CBSE Class 10 Mathematics Triangles MCQs Set 08

Question. In a right-angled triangle ABC, right angled at B, \( AB = \frac{x}{2} \), \( BC = x + 2 \) and \( AC = x + 3 \). The value of x is:
(a) 5
(b) 10
(c) 12
(d) 14
Answer: (b) 10

Question. The lengths of the diagonals of a rhombus are 16 cm and 12 cm. then, the length of the side of the rhombus is:
(a) 9 cm
(b) 10 cm
(c) 8 cm
(d) 20 cm
Answer: (b) 10 cm

Question. If D, E and F are the mid-points of sides BC, CA and AB respectively of \( \Delta ABC \), then the ratio of the areas of \( \Delta DEF \) to the area of \( \Delta ABC \) is:
(a) 1 : 4
(b) 1 : 2
(c) 1 : 3
(d) 2 : 3
Answer: (a) 1 : 4

Question. If \( \Delta ABC \sim \Delta EDF \) and \( \Delta ABC \) is not similar to \( \Delta DEF \), then which of the following is not true?
(a) \( BC \cdot EF = AC \cdot FD \)
(b) \( AB \cdot EF = AC \cdot DE \)
(c) \( BC \cdot DE = AB \cdot EF \)
(d) \( BC \cdot DE = AB \cdot FD \)
Answer: (c) \( BC \cdot DE = AB \cdot EF \)

Question. If in two triangles ABC and PQR, \( \frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ} \), then:
(a) \( \Delta PQR \sim \Delta CAB \)
(b) \( \Delta PQR \sim \Delta ABC \)
(c) \( \Delta CBA \sim \Delta PQR \)
(d) \( \Delta BCA \sim \Delta PQR \)
Answer: (a) \( \Delta PQR \sim \Delta CAB \)

Question. In the given figure, two line segments AC and BD intersect each other at point P such that \( PA = 6 \) cm, \( PB = 3 \) cm, \( PC = 2.5 \) cm, \( PD = 5 \) cm, \( \angle APB = 50^\circ \) and \( \angle CDP = 30^\circ \). Then, \( \angle PBA \) is equal to:
(a) \( 50^\circ \)
(b) \( 30^\circ \)
(c) \( 60^\circ \)
(d) \( 100^\circ \)
Answer: (d) \( 100^\circ \)

Question. In \( \Delta ABC \) and \( \Delta DEF \), \( \angle B = \angle E \), \( \angle F = \angle C \) and \( AB = 3DE \). Then, the two triangles are:
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Answer: (b) similar but not congruent

Question. If in two triangles \( \Delta DEF \) and \( \Delta PQR \), \( \angle D = \angle Q \) and \( \angle R = \angle E \), then which of the following is not true?
(a) \( \frac{EF}{PR} = \frac{DF}{PQ} \)
(b) \( \frac{DE}{PQ} = \frac{EF}{RP} \)
(c) \( \frac{DE}{QR} = \frac{DF}{PQ} \)
(d) \( \frac{EF}{RP} = \frac{DE}{QR} \)
Answer: (b) \( \frac{DE}{PQ} = \frac{EF}{RP} \)

Question. It is given that \( \Delta ABC \sim \Delta PQR \), with \( \frac{BC}{QR} = \frac{1}{3} \), then \( \frac{ar(PRQ)}{ar(BCA)} \) is equal to:
(a) 9
(b) 3
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{9} \)
Answer: (a) 9
Explanation:
It is given that \( \Delta ABC \sim \Delta PQR \)
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\( \frac{ar(PRQ)}{ar(BCA)} = \left(\frac{QR}{BC}\right)^2 = \left(\frac{QR}{BC}\right)^2 \)
\( \frac{ar(PRQ)}{ar(BCA)} = \left(\frac{3}{1}\right)^2 = 9 \) [Using eq. (i) where \( \frac{BC}{QR} = \frac{1}{3} \)]
Thus, the area of \( \Delta PRQ = 9 \) times the area of \( \Delta BCA \).

Question. If in triangles \( \Delta ABC \) and \( \Delta DEF \), \( \frac{AB}{DE} = \frac{BC}{FD} \), then they will be similar when:
(a) \( \angle B = \angle E \)
(b) \( \angle A = \angle D \)
(c) \( \angle B = \angle D \)
(d) \( \angle A = \angle F \)
Answer: (c) \( \angle B = \angle D \)

Question. If \( \Delta ABC \sim \Delta QRP \), \( \frac{ar(ABC)}{ar(PQR)} = \frac{9}{4} \), \( AB = 18 \) cm and \( BC = 15 \) cm, then PR is equal to:
(a) 10 cm
(b) 12 cm
(c) \( \frac{20}{3} \) cm
(d) 8 cm
Answer: (a) 10 cm

Question. If S is a point on side PQ of a \( \Delta PQR \) such that PS = QS = RS, then:
(a) \( PR \cdot QR = RS^2 \)
(b) \( QS^2 + RS^2 = QR^2 \)
(c) \( PR^2 + QR^2 = PQ^2 \)
(d) \( PS^2 + RS^2 = PR^2 \)
Answer: (c) \( PR^2 + QR^2 = PQ^2 \)a

Fill in the Blanks

Question. Let \( \Delta ABC \sim \Delta DEF \) and their areas be 81 cm\(^2\) and 144 cm\(^2\). If EF = 24 cm, then length of side BC is ...................... cm.
Answer: 18 cm
Explanation: Since \( \Delta ABC \sim \Delta DEF \),
\( \frac{ar(\Delta ABC)}{ar(\Delta DEF)} = \frac{BC^2}{EF^2} \)
\( \Rightarrow \frac{81}{144} = \frac{BC^2}{24^2} \)
\( \Rightarrow BC^2 = \frac{81}{144} \times 24^2 = 324 \)
\( \Rightarrow BC = 18 \) cm

Question. In \( \Delta ABC \), \( AB = 6\sqrt{3} \) cm, \( AC = 12 \) cm and \( BC = 6 \) cm, then \( \angle B = \) ...............................
Answer: \( 90^\circ \)

Question. Two triangles are similar if their corresponding sides are ...............................
Answer: Proportional (by definition)

Question. A ladder 10 m long reaches a window 8 m above the ground. The distance of the foot of the ladder from the base of the wall is ................................. m.
Answer: 6 m.
Explanation:
By Pythagoras Theorem,
\( AB^2 = AC^2 - BC^2 \)
\( = 10^2 - 8^2 \)
\( = 100 - 64 \)
\( = 36 \)
\( \Rightarrow AB = 6 \) m.
Thus, the distance of the foot of the ladder from the base of the wall is 6 m.