Area Multiple Choice Questions – Quantitative Aptitude

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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

Read Solution

Area of circle question for competition exam like GRE, GMAT, Math Olympiad, CAT, CMAT, MAT, SNAP, NMAT, SSC, SSC-CGL, SSC-CHSL, AFCAT, IBPS, Bank exam, NDA, AFCAT
radius\quad of\quad circular\quad wheel\quad =\quad 1.75\quad metres\\ \\ circumference\quad of\quad wheel\quad =\quad 2\pi r\\ \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\quad 2\quad \times \frac { 22 }{ 7 } \times \quad 1.75\quad \\\ \\ \\\ \\ No.\quad of\quad revolutions\quad =\quad \frac { total\quad distance }{ circumference\quad of\quad wheel } \quad \\\ \\ =\quad \frac { 11\quad \times \quad 1000 }{ 2\quad \times \frac { 22 }{ 7 } \times \quad 1.75\quad } \\\ \\ =\frac { 11\quad \times \quad 1000\quad \times \quad 7\quad }{ 2\quad \times \quad 22\quad \times \quad 1.75 } =\quad 1000

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
Read Solution

important area questions for bank exam

Solution
side\quad of\quad square\quad =\quad 28\quad cm\\ \\ diameter\quad of\quad circle\quad =\quad 28\quad cm\\ \\ radius\quad of\quad circle\quad =\quad 14\quad cm\\\ \\ \\\ \\ area\quad of\quad square\quad =\quad \pi { r }^{ 2 }\\\ \\ =\frac { 22 }{ 7 } \times \quad 14\quad \times \quad 14\quad \\\ \\ =44\quad \times \quad 14\quad =616\quad { cm }^{ 2 }

option (d) is right answer

(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
Read Solution

Quantitative aptitude questions for bank and SSC exam

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

area\quad of\quad square\quad =\quad { side }^{ 2 }\quad \\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =({ 12 }^{ 2 })\quad =\quad 144\quad { cm }^{ 2 }\\\ \\ \\\ \\ \ area\quad of\quad triangle = \frac { 1 }{ 2 } \times \quad base\quad \times \quad altitude\\\ \\ 144\quad =\quad \frac { 1 }{ 2 } \times \quad 12\quad \times \quad altitude\quad \\ \\ [given,\quad area\quad of\quad square\quad =\quad area\quad of\quad triangle]\\\ \\ altitude\quad =\quad 24\quad cm

option (b) is right answer

(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
Read Solution

Area of triangle question for competition exams like GRE, GMAT, Math Olympiad, SSC, SSC-CGL, SSC-CHSL, RRB, IBPS, SBI po, Sbi Clerk

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

semiperimeter\quad =\quad \frac { sides\quad of\quad triangle\quad }{ 2 } \\\ \\ =\quad \frac { 1.5+2+2.5 }{ 2 } \quad \\\ \\ =\quad \frac { 6 }{ 2 } =\quad 3\quad cm\\\ \\ \\\ \\ \ apply\quad Heron's\quad formula\quad =\quad \sqrt { s(s-a)(s-b)(s-c) } \\\ \\ =>\quad \sqrt { 3(3-1.5)(3-2)(3-2.5) } \\\ \\ =>\quad \sqrt { 3\quad \times \quad 1.5\quad \times \quad 1\quad \times \quad 0.5 } \quad \\\ \\ =>\sqrt { 2.25 } =\frac { 3 }{ 2 } { cm }^{ 2 }

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
Read Solution

Area of circle quantitative aptitude question

Solution
radius = 5 cm
Length of arc = 3.5 cm

area\quad of\quad sector\quad =\quad \frac { radius\quad \times \quad arc\quad length }{ 2 } \quad \\\ \\ =\frac { 5\quad \times \quad 3.5 }{ 2 } =8.75\quad { cm }^{ 2 }.

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 \\\ \\

Read Solution

Important quantitative aptitude questions for competition exams like GMAT, GRE, Math Olympiad, SSC, SSC-CGL, SSC-CHSL, Bank exams, SBI PO, SBI clerk, IBPS, NMAT, CAT
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
circle related quantitative aptitude questions
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
How to solve circle questions for competition exams like SSC, SSC-CGL, SSC-CHSL, NMAT, Bank exams, IBPS, RRB. Railway Ticket collector, NDA, AFCAT
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
important quantitative aptitude questions
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
how to solve aptitude questions for SSC and Banking exams

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.

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