# Area Multiple Choice Questions – Quantitative Aptitude

This post is a collection of Area Questions from Quantitative Aptitude Syllabus. All the questions have been previously asked in competition exams before, hence are extremely important for your preparation.

All the questions are in Multiple Choice format with detailed explanation. Before attempting these questions, i strongly suggest to read the theory about the Area chapter first so that you get well prepared to handle the below questions on your own.

## Area MCQ’s for Competition Exam

(01) The radius of circular wheel is 1.75 meter. The number of revolution it will make in travelling 11 km is?

(a) 800
(b) 900
(c) 1000
(d) 1200

option (c) is right

(02) The area of largest circle that can be drawn inside the square of side 28 cm is:

(a) 17248
(b) 784
(c) 8624
(d) 616

Solution

(03) If the area of triangle with base 12 cm is equal to the area of square with side 12 cm, find the altitude of the triangle

(a) 12 cm
(b) 24 cm
(c) 18 cm
(d) 36 cm

Solution
Base of triangle = 12 cm
Side of square = 12 cm

(04) The sides of a triangle are 3 cm, 4 cm and 5 cm. The area of triangle formed by joining midpoints of the triangle is?

a) 6
b) 3
c) 3/2
d) 3/4

Sides of Triangle ABC = 3 cm, 4 cm, 5 cm
Sides of Triangle DEF = 1.5 cm, 2 cm and 2.5 cm

option (c) is correct

(05) The area of a sector of a circle of radius 5 cm, formed by an arc of length 3.5 cm is?

(a) 8.5 sq cm
(b) 8.75 sq cm
(c) 7.75 sq cm
(d) 7.5 sq cm

Solution
Length of arc = 3.5 cm

option (b) is the right answer

(06) From \ a \ point\ in\ the\ interior\ of\ an\ equilateral\ triangle \\ \\ the\ perpendicular\ distance\ of\ the\ sides\ are\ \sqrt{3\ } \ cm\ ,\ 2\sqrt{3\ } \ cm\ and\ 5\sqrt{3} \ cm.\\ \\ The\ perimeter\ ( in\ cm\ ) \ of\ the\ triangle\ is\\\ \\ A)  64         B)  32\ \                    C)   48                    D) 24 \\\ \\

Let\ each\ side\ of\ equilateral\ \triangle\ ABC\ be\ x\ cm. \\\ \\ In \triangle \ ABC, \\ \\ OD= \sqrt{3} cm \\ \\ OE=2 \sqrt{3} cm \\ \\ OF=5 \sqrt{3} cm \\\ \\ area \ of \ \triangle ABC=ar \triangle BOC+ar \triangle COA+ar \triangle AOB \\\ \\ \frac{\sqrt{3}}{4} x^{2} =\frac{1}{2} \times \ [ BC\times OD+AC\times DE+AB\times OF] \\\ \\ \frac{\sqrt{3}}{4} x^{2} =\frac{1}{2} \times \ \left[ x\times \sqrt{3} +x\times 2\sqrt{3} +x\times 5\sqrt{3}\right] \\\ \\ \frac{\sqrt{3}}{4} x^{2}= \frac{1}{2} \times 8 \sqrt{3} x \\\ \\ x=16\ cm \\\ \\ Therefore , Perimeter of \triangle ABC=3 \times sides=3×16=48cm

(07) The circumference of a circle is 100 cm. The side of a square inscribed in the circle is?

A) \frac{100\sqrt{2}}{\pi }    cm                   B) \frac{50\sqrt{2}}{\pi }      cm \\\ \\ C) \frac{100}{\pi }            cm                   D) 50\sqrt{2}  cm \\\ \\ Read Solution
Circumference\ = \ 2 \pi r=100\ cm \\\ \\ Let\ radius\ of\ the\ circle\ is\ r\ cm \\\ \\ r=  \frac{50}{\pi } \ cm\\\ \\ Diagonal\ of\ square\ = Diameter\ of\ circle\ =\ 2r \\ \\ 2r = \frac{100}{\pi } \ cm \\\ \\ Let\ the\ side\ of\ the\ square\ =a\ cm \\\ \\ Now,\ diagonal\ of\ square\ = \sqrt{a^{2} +\ a^{2}} = \sqrt{2}a \\\ \\ \frac{100}{\pi } = \sqrt{2}a \\\ \\ a= \frac{100}{\sqrt{2} \pi } \\\ \\ a= \frac{100\sqrt{2}}{2\pi } \\\ \\ a= \frac{50\sqrt{2}}{\pi } cm \\\ \\ Hence\ option\ B)\ is\ correct.</p> <p>

(08) The  area of the largest triangle that can be inscribed in a semicircle of radius r cm

A)        2r\ cm^{2}                   B)        r ^{2} \ cm^{2} \\\ \\ C)        2\ cm ^{2}                   D) \frac{1}{2} r^{2} \ cm^{2} \\\ \\ Read Solution
base\ of\ triangle\ ABC\ =\ 2r\ cm \\\ \\ height\ of\ triangle\ ABC\ =\ r \ cm \\\ \\ Area\ of\ triangle\ =\ \frac{1}{2} \times base\ \times height \\\ \\ Area\ of\ triangle\ = \frac{1}{2} \times 2r\ \times r \\\ \\ Area\ of\ triangle\ =r^{2} \ cm^{2} \\\ \\ Hence\ option\ B)\ is\ correct

(09) The area of the greatest circle, which can be inscribed in a square whose perimeter is 120 cm, is 120 cm

A)        \frac{22}{7} \times ( 15)^{2} \ cm^{2}       B)        \frac{22}{7} \times \left(\frac{7}{2}\right)^{2} \ cm^{2} \\\ \\ C)        \frac{22}{7} \times \left(\frac{15}{2}\right)^{2} \ cm^{2}      D) \frac{22}{7} \times \left(\frac{9}{2}\right)^{2} \ cm^{2} \\\ \\ Read Solution
Perimeter\ =\ 120\ cm \\ \\ perimeter\ =\ 4 \times sides \\\ \\ sides= \frac{perimeter}{4} \\ \\ sides =\frac{120}{4} =30\ cm \\\ \\ radius\ of\ the\ circle= \frac{30}{2} =15\ cm \\\ \\ area\ of\ the\ circle\ = \pi r^{2} =\frac{22}{7} \times 15\times 15\ cm^{2} \\\ \\ area\ of\ circle= \frac{22}{7} \times ( 15)^{2} \ cm^{2}\\\ \\ Hence\ option A)\ is\ correct

(10) The area of the incircle of an equilateral triangle of side 42 cm is

A)        231\ cm ^{2}          B)        462\ cm ^{2} \\\ \\ C)        22 \sqrt{3\ }\ cm^{2}                 D) 924\ cm ^{2} \\\ \\ Read Solution

We draw the equilateral triangle ABC with the center O and the radius r of the incircle.
Thus, AB=BC=CA=42 cm.

From the figure, AB, BC and CA are tangents to the circle at M, N and P.
Thus, OM=ON=OP=r

Area of \triangle ABC= Area ( \triangle OAB) + Area ( \triangle OBC) + Area ( \triangle OCA)

\frac{\sqrt{3}}{4}( 42)^{2} =\frac{1}{2} \times r\times AB+\frac{1}{2} \times r\times BC+\frac{1}{2} \times r\times CA \\\ \\ \frac{\sqrt{3}}{4} \times 42\times 42=\frac{1}{2} r( AB+BC+CA) \\\ \\ 441 \sqrt{3} =\frac{1}{2} \times r( 42+42+42) \\\ \\ 441 \sqrt{3} =\frac{1}{2} \times r\times 126 \\\ \\ 441 \sqrt{3} =63r \\\ \\ r= \frac{441\sqrt{3}}{63}=7 \sqrt{3} \ cm \\\ \\ \\\ \\ The\ area\ of\ a\ circle= \pi r^{2} \\\ \\ = \frac {22} {7} \times 7\sqrt{3} \times 7\sqrt{3}\\\ \\ =154 \times 3 \\\ \\=462\ cm ^{2} \\\ \\ Hence\ option\ B)\ is\ correct.