Similarity

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 angle triangles,

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

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

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

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

It is given that △ABC ~ △EDF such that AB = 5cm, AC = 7cm, DF = 15cm and DE = 12cm. 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 = 32cm, perimeter of △PQR = 48cm and PR = 6cm, 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 6cm and which is similar to a triangle whose sides are 4cm, 7cm and 8cm.

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 given below, AB ∥ DE, AC = 3cm, CE = 7.5cm and BD = 14cm. 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 △s 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 6m 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.

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.

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.

Answer

(i) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △PQR}} = \dfrac{(1)^2}{(3)^2} \\[1em] = \dfrac{1}{9} \\[1em] = 1 : 9.$

Hence, the ratio of area of △ABC to △PQR = 1 : 9.

(ii) Let the corresponding sides be in ratio x : y.

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

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △PQR}} = \dfrac{x^2}{y^2} \\[1em] \Rightarrow \dfrac{25}{36} = \dfrac{x^2}{y^2} \\[1em] \Rightarrow \Big(\dfrac{5}{6}\Big)^2 = \Big(\dfrac{x}{y}\Big)^2 \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{5}{6} \\[1em] \Rightarrow x : y = 5 : 6.$

Hence, the ratio of corresponding sides of △ABC and △PQR = 5 : 6.

△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.

Answer

Let the length of EF be x cm.

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

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △DEF}} = \dfrac{BC^2}{EF^2} \\[1em] \Rightarrow \dfrac{9}{16} = \dfrac{(2.1)^2}{x^2} \\[1em] \Rightarrow \dfrac{9}{16} = \dfrac{2.1 \times 2.1}{x^2} \\[1em] \Rightarrow \dfrac{9}{16} = \dfrac{4.41}{x^2} \\[1em] \Rightarrow x^2 = \dfrac{4.41 \times 16}{9} \\[1em] \Rightarrow x^2 = \dfrac{70.56}{9} \\[1em] \Rightarrow x^2 = 7.84 \\[1em] \Rightarrow x = \sqrt{7.84} \\[1em] \Rightarrow x = 2.8$

Hence, the length of EF = 2.8 cm.

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

Answer

Let the area of △DEF be x sq.cm

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

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △DEF}} = \dfrac{BC^2}{EF^2} \\[1em] \Rightarrow \dfrac{54}{x} = \dfrac{3^2}{4^2} \\[1em] \Rightarrow \dfrac{54}{x} = \dfrac{9}{16} \\[1em] \Rightarrow x = \dfrac{54 \times 16}{9} \\[1em] \Rightarrow x = 6 \times 16 \\[1em] \Rightarrow x = 96.$

Hence, the area of △DEF = 96 sq.cm.

The areas 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.

Answer

Let the length of altitude of other △ be x cm

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

$\therefore \dfrac{\text{Area of first △}}{\text{Area of second △}} = \Big(\dfrac{\text{Altitude of first △}}{\text{Altitude of second △}}\Big)^2 \\[1em] \Rightarrow \dfrac{36}{25} = \Big(\dfrac{2.4}{x}\Big)^2 \\[1em] \Rightarrow \dfrac{36}{25} = \dfrac{5.76}{x^2} \\[1em] \Rightarrow x^2 = \dfrac{5.76 \times 25}{36} \\[1em] \Rightarrow x^2 = \dfrac{144}{36} \\[1em] \Rightarrow x^2 = 4 \\[1em] \Rightarrow x^2 - 4 = 0 \\[1em] \Rightarrow (x - 2)(x + 2) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 2 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = -2.$

Since length cannot be negative so, x ≠ -2.

Hence, the length of altitude of other triangle = 2 cm.

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

Answer

Considering △QOA and △POB,

∠ QOA = ∠ POB (Vertically opposite angles are equal)
∠ QAO = ∠ PBO (Both are equal to 90°)

Hence, by AA axiom △QOA ~ △POB.

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

Let the area of △QOA be x cm².

$\therefore \dfrac{\text{Area of △QOA}}{\text{Area of △POB}} = \dfrac{QO^2}{PO^2} \\[1em] \Rightarrow \dfrac{x}{120} = \dfrac{9^2}{6^2} \\[1em] \Rightarrow \dfrac{x}{120} = \dfrac{81}{36} \\[1em] \Rightarrow x = \dfrac{120 \times 81}{36} \\[1em] \Rightarrow x = \dfrac{9720}{36} \\[1em] \Rightarrow x = 270.$

Hence, the area of △QOA is 270 cm²

In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5 cm, 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.

Answer

(i) Considering △OAB and △OCD,

∠ AOB = ∠ COD (Vertically opposite angles are equal)
∠ BAO = ∠ OCD (Alternate angles are equal)

Hence, by AA axiom △OAB ~ △OCD.

(ii) Since triangles are similar hence ratio of corresponding sides are equal,

$\therefore \dfrac{AO}{OC} = \dfrac{AB}{CD} \\[1em] \Rightarrow \dfrac{10}{5} = \dfrac{6.5}{CD} \\[1em] \Rightarrow CD = \dfrac{6.5 \times 5}{10} \\[1em] \Rightarrow CD = \dfrac{32.5}{10} \\[1em] \Rightarrow CD = 3.25 \text{ cm}.$

Similarly,

$\dfrac{AO}{OC} = \dfrac{OB}{OD} \\[1em] \Rightarrow \dfrac{10}{5} = \dfrac{OB}{2.8} \\[1em] \Rightarrow OB = \dfrac{2.8 \times 10}{5} \\[1em] \Rightarrow OB = \dfrac{28}{5} \\[1em] \Rightarrow OB = 5.6 \text{ cm.}$

Hence, the length of CD = 3.25 cm and OB = 5.6 cm.

(iii) In part (i) we have proved that △OAB ~ △OCD.

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

$\therefore \dfrac{\text{Area of △OAB}}{\text{Area of △OCD}} = \dfrac{AO^2}{OC^2} \\[1em] = \dfrac{10^2}{5^2} \\[1em] = \dfrac{100}{25} \\[1em] = \dfrac{4}{1} \\[1em] = 4 : 1.$

Hence, the ratio of area of △OAB and △OCD is 4 : 1.

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.

Answer

Considering △ADE and △ABC,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

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

Let the area of △ABC be x cm².

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{DE^2}{BC^2} \\[1em] \Rightarrow \dfrac{28}{x} = \dfrac{6^2}{9^2} \\[1em] \Rightarrow \dfrac{28}{x} = \dfrac{36}{81} \\[1em] \Rightarrow x = \dfrac{28 \times 81}{36} \\[1em] \Rightarrow x = \dfrac{2268}{36} \\[1em] \Rightarrow x = 63.$

Hence, area of △ABC is 63 cm².

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

Answer

Considering △ADE and △ABC,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

Given AD : DB = 1 : 2.

$\Rightarrow \dfrac{AD}{AB - AD} = \dfrac{1}{2} \\[1em] \Rightarrow 2AD = AB - AD \\[1em] \Rightarrow 2AD + AD = AB \\[1em] \Rightarrow 3AD = AB \\[1em] \Rightarrow \dfrac{AD}{AB} = \dfrac{1}{3} \\[1em] \Rightarrow AD : AB = 1 : 3.$

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

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ADE + Area of ⏢DBCE}} = \dfrac{1^2}{3^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ADE + Area of ⏢DBCE}} = \dfrac{1}{9} \\[1em] \Rightarrow 9 \times \text{Area of △ADE} = \text{Area of △ADE + Area of ⏢DBCE} \\[1em] \Rightarrow 8 \text{ Area of △ADE} = \text{ Area of ⏢DBCE } \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{ Area of ⏢DBCE }} = \dfrac{1}{8} = 1 : 8.$

Hence, the ratio of the areas of △ADE and trapezium DBCE is 1 : 8.

In the given figure, DE || BC.

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

(ii) Given that AD = $\dfrac{1}{2}$BD, calculate DE, if BC = 4.5 cm.

(iii) If area of △ABC = 18 cm², find area of trapezium DBCE.

Answer

(i) Considering △ADE and △ABC,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

(ii) Given AD = $\dfrac{1}{2}$BD

$\Rightarrow AD = \dfrac{1}{2}(AB - AD) \\[1em] \Rightarrow 2AD = AB - AD \\[1em] \Rightarrow 2AD + AD = AB \\[1em] \Rightarrow 3AD = AB \\[1em] \Rightarrow \dfrac{AD}{AB} = \dfrac{1}{3} \\[1em] \Rightarrow AD : AB = 1 : 3.$

Since triangles ADE and ABC are similar so, ratio of their corresponding sides will be equal

$\therefore \dfrac{AD}{AB} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{DE}{4.5}\\[1em] \Rightarrow DE = \dfrac{4.5}{3} \\[1em] \Rightarrow DE = 1.5$

Hence, the length of DE = 1.5 cm.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{1^2}{3^2} \\[1em] \Rightarrow \text{Area of △ADE} = \dfrac{1}{9} \times \text{ Area of △ABC} \\[1em] \Rightarrow \text{Area of △ADE} = \dfrac{1}{9} \times 18 \\[1em] \Rightarrow \text{Area of △ADE} = 2 \text{ cm}^2$

Area of trapezium DBCE = Area of △ABC - Area of △ADE = (18 - 2) cm² = 16 cm².

Hence, the area of trapezium DBCE = 16 cm².

In the given figure, AB and DE are perpendiculars 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.

Answer

(i) Considering △DEC and △ABC,

∠ C = ∠ C (Common angles)
∠ ABC = ∠ DEC (Both angles are equal to 90°)

Hence, by AA axiom △DEC ~ △ABC.

(ii) Since △DEC ~ △ABC, so, ratio of their corresponding sides will be equal

$\therefore \dfrac{AB}{DE} = \dfrac{AC}{CD} \\[1em] \Rightarrow \dfrac{6}{4} = \dfrac{15}{CD} \\[1em] \Rightarrow CD = \dfrac{15 \times 4}{6} \\[1em] \Rightarrow CD = 10 \text{ cm}$

Hence, the length of CD = 10 cm.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △DEC}} = \dfrac{AB^2}{DE^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ABC}}{\text{Area of △DEC}} = \dfrac{6^2}{4^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ABC}}{\text{Area of △DEC}} = \dfrac{36}{16} \\[1em] \Rightarrow \dfrac{\text{Area of △ABC}}{\text{Area of △DEC}} = \dfrac{9}{4} \\[1em].$

Hence, the ratio of the area of △ABC : area of △DEC = 9 : 4.

In the adjoining figure, ABC is a triangle. DE is parallel to BC and $\dfrac{AD}{DB} = \dfrac{3}{2}.$

(i) Determine the ratio $\dfrac{AD}{AB}, \dfrac{DE}{BC}.$

(ii) Prove that △DEF is similar to △CBF. Hence, find $\dfrac{EF}{FB}.$

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

Answer

(i) We need to find $\dfrac{AD}{AB}$,

Given,

$\dfrac{AD}{DB} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{AD}{AB - AD} = \dfrac{3}{2} \\[1em] \Rightarrow 2AD = 3(AB - AD) \\[1em] \Rightarrow 2AD = 3AB - 3AD \\[1em] \Rightarrow 5AD = 3AB \\[1em] \Rightarrow \dfrac{AD}{AB} = \dfrac{3}{5}.$

Hence, the ratio $\dfrac{AD}{AB} = \dfrac{3}{5}$.

Considering △ADE and △ABC,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

Since triangles ADE and ABC are similar so the ratio of corresponding sides will be equal.

$\therefore \dfrac{DE}{BC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{DE}{BC} = \dfrac{3}{5}.$

Hence, the ratio $\dfrac{DE}{BC} = \dfrac{3}{5}$.

(ii) Considering △DEF and △CBF,

∠ DFE = ∠ BFC (Vertically opposite angles)
∠ EDF = ∠ FCB (Alternate angles are equal)

Hence, by AA axiom △DEF ~ △CBF.

Since triangles are similar hence the ratio of the corresponding sides will be equal,

$\therefore \dfrac{EF}{FB} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{EF}{FB} = \dfrac{3}{5}.$

Hence, the value of $\dfrac{EF}{FB} = \dfrac{3}{5}.$

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △DEF}}{\text{Area of △CBF}} = \dfrac{EF^2}{FB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △DEF}}{\text{Area of △CBF}} = \dfrac{3^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △DEF}}{\text{Area of △CBF}} = \dfrac{9}{25}.$

Hence, the ratio of the area of △DEF : area of △CBF = 9 : 25.

In △PQR, MN is parallel to QR and $\dfrac{PM}{MQ} = \dfrac{2}{3}.$

(i) Find $\dfrac{MN}{QR}$.

(ii) Prove that △OMN and △ORQ are similar.

(iii) Find area of △OMN : area of △ORQ.

Answer

(i) Considering △PMN and △PQR,

∠ P = ∠ P (Common angles)
∠ PMN = ∠ PQR (Corresponding angles are equal)

Hence, by AA axiom △PMN ~ △PQR.

Given,

$\dfrac{PM}{MQ} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{PM}{PQ - PM} = \dfrac{2}{3} \\[1em] \Rightarrow 3PM = 2(PQ - PM) \\[1em] \Rightarrow 3PM = 2PQ - 2PM \\[1em] \Rightarrow 3PM + 2PM = 2PQ \\[1em] \Rightarrow 5PM = 2PQ \\[1em] \Rightarrow \dfrac{PM}{PQ} = \dfrac{2}{5}.$

Since triangles are similar hence the ratio of the corresponding sides will be equal,

$\therefore \dfrac{MN}{QR} = \dfrac{PM}{PQ} = \dfrac{2}{5}.$

Hence, $\dfrac{MN}{QR} = \dfrac{2}{5}$

(ii) Considering △OMN and △ORQ,

∠ MON = ∠ QOR (Vertically opposite angles are equal)
∠ OMN = ∠ ORQ (Alternate angles are equal)

Hence, by AA axiom △OMN ~ △ORQ.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △OMN}}{\text{Area of △ORQ}} = \dfrac{MN^2}{QR^2} \\[1em] \Rightarrow \dfrac{\text{Area of △OMN}}{\text{Area of △ORQ}} = \dfrac{2^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △OMN}}{\text{Area of △ORQ}} = \dfrac{4}{25}.$

Hence, the ratio of the Area of △OMN : Area of △ORQ = 4 : 25.

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.

Answer

(i) Given,

$\dfrac{AP}{PB} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{AP}{AB - AP} = \dfrac{2}{3} \\[1em] \Rightarrow 3AP = 2(AB - AP) \\[1em] \Rightarrow 3AP = 2AB - 2AP \\[1em] \Rightarrow 3AP + 2AP = 2AB \\[1em] \Rightarrow 5AP = 2AB \\[1em] \Rightarrow \dfrac{AP}{AB} = \dfrac{2}{5}.$

Considering △APO and △ABC,

∠ A = ∠ A (Common angles)
∠ APO = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △APO ~ △ABC.

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

$\therefore \dfrac{\text{Area of △APO}}{\text{Area of △ABC}} = \dfrac{AP^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △APO}}{\text{Area of △ABC}} = \dfrac{2^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △APO}}{\text{Area of △ABC}} = \dfrac{4}{25}.$

Hence, the ratio of the area of △APO : area of △ABC = 4 : 25.

(ii) In parallelogram PBCQ opposite sides are equal,

so, PB = QC. Hence, $\dfrac{AP}{QC} = \dfrac{AP}{PB} = \dfrac{2}{3}$.

Considering △APO and △CQO,

∠ AOP = ∠ QOC (Vertically opposite angles)
∠ OAP = ∠ OCQ (Alternate angles are equal)

Hence, by AA axiom △APO ~ △CQO.

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

$\therefore \dfrac{\text{Area of △APO}}{\text{Area of △CQO}} = \dfrac{AP^2}{QC^2} \\[1em] \Rightarrow \dfrac{\text{Area of △APO}}{\text{Area of △CQO}} = \Big(\dfrac{AP}{QC}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of △APO}}{\text{Area of △CQO}} = \Big(\dfrac{2}{3}\Big)^2 = \dfrac{4}{9}.$

Hence, the ratio of the area of △APO : area of △CQO = 4 : 9.

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.

Answer

Given, AB || DC and AB = 2CD.

$\therefore \dfrac{AB}{CD} = \dfrac{2}{1}$.

Considering △AOB and △COD,

∠ AOB = ∠ COD (Vertically opposite angles)
∠ OCD = ∠ OAB (Alternate angles are equal)

Hence, by AA axiom △AOB ~ △COD.

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

$\therefore \dfrac{\text{Area of △AOB}}{\text{Area of △COD}} = \dfrac{AB^2}{CD^2} \\[1em] \Rightarrow \dfrac{\text{Area of △AOB}}{\text{Area of △COD}} = \dfrac{2^2}{1^2} \\[1em] \Rightarrow \dfrac{\text{Area of △AOB}}{\text{Area of △COD}} = \dfrac{4}{1}.$

Hence, the ratio of the area of △AOB : area of △COD = 4 : 1.

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.

Answer

(i) Given, AM = 6 cm and AN = 10 cm and area of parallelogram ABCD is 45 cm².

Area of parallelogram = base x height = CD x AM = BC x AN.

$\therefore AM \times CD = 45 \\[1em] \Rightarrow 6 \times CD = 45 \\[1em] \Rightarrow CD = \dfrac{45}{6} \\[1em] \Rightarrow CD = \dfrac{15}{2} \\[1em] \Rightarrow CD = 7.5$

In parallelogram AB = CD = 7.5 cm.

Hence, the length of AB = 7.5 cm.

(ii) Given, AM = 6 cm and AN = 10 cm and area of parallelogram ABCD is 45 cm².

Area of parallelogram = base x height = CD x AM = BC x AN.

$\therefore AN \times BC = 45 \\[1em] \Rightarrow 10 \times BC = 45 \\[1em] \Rightarrow BC = \dfrac{45}{10} \\[1em] \Rightarrow BC = 4.5$

Hence, the length of BC = 4.5 cm.

(iii) Considering △ADM and △ABN,

∠ ADM = ∠ ABN (Opposite angles of a parallelogram are equal)
∠ AMD = ∠ ANB (Both angles are equal to 90°)

Hence, by AA axiom △ADM ~ △ANB.

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

$\therefore \dfrac{\text{Area of △ADM}}{\text{Area of △ANB}} = \dfrac{AM^2}{AN^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADM}}{\text{Area of △ANB}} = \dfrac{6^2}{10^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADM}}{\text{Area of △ANB}} = \dfrac{36}{100} \\[1em] \Rightarrow \dfrac{\text{Area of △ADM}}{\text{Area of △ANB}} = \dfrac{9}{25}$

Hence, the ratio of the area of △ADM : area of △ANB = 9 : 25.

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. AEFD

(iv) area of △FEO : area of △OBC.

Answer

(i) Considering △ADB and △EFB,

∠ B = ∠ B (Common angles)
∠ DAB = ∠ FEB (Corresponding angles are equal)

Hence, by AA axiom △ADB ~ △EFB.

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

$\therefore \dfrac{AB}{BE} = \dfrac{AD}{EF} \\[1em] \Rightarrow \dfrac{EF}{AD} = \dfrac{EB}{AB} \\[1em]$

Given, $\dfrac{AE}{EB} = \dfrac{2}{3}$.

$\Rightarrow \dfrac{AB- EB}{EB} = \dfrac{2}{3} \\[1em] \Rightarrow 3(AB - EB) = 2EB \\[1em] \Rightarrow 3AB - 3EB = 2EB \\[1em] \Rightarrow 3AB = 2EB + 3EB \\[1em] \Rightarrow 3AB = 5EB \\[1em] \Rightarrow \dfrac{EB}{AB} = \dfrac{3}{5}.$

Since, $\Rightarrow \dfrac{EF}{AD} = \dfrac{EB}{AB} = \dfrac{3}{5}.$

Hence, EF : AD = 3 : 5.

(ii) Considering △ABD and △BEF,

∠ B = ∠ B (Common angles)
∠ DAB = ∠ FEB (Corresponding angles are equal)

Hence, by AA axiom △ADB ~ △BEF.

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

$\therefore \dfrac{\text{Area of △BEF}}{\text{Area of △ABD}} = \dfrac{EF^2}{AD^2} \\[1em] \Rightarrow \dfrac{\text{Area of △BEF}}{\text{Area of △ABD}} = \dfrac{3^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △BEF}}{\text{Area of △ABD}} = \dfrac{9}{25}.$

Hence, area of △BEF : area of △ABD = 9 : 25.

(iii) From (ii) we get,

$\therefore \dfrac{\text{Area of △ABD}}{\text{Area of △BEF}} = \dfrac{25}{9}$

⇒ 9 x Area of △ABD = 25 x Area of △BEF
⇒ 9 x Area of △ABD = 25 x (Area of △ABD - Area of trapezium AEFD)
⇒ 9 x Area of △ABD = 25 x Area of △ABD - 25 x Area of trapezium AEFD
⇒ 16 x Area of △ABD = 25 x Area of trapezium AEFD

$\therefore \dfrac{\text{Area of △ABD}}{\text{Area of trapezium AEFD}} = \dfrac{25}{16}$

Hence, area of △ABD : area of trapezium AEFD = 25 : 16.

(iv) Considering △FEO and △OBC,

∠ FOE = ∠ BOC (Vertically opposite angles are equal)
∠ FEO = ∠ OCB (Alternate angles are equal)

Hence, by AA axiom △FEO ~ △OBC.

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

$\therefore \dfrac{\text{Area of △FEO}}{\text{Area of △OBC}} = \Big(\dfrac{EF}{BC}\Big)^2$

Since, in parallelogram opposite sides are equal so, BC = AD.

$\Rightarrow \dfrac{\text{Area of △FEO}}{\text{Area of △OBC}} = \Big(\dfrac{EF}{AD}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of △FEO}}{\text{Area of △OBC}} = \dfrac{3^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △FEO}}{\text{Area of △OBC}} = \dfrac{9}{25}.$

Hence, the ratio of the area of △FEO : area of △OBC = 9 : 25.

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 of △CDP.

(iii) area of ||gm ABCD.

Answer

(i) Draw QN ⊥ CB as shown in the figure below:

$\Rightarrow \dfrac{\text{Area of △BPQ}}{\text{Area of △CPQ}} = \dfrac{\dfrac{1}{2} BP \times QN}{\dfrac{1}{2} PC \times QN} \\[1em] \Rightarrow \dfrac{\text{Area of △BPQ}}{\text{Area of △CPQ}} = \dfrac{BP}{PC} \\[1em] \Rightarrow \dfrac{\text{Area of △BPQ}}{\text{Area of △CPQ}} = \dfrac{1}{2} \\[1em] \therefore \text{Area of △BPQ} = \dfrac{1}{2}\text{Area of △CPQ} = \dfrac{1}{2} \times 20 = 10 \text{cm}^2.$

Hence, the area of △BPQ = 10 cm².

(ii) Considering △CDP and △BQP,

∠CPD = ∠QPB (Vertically opposite angles are equal)
∠PDC = ∠PQB (Alternate angles are equal)

Hence, by AA axiom △CDP ~ △BQP.

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

$\therefore \dfrac{\text{Area of △CDP}}{\text{Area of △BQP}} = \dfrac{PC^2}{BP^2} \\[1em] \Rightarrow \dfrac{\text{Area of △CDP}}{\text{Area of △BQP}} = \dfrac{2^2}{1^2} \\[1em] \Rightarrow \dfrac{\text{Area of △CDP}}{\text{Area of △BQP}} = \dfrac{4}{1}$

∴ Area of △CDP = 4 × Area of △BQP = 4 × 10 = 40 cm².

Hence, the area of △CDP = 40 cm².

(iii) Area of ||gm ABCD = 2 Area of △DCQ (As △DCQ and ||gm ABCD have same base and are between same parallels)

$= 2(\text{Area of △CDP + Area of △CPQ}) \\[1em] = 2(40 + 20) \\[1em] = 2 \times 60 \\[1em] = 120 \text{ cm}^2.$

Hence, the area of ||gm = 120 cm².

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.

Answer

Given, ratio of the areas of △ADE and trapezium DBCE = 4 : 5.

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of trapezium DBCE}} = \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ABC - Area of △ADE}} = \dfrac{4}{5} \\[1em] \Rightarrow 5 \text{ Area of △ADE} = 4 (\text{Area of △ABC - Area of △ADE}) \\[1em] \Rightarrow 9 \text{ Area of △ADE} = 4 \text{Area of △ABC} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{4}{9}.$

Considering △ABC and △ADE,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

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

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{DE^2}{BC^2} \\[1em] \Rightarrow \dfrac{4}{9} = \Big(\dfrac{DE}{BC}\Big)^2 \\[1em] \Rightarrow \dfrac{DE}{BC} = \sqrt{\dfrac{4}{9}} \\[1em] \Rightarrow \dfrac{DE}{BC} = \dfrac{2}{3}.$

Hence, the ratio of DE : BC is 2 : 3.

In the figure (ii) given below, AB || DC and AB = 2DC. 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.

Answer

(i) Given, AB = 2DC or, $\dfrac{AB}{DC} = \dfrac{2}{1}.$

Considering △AEB and △EDC.

∠E = ∠E (Common angles)

∠EDC = ∠EAB (Corresponding angles are equal)

Hence, by AA axiom △AEB ~ △EDC.

Since triangles are similar, hence the ratio of the corresponding sides will be equal

$\therefore \dfrac{AE}{ED} = \dfrac{AB}{DC} \\[1em] \Rightarrow \dfrac{AD + ED}{ED} = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{3 + ED}{ED} = \dfrac{2}{1} \\[1em] \Rightarrow 3 + ED = 2ED \\[1em] \Rightarrow ED = 3.$

Hence, the length of ED = 3 cm.

(ii) Since, △AEB ~ △EDC. Hence the ratio of the corresponding sides will be equal

$\therefore \dfrac{BE}{EC} = \dfrac{AB}{DC} \\[1em] \Rightarrow \dfrac{BC + EC}{EC} = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{4 + EC}{EC} = \dfrac{2}{1} \\[1em] \Rightarrow 4 + EC = 2EC \\[1em] \Rightarrow EC = 4.$