MCQ on Triangles

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Triangle is a three-sided polygon with three edges and three vertices in geometry. The sum of a triangle's interior angles equals 180 degrees is the most significant feature of a triangle and is known as the angle sum property of the triangle.

Types of triangles based on the angle:

  • Acute Angled Triangle - They measure less than 90°.
  • Obtuse Triangle: They measure greater than 90°.
  • Right Triangle: When one of the angles of a triangle is 90°.

Types of triangles based on the sides:

  • Scalene triangle: When none of the sides of a triangle are equal.
  • Isosceles triangle: When two sides of a triangle are equal.
  • Equilateral Triangle: When all three sides of a triangle are equal.

Pythagorean Theorem

The Pythagorean Theorem states that if you draw three squares, one off each of the sides of a right triangle, the area of the two smaller squares equals the area of the larger square. The hypotenuse is the longest side of the triangle. The sum of the lengths of the other two sides, squared, is always equal to the length of that side, squared.


Sample Questions

Ques. 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

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Ans. 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.

Ques. 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) 6

(C) 4

(D) 8

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Ans. C

Explanation. Since, PS is the angle bisector of angle QPR. So, by angle bisector theorem,

angle bisector of angle QPR

QS/SR = PQ/PR

⇒ 3/SR = 6/8

⇒ SR = (3 X 8)/6 cm = 4 cm

Ques. 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

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Ans: 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

Ques. 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

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Ans: B

Explanation: The far end of shadow is represented by point A, Therefore we need to Find AC. By Pythagoras theorem,

Pythagoras theorem

(18)2 + (9.6)2 = (AC)2

⇒ AC2 = 416.16

⇒ AC = 20.4 m (approx)

Ques. 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

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Ans: B

Explanation: Since,

SR || PQ,

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

Ques. 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

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Ans: C

Explanation: Since, 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

trapezium

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

Ques. 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

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Ans: A

Explanation: Let the side of the 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

Ques. 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

Click here for the answer

Ans: B

Explanation: In ΔALD, we have

BP || AD

∴ LB/BA = LP/PD

⇒ BL/AB = PL/DP

⇒ BL/DC = PL/DP [â?µ AB = DC

⇒ DP/PL = DC/BL

Ques. 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

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Ans: D

Explanation: The altitude divides the opposite side into two equal parts,

 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

Ques. 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

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Ans: 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

Also Read: 

CBSE X Related Questions

1.

Solve the following pair of linear equations by the substitution method. 
(i) x + y = 14 
    x – y = 4   

(ii) s – t = 3 
    \(\frac{s}{3} + \frac{t}{2}\) =6 

(iii) 3x – y = 3 
      9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3 
     0.4x + 0.5y = 2.3 

(v)\(\sqrt2x\) + \(\sqrt3y\)=0
    \(\sqrt3x\) - \(\sqrt8y\) = 0

(vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
    \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

      2.

      The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

      Length (in mm)

      Number of leaves

      118 - 126

      3

      127 - 135 

      5

      136 - 144

      9

      145 - 153

      12

      154 - 162

      5

      163 - 171

      4

      172 - 180

      2

      Find the median length of the leaves. 
      (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

          3.

          The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them

          Monthly consumption 
          (in units)

           Number of consumers

          65 - 85 

          4

          85 - 105

          5

          105 - 125

          13

          125 - 145

          20

          145 - 165

          14

          165 - 185

          8

          185 - 205

          4

              4.
              Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
              (i) 2, 4, 8, 16, . . . .
              (ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
              (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
              (iv) – 10, – 6, – 2, 2, . . .
              (v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
              (vi) 0.2, 0.22, 0.222, 0.2222, . . . .
              (vii) 0, – 4, – 8, –12, . . . .
              (viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
              (ix) 1, 3, 9, 27, . . . .
              (x) a, 2a, 3a, 4a, . . . .
              (xi) a, \(a^2, a^3, a^4,\)  . . . .
              (xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
              (xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
              (xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
              (xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

                  5.

                  Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

                      6.
                      Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

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