Similarity — Exercise 13.1

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):

Answer

Given,

(i) In △ABC and △PQR

$\dfrac{AB}{PQ} = \dfrac{3.2}{4} \\[1em] = \dfrac{32}{40} = \dfrac{4}{5} \\[1em] \dfrac{AC}{PR} = \dfrac{3.6}{4.5} \\[1em] = \dfrac{36}{45} = \dfrac{4}{5} \\[1em] \dfrac{BC}{QR} = \dfrac{3}{5.4} \\[1em] = \dfrac{30}{54} = \dfrac{5}{9}$

Since, the ratio of the sides of both the triangle are not same. Hence, they are not similar.

(ii) In △DEF and △LMN

∠ E = ∠ N = 40°

$\dfrac{DE}{LN} = \dfrac{4}{2} = \dfrac{2}{1} \\[1em] \dfrac{EF}{MN} = \dfrac{4.8}{2.4} = \dfrac{2}{1} \\[1em]$

Thus, by SAS rule of similarity △DEF ~ △LMN.

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?

Answer

Given two right-angled triangles,

One angle of both triangles will be equal to 90°.

Given, acute angle of both triangles are equal let it be a°.

The third angle of both triangles will be [180° - (90 + a)°].

Hence, by AA rule of similarity both triangles are similar.

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.

Answer

Given, △ABC ~ △EDF

$\therefore \dfrac{AB}{DE} = \dfrac{AC}{EF} = \dfrac{BC}{DF}. \\[1em] \text{Consider, } \dfrac{AB}{DE} = \dfrac{AC}{EF} \\[1em] \Rightarrow \dfrac{5}{12} = \dfrac{7}{EF} \\[1em] \Rightarrow EF = \dfrac{84}{5} \\[1em] \Rightarrow EF = 16.8 \text{ cm.}$

Now consider,

$\dfrac{AB}{DE} = \dfrac{BC}{DF} \\[1em] \Rightarrow \dfrac{5}{12} = \dfrac{BC}{15} \\[1em] \Rightarrow BC = \dfrac{5 \times 15}{12} \\[1em] \Rightarrow BC = \dfrac{75}{12} \\[1em] \Rightarrow BC = 6.25.$

Hence, EF = 16.8 cm and BC = 6.25 cm.

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

Answer

Given △ABC ~ △DEF, so

$\therefore \dfrac{AB}{DE} = \dfrac{AC}{DF} = \dfrac{BC}{EF} \\[1em] \text{Consider, } \dfrac{AB}{DE} = \dfrac{AC}{DF} \\[1em] \dfrac{4}{6} = \dfrac{AC}{12} \\[1em] AC = \dfrac{48}{6} \\[1em] AC = 8.$

Now consider,

$\dfrac{AB}{DE} = \dfrac{BC}{EF} \\[1em] \dfrac{4}{6} = \dfrac{BC}{9} \\[1em] BC = \dfrac{36}{6} \\[1em] BC = 6$

Perimeter of △ABC = AB + BC + AC = 4 + 6 + 8 = 18 cm.

Hence, perimeter of △ABC = 18 cm.

If △ABC ~ △PQR, perimeter of △ABC = 32 cm, perimeter of △PQR = 48 cm and PR = 6 cm, then find the length of AC.

Answer

Given, △ABC ~ △PQR

$\therefore \dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{BC}{QR} \\[1em]$

Since triangles are similar so the ratio of perimeter will be equal to ratio of sides.

$\Rightarrow \dfrac{\text{Perimeter of △ABC}}{\text{Perimeter of △PQR}} = \dfrac{AC}{PR} \\[1em] \dfrac{32}{48} = \dfrac{AC}{6} \\[1em] AC = \dfrac{192}{48} \\[1em] AC = 4.$

Hence, length of AC = 4 cm.

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.

Answer

Let △ABC ~ △DEF in which shortest side of △ABC be BC = 6cm.
△DEF, DE = 8cm, EF = 4cm and DF = 7cm.

Since, △ABC ~ △DEF so,

$\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} \\[1em] \text{Consider, } \dfrac{AB}{DE} = \dfrac{BC}{EF} \\[1em] \dfrac{AB}{8} = \dfrac{6}{4} \\[1em] AB = \dfrac{48}{4} \\[1em] AB = 12.$

Now consider,

$\dfrac{BC}{EF} = \dfrac{AC}{DF} \\[1em] \dfrac{6}{4} = \dfrac{AC}{7} \\[1em] AC = \dfrac{42}{4} \\[1em] AC = 10.5.$

Hence, the other sides of triangle are 10.5cm and 12cm.

In the figure (1) given below, AB ∥ DE, AC = 3 cm, CE = 7.5 cm and BD = 14 cm. Calculate CB and DC.

Answer

In the given figure,

AB ∥ DE, AC = 3cm, CE = 7.5cm, BD = 14cm.

From the figure,

∠ACB = ∠DCE [Vertically opposite angles]
∠BAC = ∠CED [Alternate angles]

Then, by AA rule of similarity, △ABC ~ △CDE.

So,

$\Rightarrow \dfrac{AC}{CE} = \dfrac{BC}{CD} \\[1em] \Rightarrow \dfrac{3}{7.5} = \dfrac{BC}{CD} \\[1em] \Rightarrow 7.5BC = 3CD \qquad \text{[....Eq 1]} \\[1em]$

∵ BD = 14 cm, Let BC = x cm

∴ From fig, CD = (14 - x) cm.

Putting these values of BC and CD in equation 1 we get,

$\Rightarrow 7.5x = 3(14 - x) \\[1em] \Rightarrow 7.5x = 42 - 3x \\[1em] \Rightarrow 7.5x + 3x = 42 \\[1em] \Rightarrow 10.5x = 42 \\[1em] \Rightarrow x = \dfrac{42}{10.5} \\[1em] \Rightarrow x = 4.$

∴ x = 4 and 14 - x = 10.

Hence, CB = 4cm and DC = 10cm.

In the figure (2) given below, CA ∥ BD, the lines AB and CD meet at O.

(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.

Answer

Considering △ACO and △BDO,

∠ AOC = ∠ BOD [Vertically opposite angles]
∠ A = ∠ B [Alternate angles]

Then, by AA rule of similarity, △AOC ~ △BOD.

So,

$\Rightarrow \dfrac{OA}{OB} = \dfrac{OC}{OD} = \dfrac{AC}{BD} \\[1em] \Rightarrow \text{Consider, } \dfrac{AC}{BD} = \dfrac{OA}{OB} \\[1em] \Rightarrow \dfrac{3.6}{2.4} = \dfrac{OA}{3.2} \\[1em] OA = \dfrac{3.6 \times 3.2}{2.4} \\[1em] OA = \dfrac{11.52}{2.4} \\[1em] OA = 4.8.$

Now, consider

$\dfrac{AC}{BD} = \dfrac{OC}{OD} \\[1em] \dfrac{3.6}{2.4} = \dfrac{OC}{4} \\[1em] OC = \dfrac{14.4}{2.4} \\[1em] OC = 6.$

Hence, OA = 4.8 cm and OC = 6 cm.

In the figure (i) given below, ∠P = ∠RTS. Prove that △RPQ ~ △RTS.

Answer

In △RPQ and △RTS.

∠R = ∠R ( Common )
∠P = ∠RTS (Given)

Hence, by AA rule of similarity △RPQ ~ △RTS.

In the figure (ii) given below, ∠ADC = ∠BAC. Prove that CA² = DC × BC.

Answer

In △ABC and △ADC

∠C = ∠C (Common angle for both triangle)

∠BAC = ∠ADC (Given)

Then, by AA rule of similarity, △BAC ~ △ADC.

So,

$\dfrac{CA}{DC} = \dfrac{BC}{CA} \\[1em] CA^2 = DC \times BC.$

Hence proved.

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.

Answer

(i) Given, AP = 2PB and CP = 2PD.

$\therefore \dfrac{AP}{PB} = \dfrac{2}{1} \text{ and } \dfrac{CP}{PD} = \dfrac{2}{1}.$

∠ APC = ∠ BPD [Vertically opposite angles]

So by SAS rule of similarity △ACP ~ △BDP.

Since, triangles are similar,

∴ ∠ CAP = ∠ PBD.

Since, these angles are alternate angles therefore, AC ∥ BD.

Hence, proved that △ACP ~ △BDP and AC ∥ BD.

(ii) Since triangles are similar. So,

$\dfrac{AP}{PB} = \dfrac{AC}{BD} \\[1em] \dfrac{AC}{BD} = \dfrac{2}{1} \\[1em] BD = \dfrac{AC}{2} \\[1em] BD = \dfrac{4.5}{2} \\[1em] BD = 2.25.$

Hence, BD = 2.25 cm.

In the figure (2) given below, ∠ ADE = ∠ ACB.

(i) Prove that △ABC and △AED are similar.

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

Answer

(i) From the given figure,

∠ ADE = ∠ ACB (Given)
∠ A = ∠ A. (Common)

Hence, by AA rule of similarity △ABC ~ △AED.

(ii) Since triangles are similar,

$\therefore \dfrac{BC}{DE} = \dfrac{AB}{AE} = \dfrac{AC}{AD} \\[1em]$

From figure:
AD = AB - BD = 6 - 1 = 5 cm.

Consider,

$\dfrac{AB}{AE} = \dfrac{AC}{AD} \\[1em] \dfrac{6}{3} = \dfrac{AC}{5} \\[1em] AC = \dfrac{30}{3} \\[1em] AC = 10.$

Hence, the length of AC = 10 cm.

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.

Answer

Given, ∠ PQR = ∠ PRS.

∠ P = ∠ P (common for both the triangles).

By AA rule of similarity, △PQR ~ △PRS.

Then,

$\dfrac{PQ}{PR} = \dfrac{PR}{PS} = \dfrac{QR}{SR} \\[1em] \text{Considering, } \dfrac{PQ}{PR} = \dfrac{PR}{PS} \\[1em] \dfrac{PQ}{8} = \dfrac{8}{4} \\[1em] PQ = \dfrac{64}{4} \\[1em] PQ = 16.$

Hence, the length of PQ = 16 cm.

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

Answer

Consider △ABC

Given, AB = AC

∠ B = ∠ C [Angles opposite to equal sides (Property of isosceles triangle)]

Considering △BMP and △CNP

∠ M = ∠ N = 90°.

∠ B = ∠ C.

So, by AA rule of similarity △BMP ~ △CNP.

As triangles are similar,

$\Rightarrow \dfrac{BM}{CN} = \dfrac{MP}{NP} \\[1em] \Rightarrow BM \times NP = CN \times MP.$

Hence proved.

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

Answer

Let two similar triangles be △ABC and △PQR.

We know that when triangles are similar ratio of corresponding sides are equal.

$\therefore \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR}.$

By property of ratio i.e.,

if $\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{d}{e},$ then each ratio = $\dfrac{\text{sum of antecedents}}{\text{sum of consequents}}$.

So,

$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} = \dfrac{AB + BC + AC}{PQ + QR + PR}$

Since, AB + BC + AC = Perimeter of △ABC and PQ + QR + PR = Perimeter of △PQR. So,

$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} = \dfrac{\text{Perimeter of △ABC}}{\text{Perimeter of △PQR}}.$

Hence, proved.

In the adjoining figure, ABCD is a trapezium in which AB ∥ DC. The diagonals AC and BD intersect at O. Prove that $\dfrac{AO}{OC} = \dfrac{BO}{OD}.$

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

Answer

Consider triangle AOB and COD,

∠ AOB = ∠ COD [Vertically opposite angles]

∠ OAB = ∠ OCD [Alternate angles]

So, by AA rule of similarity △AOB ~ △COD.

As triangles are similar, ratio of sides will be similar,

$\dfrac{OA}{OC} = \dfrac{OB}{OD}$

Putting values of sides from question in equation,

$\dfrac{3x - 19}{x - 3} = \dfrac{x - 4}{4} \\[1em] 4(3x - 19) = (x - 3)(x - 4) \\[1em] 12x - 76 = x^2 - 4x - 3x + 12 \\[1em] x^2 - 7x + 12 = 12x - 76 \\[1em] x^2 - 7x - 12x + 12 + 76 = 0 \\[1em] x^2 - 19x + 88 = 0 \\[1em] x^2 - 11x - 8x + 88 = 0 \\[1em] x(x - 11) - 8(x - 11) = 0 \\[1em] (x - 8)(x - 11) = 0 \\[1em] x - 8 = 0 \text{ or } x - 11 = 0 \\[1em] x = 8 \text{ or } x = 11.$

Hence, the value of x = 8 or 11.

In the figure (1) 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

Answer

(i) Consider △EFG and △CGD

∠ EGF = ∠ CGD [Vertically opposite angles]

∠ FEG = ∠ GCD [Alternate angles are equal]

So, by AA rule of similarity △EFG ~ △CGD.

Then,

$\dfrac{EG}{GC} = \dfrac{EF}{CD} \\[1em] \dfrac{5}{10} = \dfrac{EF}{18} \\[1em] EF = \dfrac{90}{10} \\[1em] EF = 9.$

Hence, the length of EF = 9 cm.

(ii) Consider △ABC and △EFC

∠ C = ∠ C [Common angles]

∠ ABC = ∠ EFC [Alternate angles are equal]

So, by AA rule of similarity △ABC ~ △EFC

Then,

$\dfrac{AC}{EC} = \dfrac{AB}{EF} \\[1em] \dfrac{AC}{EG + GC} = \dfrac{15}{9} \\[1em] \dfrac{AC}{5 + 10} = \dfrac{15}{9} \\[1em] \dfrac{AC}{15} = \dfrac{15}{9} \\[1em] AC = \dfrac{225}{9} \\[1em] AC = 25.$

Hence, the length of AC = 25 cm.

In the figure (2) given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm and BE = x and AE = y. Find the values of x and y.

Answer

Consider △AEF and △CED

∠AEF = ∠CED [Vertically opposite angles]

∠F = ∠C [Alternate angles are equal]

So, by AA rule of similarity △AEF ~ △CED

Then,

$\dfrac{AF}{CD} = \dfrac{AE}{ED} \\[1em] \dfrac{7.5}{4.5} = \dfrac{y}{3} \\[1em] y = \dfrac{22.5}{4.5} \\[1em] y = 5.$

Consider △ABE and △ACD

∠A = ∠A [Common angles]

∠ABE = ∠ACD [Alternate angles are equal]

So, by AA rule of similarity △ABE ~ △ACD.

Then,

$\dfrac{EB}{CD} = \dfrac{AE}{AD} \\[1em] \dfrac{x}{4.5} = \dfrac{y}{AE + ED} \\[1em] \dfrac{x}{4.5} = \dfrac{y}{y + 3} \\[1em] x = \dfrac{4.5 \times 5}{8} \\[1em] x = \dfrac{22.5}{8} \\[1em] x = \dfrac{225}{80} = \dfrac{45}{16} \\[1em] x = 2\dfrac{13}{16}.$

Hence, the value of x = $2\dfrac{13}{16}$ cm and y = 5 cm.

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

Answer

Given ∠A = 90°,

or, ∠BAD + ∠DAC = 90° .....(i)

Now, consider △ADC

∠ADC = 90°

or, ∠DCA + ∠DAC = 90° .....(ii)

From equation (i) and equation (ii)

We have,

∠BAD + ∠DAC = ∠DCA + ∠DAC

∠BAD = ∠DCA .....(iii)

So, from △BDA and △ADC

∠BDA = ∠ADC = 90°
∠BAD = ∠DCA [From equation (iii)]

So, by AA rule of similarity △BDA ~ △ADC.

Since, corresponding sides of similar triangles are proportional,

$\therefore \dfrac{BD}{AD} = \dfrac{AD}{DC} = \dfrac{AB}{AC} \\[1em] \text{Considering, } \dfrac{BD}{AD} = \dfrac{AD}{DC} \\[1em] AD^2 = BD \times CD \\[1em] AD^2 = 2 \times 8 \\[1em] AD^2 = 16 \\[1em] AD^2 - 16 = 0 \\[1em] AD^2 - 4^2 = 0 \\[1em] (AD - 4)(AD + 4) = 0 \\[1em] AD - 4 = 0 \text{ and } AD + 4 = 0 \\[1em] AD = 4 \text{ and } AD = -4.$

Since, length cannot be negative hence, AD ≠ -4.

Hence, the length of AD = 4 cm.

A 15 meters 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 meters long. Find the height of the telephone pole.

Answer

Let AB be tower and CD be pole.

BE = Shadow of tower and
DE = Shadow of telephone pole.

Considering △ABE and △CDE

∠ABE = ∠CDE (Both are equal to 90°)
∠AEB = ∠CED [Common angles]

So, by AA rule of similarity △ABE ~ △CDE. Hence, the ratio of corresponding sides will be equal.

$\therefore \dfrac{AB}{CD} = \dfrac{BE}{DE} \\[1em] \Rightarrow \dfrac{15}{CD} = \dfrac{24}{16} \\[1em] \Rightarrow CD = \dfrac{15 \times 16}{24} \\[1em] \Rightarrow CD = \dfrac{240}{24} \\[1em] \Rightarrow CD = 10.$

Hence, the height of telephone pole is 10 m.

A street light bulb is fixed on a pole 6 m above the level of street. If a woman of height 1.5 m casts a shadow of 3 m, find how far she is away from the base of the pole?

Answer

Let AB be the pole and DE be the woman as shown in the figure below:

Height of pole (AB) = 6 m
and height of a woman (DE) = 1.5 m

Here shadow EF = 3 m

Let BE(Distance of woman from pole) = x meters.

Considering △ABF and △EFD

∠ABF = ∠DEF (Both are equal to 90°)
∠F = ∠F [Common angles]

So, by AA rule of similarity △ABF ~ △EFD. Hence, the ratio of corresponding sides will be equal.

$\therefore \dfrac{BF}{EF} = \dfrac{AB}{DE} \\[1em] \Rightarrow \dfrac{3 + x}{3} = \dfrac{6}{1.5} \\[1em] \Rightarrow \dfrac{3 + x}{3} = \dfrac{60}{15} \\[1em] \Rightarrow 15(3 + x) = 180 \\[1em] \Rightarrow 45 + 15x = 180 \\[1em] \Rightarrow 15x = 180 - 45 \\[1em] \Rightarrow 15x = 135 \\[1em] \Rightarrow x = \dfrac{135}{15} \\[1em] \Rightarrow x = 9.$

Hence, woman is 9 m away from the pole.