Similarity of Triangles — Analytical & Application Based Questions

Given a triangle ABC, and D is a point on BC such that BD = 4 cm and DC = x cm. If ∠BAD = ∠C and AB = 8 cm, then,

(a) prove that triangle ABD is similar to triangle CBA.

(b) find the value of 'x'.

Answer

(a) In △ ABD and △ CBA,

⇒ ∠ABD = ∠CBA (Common angle)

⇒ ∠BAD = ∠ACB (Given)

∴ △ ABD ~ △ CBA (By A.A. axiom)

Hence, proved that △ ABD ~ △ CBA.

(b) We know that,

Corresponding sides of similar triangle are proportional.

$\therefore \dfrac{AB}{BC} = \dfrac{BD}{BA} \\[1em] \Rightarrow \dfrac{8}{x + 4} = \dfrac{4}{8} \\[1em] \Rightarrow 4(x + 4) = 8 \times 8 \\[1em] \Rightarrow 4x + 16 = 64 \\[1em] \Rightarrow 4x = 64 - 16 \\[1em] \Rightarrow 4x = 48 \\[1em] \Rightarrow x = \dfrac{48}{4} = \text{ 12 cm}.$

Hence, x = 12 cm.

ABCD is a rectangle in which side BC is twice side AB. If △ACQ ~ △BAP, find area of △BAP : area of △ACQ.

Answer

Given,

ABCD is a rectangle where side BC is twice side AB.

⇒ BC = 2AB

In right angled triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (2AB)²

⇒ AC² = AB² + 4AB²

⇒ AC² = 5AB²

⇒ AC = $\sqrt{5}$ AB.

We know that,

The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.

$\therefore \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{AC^2} \\[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{(\sqrt{5}BA)^2} \\[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{5BA^2} \\[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{1}{5} \\[1em]$

Area of △BAP : Area of △ACQ = 1 : 5.

Hence, Area of △BAP : Area of △ACQ = 1 : 5.

While preparing a PowerPoint presentation, ∆ ABC is enlarged along the side BC to ∆ AB'C', as shown in the diagram, such that BC ∶ B'C' is 3 ∶ 5. Find :

(a) AB ∶ BB'

(b) length AB, if BB' = 4 cm.

(c) Is ∆ ABC ~ ∆ AB'C' ? Justify your answer.

(d) ar (∆ ABC) : ar (quad. BB'C'C).

Answer

Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.

∴ ∆ ABC and ∆ AB'C' are similar triangles.

(a) We know that,

Ratio of corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{AB'} = \dfrac{BC}{B'C'} \\[1em] \Rightarrow \dfrac{AB}{AB'} = \dfrac{3}{5}$

Let AB = 3x and AB' = 5x

From figure,

⇒ AB' = AB + BB'

⇒ 5x = 3x + BB'

⇒ BB' = 5x - 3x = 2x.

⇒ AB : BB' = 3x : 2x = 3 : 2.

Hence, AB : BB' = 3 : 2.

(b) As,

$\Rightarrow \dfrac{AB}{BB'} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{AB}{4} = \dfrac{3}{2} \\[1em] \Rightarrow AB = \dfrac{3}{2} \times 4 \\[1em] \Rightarrow AB = 6 \text{ cm}.$

Hence, AB = 6 cm.

(c) Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.

∴ BC || B'C'

In ∆ ABC and ∆ AB'C',

⇒ ∠BAC = ∠B'AC' (Common angle)

⇒ ∠ABC = ∠AB'C' (Corresponding angle are equal)

∴ ∆ ABC ~ ∆ AB'C' (By A.A. axiom).

Hence, proved that ∆ ABC ~ ∆ AB'C'.

(d) We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\therefore \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\dfrac{AB}{AB'}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\dfrac{3}{5}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \dfrac{9}{25}.$

Let area of ∆ ABC = 9a and area of ∆ AB'C' = 25a.

Area of quadrilateral BB'C'C = Area of ∆ AB'C' - Area of ∆ ABC = 25a - 9a = 16a.

∴ Area of ∆ ABC : Area of quadrilateral BB'C'C = 9a : 16a = 9 : 16.

Hence, area of ∆ ABC : area of quadrilateral BB'C'C = 9 : 16.

In the given figure (not drawn to scale), BC is parallel to EF, CD is parallel to FG, AE : EB = 2 : 3, ∠BAD = 70°, ∠ACB = 105°, ∠ADC = 40° and AC is bisector of ∠BAD.

(a) Prove Δ AEF ~ Δ AGF

(b) Find :

(i) AG : AD

(ii) area of Δ ACB: area Δ ACD

(iii) area of quadrilateral ABCD: area of Δ ACB.

Answer

(a) In Δ AFE,

⇒ ∠AFE = ∠ACB = 105° (Corresponding angle are equal)

⇒ ∠EAF = $\dfrac{∠BAD}{2} = \dfrac{70°}{2}$ = 35° (AC is the bisector of ∠BAD)

By angle sum property of triangle,

⇒ ∠AFE + ∠EAF + ∠AEF = 180°

⇒ 105° + 35° + ∠AEF = 180°

⇒ 140° + ∠AEF = 180°

⇒ ∠AEF = 180° - 140° = 40°.

∠AGF = ∠ADC = 40° (Corresponding angles are equal)

In Δ AGF,

⇒ ∠GAF = $\dfrac{∠BAD}{2} = \dfrac{70°}{2}$ = 35° (AC is the bisector of ∠BAD)

In Δ AEF and Δ AGF,

⇒ ∠EAF = ∠GAF = 35° (AC being the bisector)

⇒ ∠AEF = ∠AGF = 40° (Proved above)

∴ Δ AEF ~ Δ AGF (By A.A. axiom)

Hence, proved that Δ AEF ~ Δ AGF.

(b) In Δ ACD,

⇒ ∠CAD = 35°

⇒ ∠ADC = 40°

By angle sum property of triangle,

⇒ ∠CAD + ∠ADC + ∠DCA = 180°

⇒ 35° + 40° + ∠DCA = 180°

⇒ ∠DCA + 75° = 180°

⇒ ∠DCA = 180° - 75° = 105°.

In Δ ACD and Δ ACB,

⇒ ∠ACB = ∠ACD = 105° (Proved above)

⇒ ∠BAC = ∠DAC = 35° (Proved above)

∴ Δ ACD ~ Δ ACB (By A.A. axiom)

(i) Given,

⇒ AE : EB = 2 : 3

Let AE = 2x and EB = 3x.

∴ AB = AE + EB = 2x + 3x = 5x.

We know that,

Ratio of corresponding sides are proportional.

$\therefore \dfrac{AE}{AG} = \dfrac{AB}{AD} \\[1em] \Rightarrow \dfrac{AG}{AD} = \dfrac{AE}{AB} \\[1em] \Rightarrow \dfrac{AG}{AD} = \dfrac{2x}{5x} = \dfrac{2}{5}.$

Hence, AG : AD = 2 : 5.

(ii) In △ ABC and △ ADC,

⇒ ∠BAC = ∠DAC (Both equal to 35°)

⇒ ∠ACB = ∠ACD (Both equal to 105°)

⇒ ∠ABC = ∠ADC (Both equal to 40°)

∴ △ ABC and △ ADC are congruent.

We know that,

Area of congruent triangles are equal.

Let area of △ ABC and area of △ ADC = x.

∴ Area of △ ABC : Area of △ ADC = x : x = 1 : 1.

Hence, area of △ ABC : area of △ ADC = 1 : 1.

(iii) From figure,

Area of quadrilateral ABCD = Area of △ ABC + Area of △ ADC = x + x = 2x.

∴ Area of quadrilateral ABCD : Area of △ ACB = 2x : x = 2 : 1.

Hence, area of quadrilateral ABCD : area of △ ACB = 2 : 1.

In the figure given below (not drawn to scale), AD ∥ GE ∥ BC, DE = 18 cm, EC = 3 cm, AD = 35 cm. Find :

(a) AF : FC

(b) length of EF

(c) area(trapezium ADEF) : area(Δ EFC)

(d) BC ∶ GF

Answer

(a) In △ ACD and △ FCE,

⇒ ∠ACD = ∠FCE (Common angles)

⇒ ∠ADC = ∠FEC (Corresponding angles are equal)

∴ △ ACD ~ △ FCE (By A.A. axiom)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \dfrac{FC}{AC} = \dfrac{EC}{CD} \\[1em] \Rightarrow \dfrac{FC}{AC} = \dfrac{EC}{EC + ED} \\[1em] \Rightarrow \dfrac{FC}{AC} = \dfrac{3}{3 + 18} \\[1em] \Rightarrow \dfrac{FC}{AC} = \dfrac{3}{21} \\[1em] \Rightarrow \dfrac{FC}{AC} = \dfrac{1}{7}.$

Let FC = x and AC = 7x.

From figure,

AF = AC - FC = 7x - x = 6x.

$\therefore \dfrac{AF}{FC} = \dfrac{6x}{x} = \dfrac{6}{1}$ = 6 : 1.

Hence, AF : FC = 6 : 1.

(b) Since, △ ACD ~ △ FCE

$\therefore \dfrac{EF}{AD} = \dfrac{EC}{CD} \\[1em] \Rightarrow \dfrac{EF}{35} = \dfrac{3}{21} \\[1em] \Rightarrow EF = \dfrac{3}{21} \times 35 \\[1em] \Rightarrow EF = \dfrac{105}{21} = 5 \text{ cm}.$

Hence, EF = 5 cm.

(c) We know that,

The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.

$\therefore \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \Big(\dfrac{DC}{EC}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \Big(\dfrac{21}{3}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \dfrac{441}{9} \\[1em] \Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \dfrac{49}{1}.$

Let area of △ ADC = 49a and area of △ FCE = a.

Area of trapezium ADEF = Area of △ ADC - Area of △ FCE = 49a - a = 48a.

$\therefore \dfrac{\text{Area of trapezium ADEF}}{\text{Area of △ FCE}} = \dfrac{48a}{a} = \dfrac{48}{1}.$

Hence, area of trapezium ADEF : area of △ EFC = 48 : 1.

(d) In △ AGF and △ ABC,

⇒ ∠AGF = ∠ABC (Corresponding angles are equal)

⇒ ∠GAF = ∠BAC (Common angles)

∴ △ AGF ~ △ ABC (By A.A. axiom)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \dfrac{AC}{AF} = \dfrac{BC}{GF}$

From part (a),

⇒ FC = x and AC = 7x

⇒ AF = AC - FC = 7x - x = 6x.

$\Rightarrow \dfrac{BC}{GF} = \dfrac{7x}{6x} = \dfrac{7}{6}.$

Hence, BC : GF = 7 : 6.