Triangles: Case Study PYQs with Answers

Figures A and C

In Figure A, all corresponding angles are equal.

The corresponding side ratios are 1.2/2.4 = 2.5/5 = 2.4/4.8 = 2/4 = 1/2.

Hence, the quadrilaterals in Figure A are similar.

The corresponding sides of the triangles in Figure C are equal, so the triangles are congruent and therefore similar.

Figure C

In Figure C, the corresponding sides of the two triangles are 2 cm, 3 cm and 4.5 cm.

The triangles are congruent by SSS congruence.

Congruent triangles are similar; similar triangles need not be congruent.

Congruent triangles have equal corresponding angles.

Therefore, they are similar by the AA similarity criterion.

Consider two equilateral triangles of sides 2 cm and 4 cm.

Their corresponding angles are all 60°, so they are similar.

Their corresponding sides are unequal, so they are not congruent.

One pair of corresponding sides must be equal.

Similar triangles have all corresponding sides in the same ratio.

If one pair of corresponding sides is equal, this common ratio is 1.

Therefore, all corresponding sides are equal.

The triangles are congruent by SSS congruence.

△OAC ∼ △OBD

AC ∥ BD.

∠OAC = ∠OBD, since they are alternate interior angles.

∠AOC = ∠BOD, since they are vertically opposite angles.

Therefore, △OAC ∼ △OBD by AA similarity.

OA/AC = OB/BD

Since △OAC ∼ △OBD, corresponding sides are proportional.

$\frac{\text{(}\text{OA}\text{)/(}}{\text{OB}\text{)}}$ = $\frac{\text{(}\text{AC}\text{)/(}}{\text{BD}\text{)}}$.

OA × BD = OB × AC.

Therefore, $\frac{\text{(}\text{OA}\text{)/(}}{\text{AC}\text{)}}$ = $\frac{\text{(}\text{OB}\text{)/(}}{\text{BD}\text{)}}$.

x = 2; OC = 25

By similarity, $\frac{\text{(}\text{OA}\text{)/(}}{\text{OB}\text{)}}$ = $\frac{\text{(}\text{OC}\text{)/(}}{\text{OD}\text{)}}$.

$\frac{\text{(}\text{3x + 4}\text{)/(}}{\text{x}\text{)}}$ = $\frac{\text{(}\text{3x + 19}\text{)/(}}{\text{x + 3}\text{)}}$.

(3x + 4)(x + 3) = x(3x + 19).

3x² + 13x + 12 = 3x² + 19x.

6x = 12.

x = 2.

OC = 3(2) + 19 = 25.

BD/AC = 1/5

By similarity, $\frac{\text{(}\text{OA}\text{)/(}}{\text{OB}\text{)}}$ = $\frac{\text{(}\text{OC}\text{)/(}}{\text{OD}\text{)}}$.

$\frac{\text{(}\text{3x + 4}\text{)/(}}{\text{x}\text{)}}$ = $\frac{\text{(}\text{3x + 19}\text{)/(}}{\text{x + 3}\text{)}}$.

(3x + 4)(x + 3) = x(3x + 19).

6x = 12, so x = 2.

OA = 3(2) + 4 = 10 and OB = 2.

$\frac{\text{(}\text{BD}\text{)/(}}{\text{AC}\text{)}}$ = $\frac{\text{(}\text{OB}\text{)/(}}{\text{OA}\text{)}}$ = $\frac{\text{(}\text{2}\text{)/(}}{\text{10}\text{)}}$ = $\frac{\text{(}\text{1}\text{)/(}}{\text{5}\text{)}}$.

80 m

The vertical objects and their shadows form right triangles.

At the same time, the Sun’s rays make equal angles with level ground, so the triangles are similar by AA.

Let the height of the tower be h metres.

$\frac{\text{(}\text{h}\text{)/(}}{\text{40}\text{)}}$ = $\frac{\text{(}\text{12}\text{)/(}}{\text{6}\text{)}}$.

h = 40 × 2 = 80 m.

24 m

Let the height of Ramesh’s house be h metres.

The height-to-shadow ratio is the same for these similar triangles.

$\frac{\text{(}\text{h}\text{)/(}}{\text{12}\text{)}}$ = $\frac{\text{(}\text{12}\text{)/(}}{\text{6}\text{)}}$.

h = 24 m.

18 m

The tower is 80 m high and Ramesh’s house is 24 m high.

Let the required shadow length be s metres.

$\frac{\text{(}\text{24}\text{)/(}}{\text{s}\text{)}}$ = $\frac{\text{(}\text{80}\text{)/(}}{\text{60}\text{)}}$.

s = $\frac{\text{(}\text{24 × 60}\text{)/(}}{\text{80}\text{)}}$ = 18 m.

7.2 m

The tower is 80 m high and Raghu’s house is 12 m high.

Let the required shadow length be s metres.

$\frac{\text{(}\text{12}\text{)/(}}{\text{s}\text{)}}$ = $\frac{\text{(}\text{80}\text{)/(}}{\text{48}\text{)}}$.

s = $\frac{\text{(}\text{12 × 48}\text{)/(}}{\text{80}\text{)}}$ = 7.2 m.

AC = 32/3 cm

AB = AD + DB = 3 + 5 = 8 cm.

Since DE ∥ BC, △ADE ∼ △ABC by AA similarity.

$\frac{\text{(}\text{AD}\text{)/(}}{\text{AB}\text{)}}$ = $\frac{\text{(}\text{AE}\text{)/(}}{\text{AC}\text{)}}$.

$\frac{\text{(}\text{3}\text{)/(}}{\text{8}\text{)}}$ = $\frac{\text{(}\text{4}\text{)/(}}{\text{AC}\text{)}}$.

AC = $\frac{\text{(}\text{32}\text{)/(}}{\text{3}\text{)}}$ cm.

85°

In △ADE, ∠AED = 180° − 50° − 45° = 85°.

Since DE ∥ BC, ∠ACB = ∠AED by corresponding angles.

Therefore, ∠ACB = 85°.

DE/BC = 2/5

AB = AD + DB = 4 + 6 = 10 cm.

Since △ADE ∼ △ABC, $\frac{\text{(}\text{DE}\text{)/(}}{\text{BC}\text{)}}$ = $\frac{\text{(}\text{AD}\text{)/(}}{\text{AB}\text{)}}$.

$\frac{\text{(}\text{DE}\text{)/(}}{\text{BC}\text{)}}$ = $\frac{\text{(}\text{4}\text{)/(}}{\text{10}\text{)}}$ = $\frac{\text{(}\text{2}\text{)/(}}{\text{5}\text{)}}$.

AB/AC = 3/5

By the Basic Proportionality Theorem, $\frac{\text{(}\text{AD}\text{)/(}}{\text{DB}\text{)}}$ = $\frac{\text{(}\text{AE}\text{)/(}}{\text{EC}\text{)}}$.

$\frac{\text{(}\text{3}\text{)/(}}{\text{6}\text{)}}$ = $\frac{\text{(}\text{5}\text{)/(}}{\text{EC}\text{)}}$.

EC = 10 cm.

AB = 3 + 6 = 9 cm.

AC = 5 + 10 = 15 cm.

$\frac{\text{(}\text{AB}\text{)/(}}{\text{AC}\text{)}}$ = $\frac{\text{(}\text{9}\text{)/(}}{\text{15}\text{)}}$ = $\frac{\text{(}\text{3}\text{)/(}}{\text{5}\text{)}}$.

θ = α

RT is perpendicular to AB.

α + ∠PRT = 90°.

θ + ∠CRT = 90°.

Given ∠PRT = ∠CRT, their complementary angles are equal.

Therefore, θ = α.

△PQR ∼ △CBR

∠PQR = ∠CBR = 90°.

From part (i), ∠PRQ = ∠CRB.

Therefore, △PQR ∼ △CBR by AA similarity.

x = 36.4 cm

AQ = PS = 9 cm.

QR = AB − BR − AQ = 65 − x − 9 = 56 − x cm.

Since △PQR ∼ △CBR, $\frac{\text{(}\text{PQ}\text{)/(}}{\text{CB}\text{)}}$ = $\frac{\text{(}\text{QR}\text{)/(}}{\text{BR}\text{)}}$.

$\frac{\text{(}\text{35}\text{)/(}}{\text{65}\text{)}}$ = $\frac{\text{(}\text{56 − x}\text{)/(}}{\text{x}\text{)}}$.

35x = 65(56 − x).

100x = 3640.

x = 36.4 cm.

x = 36.4 cm

QR = 65 − x − 9 = 56 − x cm.

Area △PQR = (1/2) × 35 × (56 − x).

Area △CBR = (1/2) × 65 × x.

Using the area relation given in the question, $\frac{\text{(}\text{35(56 − x)}\text{)/(}}{\text{65x}\text{)}}$ = $\frac{\text{(}\text{35²}\text{)/(}}{\text{65²}\text{)}}$.

$\frac{\text{(}\text{56 − x}\text{)/(}}{\text{x}\text{)}}$ = $\frac{\text{(}\text{35}\text{)/(}}{\text{65}\text{)}}$.

65(56 − x) = 35x.

100x = 3640.

x = 36.4 cm.

△DPQ ∼ △DEF

PQ ∥ EF.

∠DPQ = ∠DEF, by corresponding angles.

∠PDQ = ∠EDF, the common angle at D.

Therefore, △DPQ ∼ △DEF by AA similarity.

PQ/EF = 5/12

DE = DP + PE = 50 + 70 = 120 cm.

Since △DPQ ∼ △DEF, $\frac{\text{(}\text{PQ}\text{)/(}}{\text{EF}\text{)}}$ = $\frac{\text{(}\text{DP}\text{)/(}}{\text{DE}\text{)}}$.

$\frac{\text{(}\text{PQ}\text{)/(}}{\text{EF}\text{)}}$ = $\frac{\text{(}\text{50}\text{)/(}}{\text{120}\text{)}}$ = $\frac{\text{(}\text{5}\text{)/(}}{\text{12}\text{)}}$.

Perimeter ratio = 5/2

2AB = 5DE gives $\frac{\text{(}\text{AB}\text{)/(}}{\text{DE}\text{)}}$ = $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$.

Since △ABC ∼ △DEF, $\frac{\text{(}\text{AB}\text{)/(}}{\text{DE}\text{)}}$ = $\frac{\text{(}\text{BC}\text{)/(}}{\text{EF}\text{)}}$ = $\frac{\text{(}\text{AC}\text{)/(}}{\text{DF}\text{)}}$ = $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$.

AB = (5/2)DE, BC = (5/2)EF and AC = (5/2)DF.

$\frac{\text{(}\text{AB + BC + AC}\text{)/(}}{\text{DE + EF + DF}\text{)}}$ = $\frac{\text{(}\text{(5/2)(DE + EF + DF)}\text{)/(}}{\text{DE + EF + DF}\text{)}}$.

Therefore, the ratio of the perimeters is the constant 5/2.

△ABM ∼ △DEN, using the parent-triangle similarity intended by the official scheme.

The printed OR does not repeat the condition △ABC ∼ △DEF. The official marking scheme uses this similarity; medians alone would not prove the claim.

Using △ABC ∼ △DEF, $\frac{\text{(}\text{AB}\text{)/(}}{\text{DE}\text{)}}$ = $\frac{\text{(}\text{BC}\text{)/(}}{\text{EF}\text{)}}$ and ∠ABC = ∠DEF.

Since AM and DN are medians, BM = BC/2 and EN = EF/2.

$\frac{\text{(}\text{BM}\text{)/(}}{\text{EN}\text{)}}$ = $\frac{\text{(}\text{BC}\text{)/(}}{\text{EF}\text{)}}$ = $\frac{\text{(}\text{AB}\text{)/(}}{\text{DE}\text{)}}$.

As M lies on BC and N lies on EF, ∠ABM = ∠DEN.

Therefore, △ABM ∼ △DEN by SAS similarity.