Chapter 6 Triangles Class 10 MCQs | NCERT Maths Chapter-6 – Edu grown
1. O is the point of intersection of two equal chords ABand 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 ∠AOC = ∠DOB = 450
Now in triangles OAC and ODB
AO/OB = CO/OD
And ∠AOC = ∠DOB = 450
So triangles are isosceles and similar
2. 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
3. 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,
QS/SR = PQ/PR
⇒ 3/SR = 6/8
⇒ SR = (3 X 8)/6 cm = 4 cm
4. The lengths of the diagonals of a rhombus are 16 cm and 12cm. Then, thelength 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
(16/2)2 + (12/2)2 = Side2
⇒ 82 + 62 = Side2
⇒ 64 + 36 = Side2
⇒ Side = 10 cm
5. 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,
Therefore we need to Find AC
By Pythagoras theorem,
(18)2 + (9.6)2 = (AC)2
⇒ AC2 = 416.16
⇒ AC = 20.4 m (approx)
6. 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, ∠OSR= ∠OQP (alternate interior angles)
Also ∠SOR= ∠POQ (vertically opposite angles)
So triangles SOR and POQ are similar,
Therefore,
ar(POQ)/ar(SOR) = (PQ/SR)2
ar(POQ)/ar(SOR) = (3 SR/SR)2
ar(POQ)/ar(SOR) = 9/1
7. ABCD is a trapezium in which AB|| DC and P, Q are points on ADand BC respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm andQC = 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
DP/AP = OD/OB
In triangle BDC
BQ/QC = OB/OD
This implies
DP/AP = QC/BQ
18/AP = 15/35
AP = (18 x 35)/15
AP = 42
Therefore, AD = AP + DP = 42 + 18 = 60cm
8. 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.
ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of smaller triangle)2
100/36 = (20/x)2
x = √144
X = 12 cm
9. In the figure if ∠ACB = ∠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
∠A=∠A
∠ACB = ∠CDA
Therefore triangle ACB and ADC are similar,
Hence
AC/AD = AB/AC
AC2 = AD X AB
82 = 3 x AB
⇒ AB = 64/3
This implies,
BD = 64/3 – AD
⇒ BD = 55/3
10. 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 ΔALD, we have
BP || AD
∴ LB/BA = LP/PD
⇒ BL/AB = PL/DP
⇒ BL/DC = PL/DP [∵ AB = DC
⇒ DP/PL = DC/BL
11. 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
∴ AD/DB = AE/EC (By Thale’s Theorem)
⇒ x/x – 2 = (x + 2)/(x – 1)
⇒ x (x – 1) = (x – 2)(x + 2)
⇒ x2 – x = x2 – 4
⇒ x = 4
12. 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
AB2 = AD2 + BD2
82 = AD2 + 42
AD2 = 64 – 16
AD2 = 48
AD = 4√3 cm
13. If ΔABC ~ ΔDEF, 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,
ΔABC ~ ΔDEF,
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
⇒ 4/6 = BC/9
⇒ BC = 6 cm
And
4/6 = AC/12
⇒ AC = 8 cm
Perimeter = AB + BC + CA
= 4 + 6 + 8
= 18 cm
14. 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
AC2 = AB2 + BC2
⇒52 = AB2 + 42
⇒ AB = 3m
⇒ DB = AB – AD = 3 – 1.6 = 1.4m
In right angled ΔEBD
ED2 = EB2 + BD2
⇒ 52 = EB2 + (1.4)2
⇒ EB = 4.8m
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.
15. 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(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of larger triangle)2
ar(Larger Triangle)/48 = (3/2)2
ar(Larger Triangle) = (9 x 48 )/4
ar(Larger Triangle) = 108 cm2
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Quick Revision Notes : Triangles
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MCQs: Triangles
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