CBSE Class 10 Triangles Sure Shot Questions Set B

Question. O is the point of intersection of two equal chords AB and CD such that OB = OD, then triangles OAC and ODB are
(a) Equilateral but not similar
(b) Isosceles but not similar
(c) Equilateral and similar
(d) Isosceles and similar
Answer: (D)
Explanation: Since O is the point of intersection of two equal chords AB and CD such that OB = OD,
As chords are equal and OB = OD, so AO will also be equal to OC
Also \( \angle AOC = \angle DOB = 45^\circ \)
Now in triangles OAC and ODB
\( \frac{AO}{OB} = \frac{CO}{OD} \)
And \( \angle AOC = \angle DOB = 45^\circ \)
So triangles are isosceles and similar

Question. D and E are respectively the midpoints on the sides AB and AC of a triangle ABC and BC = 6 cm. If DE || BC, then the length of DE (in cm) is
(a) 2.5
(b) 3
(c) 5
(d) 6
Answer: B
Explanation: By midpoint theorem,
If D and E are respectively the midpoints on the sides AB and AC of a triangle ABC, DE||BC and BC = 6 cm
So, DE will be half of BC i.e. 3cm

Question. In triangle PQR, if PQ = 6 cm, PR = 8 cm, QS = 3 cm, and PS is the bisector of angle QPR, what is the length of SR?
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (B)
Explanation: Since, PS is the angle bisector of angle QPR
So, by angle bisector theorem,
\( \frac{QS}{SR} = \frac{PQ}{PR} \)
\( \Rightarrow \frac{3}{SR} = \frac{6}{8} \)
\( \Rightarrow SR = \frac{(3 \times 8)}{6} \text{ cm} = 4 \text{ cm} \)

Question. The lengths of the diagonals of a rhombus are 16 cm and 12cm. Then, the length of the side of the rhombus is
(a) 9 cm
(b) 10 cm
(c) 8 cm
(d) 20 cm
Answer: (B)
Explanation: The diagonals of rhombus bisect each other at right angle, so side of rhombus is the hypotenuse for the triangles formed.
Therefore,
By Pythagoras theorem
\( (\frac{16}{2})^2 + (\frac{12}{2})^2 = \text{Side}^2 \)
\( \Rightarrow 8^2 + 6^2 = \text{Side}^2 \)
\( \Rightarrow 64 + 36 = \text{Side}^2 \)
\( \Rightarrow \text{Side} = 10 \text{ cm} \)

Question. A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.
(a) 25.6
(b) 20.4
(c) 23.7
(d) 32.5
Answer: (B)
Explanation: According to given question, the far end of shadow is represented by point A, There fore we need to Find AC
By Pythagoras theorem,
\( (18)^2 + (9.6)^2 = (AC)^2 \)
\( \Rightarrow AC^2 = 416.16 \)
\( \Rightarrow AC = 20.4 \text{ m (approx)} \)

Question. Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3 RS, Then the ratio of areas of triangles POQ and ROS is:
(a) 1:9
(b) 9:1
(c) 3:1
(d) 1:3
Answer: (B)
Explanation: According to given Question Since SR || PQ,
So, \( \angle OSR = \angle OQP \) (alternate interior angles)
Also \( \angle SOR = \angle POQ \) (vertically opposite angles)
So triangles SOR and POQ are similar,
Therefore,
\( \frac{ar(POQ)}{ar(SOR)} = (\frac{PQ}{SR})^2 \)
\( \frac{ar(POQ)}{ar(SOR)} = (\frac{3 SR}{SR})^2 \)
\( \frac{ar(POQ)}{ar(SOR)} = \frac{9}{1} \)

Question. ABCD is a trapezium in which AB|| DC and P, Q are points on AD and BC respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.
(a) 55cm
(b) 57cm
(c) 60cm
(d) 62cm
Answer: (C)
Explanation: According to question ABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm,
In triangle ABD
\( \frac{DP}{AP} = \frac{OD}{OB} \)
In triangle BDC
\( \frac{BQ}{QC} = \frac{OB}{OD} \)
This implies
\( \frac{DP}{AP} = \frac{QC}{BQ} \)
\( \frac{18}{AP} = \frac{15}{35} \)
\( AP = (18 \times 35)/15 \)
\( AP = 42 \)
Therefore, \( AD = AP + DP = 42 + 18 = 60\text{cm} \)

Question. Areas of two similar triangles are 36 cm2 and 100 cm2. If the length of a side of the larger triangle is 20 cm, then the length of the corresponding side of the smaller triangle is:
(a) 12cm
(b) 13cm
(c) 14cm
(d) 15cm
Answer: (A)
Explanation: Let the side of smaller triangle be x cm.
\( \frac{ar(\text{Larger Triangle})}{ar(\text{Smaller Triangle})} = (\frac{\text{side of larger triangle}}{\text{side of smaller triangle}})^2 \)
\( \frac{100}{36} = (20/x)^2 \)
\( x = \sqrt{144} \)
\( X = 12 \text{ cm} \)

Question. In the figure if \( \angle ACB = \angle CDA \), AC = 8 cm and AD = 3 cm, find BD.
(a) 53/3 cm
(b) 55/3 cm
(c) 64/3 cm
(d) 35/7 cm
Answer: (B)
Explanation: In triangle ACB and ADC
\( \angle A = \angle A \)
\( \angle ACB = \angle CDA \)
There fore triangle ACB and ADC are similar,
Hence
\( \frac{AC}{AD} = \frac{AB}{AC} \)
\( AC^2 = AD \times AB \)
\( 8^2 = 3 \times AB \)
\( \Rightarrow AB = 64/3 \)
This implies,
\( BD = 64/3 – AD \)
\( \Rightarrow BD = 55/3 \)

Question. If ABCD is parallelogram, P is a point on side BC and DP when produced meets AB produced at L, then select the correct option
(a) DP/BL = DC/PL
(b) DP/PL = DC/BL
(c) DP/PL = BL/DC
(d) DP/PL = AB/DC
Answer: (B)
Explanation: In \( \Delta ALD \), we have
BP || AD
\( \therefore LB/BA = LP/PD \)
\( \Rightarrow BL/AB = PL/DP \)
\( \Rightarrow BL/DC = PL/DP \) [\( \because AB = DC \)]
\( \Rightarrow DP/PL = DC/BL \)

Question. In the figure given below DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, the value of x is:
(a) 4
(b) 8
(c) 16
(d) 32
Answer: (A)
Explanation: In triangle ABC, we have DE || BC
\( \therefore AD/DB = AE/EC \) (By Thale’s Theorem)
\( \Rightarrow x/x - 2 = (x + 2)/(x - 1) \)
\( \Rightarrow x (x - 1) = (x - 2)(x + 2) \)
\( \Rightarrow x^2 – x = x^2 – 4 \)
\( \Rightarrow x = 4 \)

Question. The length of altitude of an equilateral triangle of side 8cm is
(a) √3 cm
(b) 2√3 cm
(c) 3√3 cm
(d) 4√3 cm
Answer: (D)
Explanation: The altitude divides the opposite side into two equal parts, Therefore, BD = DC = 4 cm
In triangle ABD
\( AB^2 = AD^2 + BD^2 \)
\( 8^2 = AD^2 + 4^2 \)
\( AD^2 = 64 – 16 \)
\( AD^2 = 48 \)
\( AD = 4\sqrt{3} \text{ cm} \)

Question. If \( \Delta ABC \sim \Delta DEF \), AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of ABC.
(a) 18 cm
(b) 20 cm
(c) 21 cm
(d) 22 cm
Answer: (A)
Explanation: According to question, \( \Delta ABC \sim \Delta DEF \), AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm,
Therefore,
\( AB/DE = BC/EF = AC/DF \)
\( 4/6 = BC/9 = AC/12 \)
\( \Rightarrow 4/6 = BC/9 \)
\( \Rightarrow BC = 6 \text{ cm} \)
And
\( 4/6 = AC/12 \)
\( \Rightarrow AC = 8 \text{ cm} \)
Perimeter = AB + BC + CA = 4 + 6 + 8 = 18 cm

Question. A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then the distance by which the top of the ladder would slide upwards on the wall is:
(a) 2 m
(b) 1.2 m
(c) 0.8 m
(d) 0.5 m
Answer: (C)
Explanation: Let AC be the ladder of length 5m and BC = 4m be the height of the wall where ladder is placed. If the foot of the ladder is moved 1.6m towards the wall i.e. AD = 1.6 m, then the ladder is slided upward to position E i.e. CE = x m.
In right triangle ABC
\( AC^2 = AB^2 + BC^2 \)
\( \Rightarrow 5^2 = AB^2 + 4^2 \)
\( \Rightarrow AB = 3\text{m} \)
\( \Rightarrow DB = AB – AD = 3 – 1.6 = 1.4\text{m} \)
In right angled \( \Delta EBD \)
\( ED^2 = EB^2 + BD^2 \)
\( \Rightarrow 5^2 = EB^2 + (1.4)^2 \)
\( \Rightarrow EB = 4.8\text{m} \)
EC = EB – BC = 4.8 – 4 = 0.8m
Hence the top of the ladder would slide upwards on the wall at distance 0.8 m.

Question. Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm2, then the area of the larger triangle is:
(a) 108 m2
(b) 107 m2
(c) 106 m2
(d) 230 m2
Answer: (A)
Explanation: According to given Question
\( ar(\text{Larger Triangle})/ar(\text{Smaller Triangle}) = (\text{side of larger triangle}/\text{side of larger triangle})^2 \)
\( ar(\text{Larger Triangle})/48 = (3/2)^2 \)
\( ar(\text{Larger Triangle}) = (9 \times 48 )/4 \)
\( ar(\text{Larger Triangle}) = 108 \text{ cm}^2 \)