Area Questions and Answers – Quadrilaterals

This set of Aptitude Questions and Answers (MCQs) focuses on “Quadrilaterals”.

1. Which of the following options correctly justifies the expression half of the product of the sum of the lengths of parallel sides and the parallel distance between them?
a) Area of triangle
b) Area of parallelogram
c) Area of trapezium
d) Area of rectangle
View Answer

Answer: c
Explanation: Formula for the area of a trapezium = half of the sum of the lengths of parallel sides * perpendicular distance between them.

2. The diagonal of a quadrilateral shaped field is 20 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 10 m. What will be the area (in sq. m) of this field?
a) 180
b) 200
c) 220
d) 360
View Answer

Answer: a
Explanation: Given,
d = 20 m, h1 = 8 m, h2 = 10 m
Area = (1 / 2) * d * (h1 + h2)
= (1 / 2) * 20 * (8 + 10)
= (1 / 2) * 20 * (18)
= 180 m2

3. What will be the area (in sq. cm) of a rhombus whose side is 8 cm and altitude is 6 cm?
a) 36
b) 48
c) 67
d) 59
View Answer

Answer: b
Explanation: Given,
l = 8 cm, d = 6 cm
Area of rhombus = length of side * altitude
Area = l * d
= 8 cm * 6 cm
= 48 cm2
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4. If a triangle and a parallelogram are on the same base and between same horizontals, then what is the ratio of three times the area of the triangle to two times the area of parallelogram?
a) 3:4
b) 4:3
c) 3:7
d) 7:4
View Answer

Answer: a
Explanation:
Area of triangle = (1 / 2) * b1 * h1
Area of parallelogram = b2 * h2
According to the question,
b1 = b2 = b and h1 = h2 = h
Thus, the ratio of area of triangle to area of parallelogram
➩ 3 * (1 / 2) * b * h : 2 * b * h
➩ (3 / 2) : 2
➩ 3 : 4

5. What will be the area (in sq. m) of this rhomboid field which has one of its diagonal measuring 10 m and the other 12 m?
a) 150
b) 180
c) 120
d) 240
View Answer

Answer: c
Explanation: Given,
d1 = 10 m, d2 = 12 m
Area of rhombus = d1 * d2
= 10 * 12 m2
= 120 m2

6. What will be the of a trapezoid field having altitude 10 m and length of opposite parallel sides as 6 m and 12 m?
a) 150 m2
b) 180 m2
c) 110 m2
d) 90 m2
View Answer

Answer: d
Explanation: Given,
h = 10m, a = 6m, b = 12m
Area of trapezium = (1 / 2) * (a + b) * h
= (1 / 2) * (6 + 12) * 10
= (1 / 2) * (18) * 10
= 90 m2

7. If the base of a parallelogram is 11 cm and height is 12 cm. Which of the following options represents the correct value of area of the parallelogram?
a) 132 cm2
b) 66 cm2
c) 33 cm2
d) 264 cm2
View Answer

Answer: a
Explanation: Given,
b = 11 cm, h = 12 cm
Area of parallelogram = b * h
= 11 * 12
= 132 cm2
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8. What will be the area (in sq. cm) of a kite having smaller diagonal as 6 cm and longer diagonal as 13 cm?
a) 65
b) 56
c) 39
d) 78
View Answer

Answer: c
Explanation: Given,
p = 6 cm, q = 13 cm.
Area of a kite = (p * q) / 2
Area = (6 * 13) / 2
= 39 cm2

9. If the height of a trapezium is 9 cm and the sum of parallel sides is 17 cm, then, what will be the area (in sq. cm) of the trapezium?
a) 153
b) 76.5
c) 112
d) 132
View Answer

Answer: b
Explanation: Given,
h = 9 cm, (a + b) = 17 cm
Area of trapezium = (1 / 2) * h * (a + b)
= (1 / 2) * 9 * 17
= 76.5 cm2
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10. If the side of a square is increased to 1.5 times the original, then, by how times is the area increased?
a) 2.25 times
b) 4.50 times
c) 1.5 times
d) 3.00 times
View Answer

Answer: a
Explanation: Let the original length = a
New length = 1.5 a
Original area = a2
New area = (1.5 a)2
= 2.25 a2
Thus, on comparing new area is 2.25 times the original area.

To practice all aptitude questions, please visit “1000+ Quantitative Aptitude Questions”, “1000+ Logical Reasoning Questions”, and “Data Interpretation Questions”.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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