CBSE Class 10 Mathematics Triangles Assignment Set 17

Exercise-1 Objective: Introduction to Similar Triangles and Thales Theorem

Teacher's Note:
1. BD : DC = 5 : 3.

Teacher's Note:
1. Both ratios equal AP : AC.

Teacher's Note:
1. AE is one third of AD.

Teacher's Note:
1. Apply the angle bisector theorem twice and multiply.

Exercise-1 Objective: Criteria for Similarity of Triangles

B-1. The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm, the corresponding side of the other triangle is
(A) 6.2 cm   (B) 3.4 cm   (C) 5.4 cm   (D) 8.4 cm
Answer:
Ans. (C) 5.4 cm
Corresponding sides are in the ratio of the perimeters: \( \dfrac{x}{9} = \dfrac{15}{25} \Rightarrow x = 5.4 \) cm.
In simple words: The ratio of perimeters equals the ratio of corresponding sides.

Teacher's Note:
1. Side ratio equals perimeter ratio.

B-2. Two triangles ABC and PQR are similar. If BC : CA : AB = 2 : 3 : 4, then \( \dfrac{QR}{PR} \) is
(A) \( \dfrac{2}{5} \)   (B) \( \dfrac{1}{2} \)   (C) \( \dfrac{1}{\sqrt{2}} \)   (D) \( \dfrac{2}{3} \)
Answer:
Ans. (D) \( \dfrac{2}{3} \)
QR corresponds to BC and PR to CA, so \( \dfrac{QR}{PR} = \dfrac{BC}{CA} = \dfrac{2}{3} \).
In simple words: Letters in the similarity statement give the correspondence.

Teacher's Note:
1. Match QR with BC and PR with CA.

B-4. The perimeters of two similar triangles ABC and LMN are 60 cm and 48 cm respectively. If LM = 8 cm, the length of AB is
(A) 10 cm   (B) 8 cm   (C) 5 cm   (D) 6 cm
Answer:
Ans. (A) 10 cm
AB corresponds to LM, so \( \dfrac{AB}{LM} = \dfrac{60}{48} \Rightarrow AB = 8 \times \dfrac{5}{4} = 10 \) cm.
In simple words: The ratio of perimeters is the ratio of sides.

Teacher's Note:
1. Scale factor 5/4.

Teacher's Note:
1. Find the two missing sides from the 30-60-90 triangles.

B-6. If in \( \triangle ABC \) and \( \triangle DEF \), \( \dfrac{AB}{DE} = \dfrac{BC}{FD} \), then they will be similar if
(A) \( \angle B = \angle E \)   (B) \( \angle A = \angle D \)   (C) \( \angle B = \angle D \)   (D) \( \angle A = \angle F \)
Answer:
Ans. (C) \( \angle B = \angle D \)
The sides AB, BC include \( \angle B \); the sides DE, FD include \( \angle D \). For SAS similarity the included angles must be equal: \( \angle B = \angle D \).
In simple words: Find the vertex common to both sides in each ratio.

Teacher's Note:
1. Equal included angles complete the SAS test.

B-7. A vertical stick 30 m long casts a shadow 15 m long on the ground. At the same time, a tower casts a shadow 75 m long on the ground. The height of the tower is
(A) 150 m   (B) 100 m   (C) 25 m   (D) 200 m
Answer:
Ans. (A) 150 m
At the same time the ratio of height to shadow is the same: \( \dfrac{h}{75} = \dfrac{30}{15} \Rightarrow h = 150 \) m.
In simple words: Sun rays make equal angles, so the triangles are similar.

Teacher's Note:
1. Height is twice the shadow.

B-8. If the ratio of the corresponding sides of two similar triangles is 2 : 3, then the ratio of their corresponding altitudes is
(A) 3 : 2   (B) 16 : 81   (C) 4 : 9   (D) 2 : 3
Answer:
Ans. (D) 2 : 3
Corresponding altitudes of similar triangles are in the same ratio as the corresponding sides: 2 : 3.
In simple words: Areas scale as the square, lengths scale as the ratio itself.

Teacher's Note:
1. Altitudes scale like sides.

Teacher's Note:
1. The two triangles at O are similar.

Exercise-1 Objective: Areas of Similar Triangles

C-1. In a triangle ABC, a straight line parallel to BC intersects AB and AC at D and E respectively. If the area of ADE is one-fifth of the area of ABC and BC = 10 cm, then DE equals
(A) 2 cm   (B) \( 2\sqrt{5} \) cm   (C) 4 cm   (D) 5 cm
Answer:
Ans. (B) \( 2\sqrt{5} \) cm
\( \dfrac{ar(ADE)}{ar(ABC)} = \dfrac{DE^2}{BC^2} = \dfrac{1}{5} \Rightarrow DE^2 = \dfrac{100}{5} = 20 \).
\( DE = 2\sqrt{5} \) cm.
In simple words: The side ratio is the square root of the area ratio.

Teacher's Note:
1. Area ratio 1 : 5 gives side ratio 1 : √5.

C-2. \( \triangle ABC \sim \triangle PQR \). M is the mid-point of BC and N is the mid-point of QR. If the area of \( \triangle ABC \) is 100 sq. cm, the area of \( \triangle PQR \) is 144 sq. cm and AM = 4 cm, then PN is
(A) 4.8 cm   (B) 12 cm   (C) 4 cm   (D) 5.6 cm
Answer:
Ans. (A) 4.8 cm
Ratio of medians = ratio of sides \( = \sqrt{\dfrac{100}{144}} = \dfrac{5}{6} \).
\( \dfrac{AM}{PN} = \dfrac{5}{6} \Rightarrow PN = \dfrac{6 \times 4}{5} = 4.8 \) cm.
In simple words: Corresponding medians of similar triangles are in the ratio of the sides.

Teacher's Note:
1. Medians scale like sides.

Teacher's Note:
1. One third of each side, one ninth of the area.

Teacher's Note:
1. MN corresponds to NP; square the ratio 3 : 2.

Teacher's Note:
1. DE : BC = 5 : 9.

Exercise-1 Objective: Pythagoras Theorem

D-1. In a triangle ABC, if \( \angle B = 90^\circ \) and D is the point on BC such that BD = 2DC, then
(A) \( AC^2 = AD^2 + 3CD^2 \)   (B) \( AC^2 = AD^2 + 5CD^2 \)   (C) \( AC^2 = AD^2 + 7CD^2 \)   (D) \( AC^2 = AB^2 + 5BD^2 \)
Answer:
Ans. (B) \( AC^2 = AD^2 + 5CD^2 \)
Let DC = k, BD = 2k, BC = 3k. \( AC^2 = AB^2 + 9k^2 \) and \( AD^2 = AB^2 + 4k^2 \).
\( AC^2 - AD^2 = 5k^2 = 5CD^2 \), so \( AC^2 = AD^2 + 5CD^2 \).
In simple words: Take DC = k to keep the algebra short.

Teacher's Note:
1. Subtract the two Pythagoras relations.

D-2. In an isosceles \( \triangle ABC \), if AC = BC and \( AB^2 = 2AC^2 \), then \( \angle C \) is equal to
(A) 45°   (B) 60°   (C) 30°   (D) 90°
Answer:
Ans. (D) 90°
\( AB^2 = AC^2 + AC^2 = AC^2 + BC^2 \), so by the converse of the Pythagoras theorem \( \angle C = 90^\circ \).
In simple words: AC = BC, so 2AC² = AC² + BC².

Teacher's Note:
1. The side AB is opposite the angle C.

Teacher's Note:
1. Use the obtuse-angle form of the cosine rule via Pythagoras.

D-4. P and Q are the mid-points of the sides AB and BC respectively of the triangle ABC, right-angled at B. Then
(A) \( AQ^2 + CP^2 = AC^2 \)   (B) \( AQ^2 + CP^2 = \dfrac{4}{5}AC^2 \)   (C) \( AQ^2 + CP^2 = \dfrac{5}{4}AC^2 \)   (D) \( AQ^2 + CP^2 = \dfrac{3}{5}AC^2 \)
Answer:
Ans. (C) \( AQ^2 + CP^2 = \dfrac{5}{4}AC^2 \)
\( AQ^2 = AB^2 + \dfrac{BC^2}{4} \) and \( CP^2 = BC^2 + \dfrac{AB^2}{4} \).
Sum \( = \dfrac{5}{4}(AB^2 + BC^2) = \dfrac{5}{4}AC^2 \).
In simple words: The medians AQ and CP are hypotenuses of right triangles at B.

Teacher's Note:
1. Each median adds a quarter of the other leg's square.

Teacher's Note:
1. Express each square through AB² and t².

Teacher's Note:
1. CA² − AB² = CD² − BD².