Area Questions and Answers – Polygons

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

1. Which of the following options does not describes the interior angles of a convex polygon correctly?
a) Obtuse angle
b) Acute angle
c) Reflex angle
d) Right angle
View Answer

Answer: c
Explanation: A polygon is called a convex polygon if all the interior angles are less than 180°. This means that the angle can be acute, right and obtuse. Thus, reflex angle is the wrong option.

2. How many sides does a polygon having sum of interior angles as 2340 will have?
a) 15
b) 16
c) 13
d) 12
View Answer

Answer: a
Explanation: Given,
Sum of interior angles = 2340
Let the number of sides be n.
➩ 2340 = (n – 2) * 180
➩ 2340 = 180n – 360
➩ 2340 + 360 = 180n
➩ 2700 = 180n
➩ 2700 / 180 = n
➩ n = 15

3. Which of the following is the correct option with respect to the area of a regular pentagon of side z?
a) 1 / 5 * (5(5 + 2√5))1/2 * z2
b) \(\frac {1}{4}\) (5(5 + 2√5))1/2 * z2
c) \(\frac {1}{4}\) (5 + 2√5)1/2 * z2
d) \(\frac {1}{4}\) (25(5 + 2√5))1/2 * z5
View Answer

Answer: b
Explanation: Given,
Side of regular pentagon = z.
Correct formula for area of a regular pentagon = \(\frac {1}{4}\) (5(5 + 2√5))1/2 * z2.
advertisement
advertisement

4. What will be the measure of each interior angle of a regular hexagon in degrees?
a) 120
b) 140
c) 80
d) 60
View Answer

Answer: a
Explanation: As we know,
Sum of interior angles of polygon of side n = (n – 2) * 180
For a regular hexagon n = 6
Sum of interior angles = (6 – 2) * 180
= 4 * 180
= 720.
Now the hexagon will have 6 equal interior angles thus,
Each interior angle = 720 / 6
= 120 degrees

5. Which of the following options represents the correct expression for calculating the area of a regular hexagon of side a?
a) (3√3 / 2) * a3
b) (6√3 / 2) * a2
c) (3√3 / 2) * a2
d) (3√9 / 2) * a3
View Answer

Answer: c
Explanation: Given,
The side of regular hexagon = a
Correct formula for area of regular hexagon = (3√3 / 2) * a2.

6. If the base of a rhombus is 8 cm and its altitude is 5 cm the, what will be the area (in sq. cm) of this rhombus?
a) 60
b) 30
c) 20
d) 40
View Answer

Answer: d
Explanation: Given,
l = 8 cm, h = 5 cm
Area of rhombus = l * h
= 8 * 5
= 40 cm2

7. A four sided polygon PQRS is such that it is formed of two equilateral triangles PQS and QRS joined together. The area of a PQRS is 8√3 cm2, then what will be the perimeter (in cm) of PQRS?
a) 16
b) 32
c) 12
d) 20
View Answer

Answer: a
Explanation: Given,
Area of PQRS = 8√3 cm2 = 2 * Area of PQS = 2 * Area of QRS
Let the length of side of PQRS = x cm
Area of PQS = √3 / 4 * x2
On equating
➩ 2 * √3 / 4 * x2 = 8√3
➩ √3 / 4 * x2 = 4√3
➩ x2 = 4√3 * 4 / √3
➩ x2 = 16 cm2
➩ x = 4 cm.
Perimeter of PQRS = 4 * x
= 4 * 4 cm
= 16 cm
advertisement

8. By what percentage is the area of a regular pentagon increased if its side is doubled?
a) 600%
b) 500%
c) 300%
d) 400%
View Answer

Answer: c
Explanation: Let the original length = x
Original area = \(\frac {1}{4}\) (5(5 + 2√5))1/2 * x2
New length = 2x
New area = \(\frac {1}{4}\) (5(5 + 2√5))1/2 * (2x)2
= \(\frac {1}{4}\) (5(5 + 2√5))1/2 * 4x2
% increase in area = difference in area / original area * 100
= 300%.

9. What will be the area (in sq. cm) of an equilateral triangle of side 1 cm?
a) √3 / 2
b) √3 / 4
c) √3 / 8
d) 3√3 / 4
View Answer

Answer: b
Explanation: Given,
a = 1cm
Area = √3 / 4 * a2
= √3 / 4 * (1)2
= √3 / 4 cm2
advertisement

10. What will be the area (in sq. m) of a hexagonal field having side as 2m?
a) 6√3
b) 3√3
c) 3√3 / 2
d) 3√3 / 4
View Answer

Answer: a
Explanation: Given,
a = 2m
Area = 3√3 / 2 * a2
= 3√3 / 2 * (2)2
= 3√3 / 2 * 4
= 6√3 m2

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

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
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.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.