] \Rightarrow (x + 3)(2x + 3) = 2x(x + 5) \\[1em] \Rightarrow 2x^2 + 3x + 6x + 9 = 2x^2 + 10x \\[1em] \Rightarrow 2x^2 - 2x^2 + 9 = 10x - 9x \\[1em] \Rightarrow x = 9 \\[1em]$

Hence, the value of x = 9.

E and F are points on the sides PQ and PR respectively of a △PQR. For each of the following cases, state whether EF || QR.

(i) PE = 3.9 cm, EQ = 3 cm, PF = 8cm and RF = 9 cm.

(ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

Answer

(i) So, according to question two triangles will be formed PQR and PEF.

Comparing ratios of corresponding side we get,

$\Rightarrow \dfrac{PE}{PQ} \text{ and } \dfrac{PF}{PR} \\[1em] \Rightarrow \dfrac{PE}{PE + EQ} \text{ and } \dfrac{PF}{PF + FR} \\[1em] \Rightarrow \dfrac{3.9}{3.9 + 3} \text{ and } \dfrac{8}{8 + 9} \\[1em] \Rightarrow \dfrac{3.9}{6.9} \text{ and } \dfrac{8}{17} \\[1em] \Rightarrow \dfrac{39}{69} \text{ and } \dfrac{8}{17}.$

Since, both ratios are different hence, EF and QR are not parallel.

(ii) The triangles are shown in the figure below:

Comparing ratios of corresponding side we get,

$\Rightarrow \dfrac{PE}{PQ} \text{ and } \dfrac{PF}{PR} \\[1em] \Rightarrow \dfrac{0.18}{1.28} \text{ and } \dfrac{0.36}{2.56} \\[1em] \Rightarrow \dfrac{18}{128} \text{ and } \dfrac{36}{256} \\[1em] \Rightarrow \dfrac{9}{64} \text{ and } \dfrac{9}{64}.$

Since, both ratios are same hence by converse of basic proportionality theorem triangles are similar and so EF and QR are parallel.

A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5cm, BR = 6 cm and PB = 4 cm. Is AB || QR ? Give reasons for your answer.

Answer

Checking the ratios in order to check the similarity of triangles PAB and PQR.

$\Rightarrow \dfrac{PA}{PQ} \text{ and } \dfrac{PB}{PR} \\[1em] \Rightarrow \dfrac{PA}{PQ} \text{ and } \dfrac{PB}{PB + BR} \\[1em] \Rightarrow \dfrac{5}{12.5} \text{ and } \dfrac{4}{4 + 6} \\[1em] \Rightarrow \dfrac{50}{125} \text{ and } \dfrac{4}{10} \\[1em] \Rightarrow \dfrac{2}{5} \text{ and } \dfrac{2}{5} \\[1em]$

Since, both ratios $\Big(\dfrac{PA}{PQ} = \dfrac{PB}{PR}\Big)$ are same hence by converse of basic proportionality theorem triangles are similar and so AB and QR are parallel.

In the figure (i) given below, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm.

Answer

Considering △ABC and △BDE,

∠B = ∠B (Common angle)
∠DEB = ∠ACB (Corresponding angles)

So, by AA rule of similarity △ABC ~ △BDE. Since ratio of corresponding sides is same,

$\Rightarrow \dfrac{BE}{BC} = \dfrac{BD}{BA} \\[1em] \Rightarrow \dfrac{BE}{BE + EC} = \dfrac{BD}{BA} \\[1em] \Rightarrow \dfrac{4}{4 + 2} = \dfrac{BD}{AB}\\[1em] \Rightarrow \dfrac{4}{6} = \dfrac{BD}{AB} \qquad \text{[....Eq 1]} \\[1em]$

Considering △ABL and △BDC,

∠B = ∠B (Common angle)
∠BDC = ∠BAL (Corresponding angles)

So, by AA rule of similarity △ABL ~ △BDC. Since ratio of corresponding sides is same,

$\Rightarrow \dfrac{BD}{BA} = \dfrac{BC}{BL} \\[1em] \Rightarrow \dfrac{BD}{BA} = \dfrac{BE + EC}{BE + EC + CL} \\[1em] \Rightarrow \dfrac{BD}{BA} = \dfrac{4 + 2}{4 + 2 + CL} \\[1em] \Rightarrow \dfrac{BD}{BA} = \dfrac{6}{6 + CL} \qquad \text{[....Eq 2]} \\[1em]$

Solving Eq 1 and Eq 2 we get,

$\Rightarrow \dfrac{6}{6 + CL} = \dfrac{4}{6} \\[1em] \Rightarrow 36 = 4(6 + CL) \\[1em] \Rightarrow 24 + 4CL = 36 \\[1em] \Rightarrow 4CL = 36 - 24 \\[1em] \Rightarrow 4CL = 12 \\[1em] \Rightarrow CL = 3.$

Hence, the length of CL is 3 cm.

In the figure (ii) given below, ∠D = ∠E and $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. Prove that ABC is an isosceles triangle.

Answer

Given, ∠D = ∠E

So, AD = AE [Sides opposite to equal angles]

Given, $\dfrac{AD}{DB} = \dfrac{AE}{EC} \qquad \text{[....Eq 1]}$

Hence, by basic proportionality theorem, DE is parallel to BC.

As AD = AE so in order to satisfy Eq 1, DB = EC.

AB = AD + DB = AE + EC

and AC = AE + EC.

Hence, AB = AC which means ABC is an isosceles triangle.

In the adjoining figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Answer

Consider △POQ

AB || PQ ....[ Given ]

So, By basic proportionality theorem,

$\dfrac{OA}{AP} = \dfrac{OB}{BQ} \qquad \text{[....Eq 1]}$

Then consider △OPR

AC || PR ....[ Given ]

So, By basic proportionality theorem,

$\dfrac{OA}{AP} = \dfrac{OC}{CR} \qquad \text{[....Eq 2]}$

Comparing Eq 1 and Eq 2 we get,

$\Rightarrow \dfrac{OB}{BQ} = \dfrac{OC}{CR}$

Hence, by basic proportionality theorem BC || QR.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem prove that $\dfrac{AO}{BO} = \dfrac{CO}{DO}.$

Answer

Trapezium ABCD is shown in the figure below:

Consider △OAB and △OCD,

∠AOB = ∠COD [Vertically opposite angles are equal]
∠OBA = ∠ODC [Alternate angles are equal]
∠OAB = ∠OCD [Alternate angles are equal]

Therefore, by AA rule of similarity △OAB ~ △OCD,

$\Rightarrow \dfrac{AO}{CO} = \dfrac{BO}{DO} \\[1em] \Rightarrow \dfrac{AO}{BO} = \dfrac{CO}{DO} \text{ (On cross-multiplication)}$

Hence, proved that $\dfrac{AO}{BO} = \dfrac{CO}{DO}.$

In the adjoining figure, AD is bisector of ∠BAC. If AB = 6 cm, AC = 4 cm and BD = 3 cm, find BC.

Answer

We know that,

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

$\therefore \dfrac{BD}{DC} = \dfrac{AB}{AC}$.

Putting values in above equation we get,

$\Rightarrow \dfrac{3}{DC} = \dfrac{6}{4} \\[1em] \Rightarrow DC = \dfrac{3 \times 4}{6} \\[1em] \Rightarrow DC = \dfrac{12}{6} \\[1em] \Rightarrow DC = 2 \text{ cm}.$

BC = BD + DC = 3 + 2 = 5 cm.

Hence, the length of BC = 5 cm.