# Area Questions – Quantitative Aptitude

(01) A circle is inscribed in a square, An equilateral triangle of side 4 \sqrt{3} cm is inscribed in that circle. The length of the diagonal of the square (in cm) is

A) 4 \sqrt{2}\\ \\ B) 8 \\ \\ C) 8 \sqrt{2} \\ \\ D) 16 \\\ \\ Read Solution

Side of equilateral triangle=4 \sqrt{3} \ cm

Circumradius of triangle= \frac{side}{\sqrt{3}} \\ \\ = \frac{4\sqrt{3}}{\sqrt{3}} =4\ cm

We know that,
side of square= 2 * circumradius
=> 2 * 4 = 8

To find diagonal of square
Applying Pythagoras theorem,
Diagonal^{2}= 8^{2} +8^{2} \\\ \\ => 64+64 \ = 128 \ = 8\sqrt{2} \ cm

Hence option, C) is correct

(02) The area of circle whose radius is 6 cm is trisected by two concentric circles. The radius of the smallest circle is
A) 2 \sqrt{3} \ cm \\\ \\ B) 2 \sqrt{6} \ cm \\\ \\ C) 2 \ cm \\\ \\ D) 3 \ cm \\\ \\ Read Solution

Area of C circle= \pi r^{2} \\\ \\ ⟹ {\pi ( 6)^{2}} \\\ \\ ⟹ 36 \pi \ cm^{2}

Since given circle is trisected = divided into 3 equal parts

Area of each part= \frac{36\pi }{3}=12π cm²

area of each circle= \pi r^{2} \\\ \\ 12 \pi =\pi r^{2} \\\ \\  r ^{2}=12 \\\ \\ r=2 \sqrt{3} \ cm

Hence option A) is correct.

(03) A circle is inscribed in an equilateral triangle of side 8 cm. The area of the portion between the triangle and the circle is
A)  11 \ cm^{2} \\\ \\ B)  10.95 \ cm^{2} \\\ \\ C) 10 \ cm^{2} \\\ \\ D) 10.50 \ cm^{2} \\\ \\ Read Solution

given side of triangle=8 cm

inradius of circle= \frac{side}{2\sqrt{3}}\\\ \\ =\frac{8}{2\sqrt{3}} \\\ \\ =\frac{4}{\sqrt{3}}

area of circle= \pi r^{2} \\\ \\ = {\pi \left(\frac{4}{\sqrt{3}}\right)^{2}} \\\ \\ = \frac{22}{7} \times \frac{16}{3}

required area=area of equilateral triangle – area of circle

= \frac{\sqrt{3}}{4}( 8)^{2}- \frac{22}{7} \times \frac{16}{3} \\\ \\ = \frac{\sqrt{3}}{4} \times \ 64 - \frac{22}{7} \times \frac{16}{3} \\\ \\ = \ 16 \sqrt{3} -16.76 \\\ \\ =16 \times 1.73-16.76\ \\\ \\ = 27.68 - 16.76 =10.92 cm ^{2}\\\ \\ Hence 10.92\approx 10.95\

Thus option B) is correct

(04) The circumference of a circle is 11 cm and the angle of a sector of the circle is 60 \degree . The area of the sector is

A) 1 \frac{29}{48} \ \ cm^{2} \\\ \\ B) 2 \frac{29}{48} \ \ cm^{2} \\\ \\ C) 1 \frac{27}{48} \ \ cm^{2} \\\ \\ D) 2 \frac{27}{48} \ \ cm^{2} \\\ \\ Read Solution

Circumference=11 cm

2 \pi r=11 \\\ \\ 2 \frac{22}{7} r=11 \\\ \\ r= \frac{7}{4} \ cm \\\ \\

area of sector= \frac{\theta }{360\degree } \times \pi r^{2} \\\ \\ = \frac{60}{360} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \\\ \\ = \frac{77}{48} =1\frac{29}{48} \ cm^{2}

Hence option A) is correct.

(05) The sides of a triangle are 6 cm, 8 cm, and 10 cm. The area of the greatest square that can be inscribed in it, is

A) 18 sq. cm

B) 15 sq. cm

C)  \frac{2304}{49} sq. cm

D) \frac{576}{49} sq. cm

Solution

Since the given sides of triangle denotes that it is right angled triangle.perpendicular at AC.

side\ of\ square= \frac{perpendicular\times base}{perpendicular\ +\ base}\\\ \\ =\frac{8\times 6}{8+6}\\\ \\ =\frac{48}{14}\\\ \\ =\frac{24}{7} \\\ \\ \\\ \\ area\ of\ square= \frac{24}{7} \times \frac{24}{7} =\frac{576}{49} \ cm^{2} \\\ \\

Hence, option D) is correct.

(06) The length of a side of an equilateral triangle is 8 cm. The area of the region lying between the circumcircle and the incircle of the triangle is

A) 50 \frac{1}{7}\ cm ^{2} \\\ \\ B) 50 \frac{2}{7}\ cm ^{2}  \\\ \\  C) 75\frac{1}{7}\ cm ^{2} \\\ \\  D) 75\frac{2}{7}\ cm ^{2}\\ \\ Read Solution
Radius\ of\ circumcircle\ = \frac{8}{\sqrt{3}}cm \\\ \\ Radius\ of\ incircle\ = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}}\ cm \\\ \\ Area\ bounded\ by\ the\ circle\ = \pi \left(( radius\ circumcircle)^{2}\\ \\ - ( radius\ incircle)^{2}\right) \\\ \\ \ = \pi \left(\left(\frac{8}{\sqrt{3}}\right)^{2} - \left(\frac{4}{\sqrt{3}}\right)^{2}\right)\\\ \\ \ = \frac{22}{7} \times \left(\frac{64}{3} - \frac{16}{3}\right) \\\ \\ = \frac{22}{7} \times \left( \frac{64-16}{3}\right) \\\ \\ =\frac{22}{7} \times \frac{48}{3} \\\ \\ = \frac{22\times 16}{7} =50 \frac{2}{7} \ cm^{2}

Hence, option B) is correct.

(07) One acute angle of a right angled triangle is double the other.If the length of its hypotenuse is 10 cm, then its area is ?

A) \frac{25}{2}\sqrt{3} cm^{2} \\\ \\ B) 25\ cm ^{2} \\\ \\ C)25 \sqrt{3}\ cm ^{2}\\\ \\ D) \frac{75}{2} cm^{2} \\\ \\  Read Solution
The\ angles\ of\ the\ given\ triangle\ are\ 90 \degree \\ \\ ,30 \degree and 60 \degree \\\ \\ sin 30 \degree = \frac{perpendicular}{hypotenuse} \\\ \\ we\ know\ that\ sin 30 \degree =\frac{1}{2}\\\ \\ Put\ this\ value\ in\ above\ formula\ we\ get \\\ \\ \frac{1}{2} =\frac{P}{10} \\\ \\ P= \frac{10}{2} =\ 5\ cm \\\ \\ Apply\ Pythagoras\ theorem\\\ \\ Hypotenuse ^{2}=perpendicular ^{2} +base^{2} \\\ \\ base ^{2}=hypotenuse ^{2} -perpendicular^{2} \\\ \\ base= \sqrt{hypotenuse^{2} -perpendicular^{2}} \\\ \\ base= \sqrt{10^{2} -5^{2}} \\\ \\ base= \sqrt{100-25} \ =\sqrt{75} =5\sqrt{3} \ cm \\\ \\ now\ calculate\ area \\\ \\ area= \frac{1}{2} \times base\times height \\\ \\ = \frac{1}{2} \times 5\sqrt{3} \times 5=\frac{25\sqrt{3}}{2}

hence option A) is correct.

(08) An equilateral triangle of side 6 cm has its corners cut off to form a regular hexagon. Area of this regular hexagon will be

A) 3 \sqrt{3} \\\ \\ B) 3 \sqrt{6} \\\ \\ C) 6 \sqrt{3} \\\ \\ D) \frac{5\sqrt{3}}{2}\\ \\ Read Solution

Solution
Given each sides of equilateral triangle ABC=6 cm
AD = DI = IB = 2 cm

Similarly,
AE = EF = FC = 2 cm
BH = GH = HC = 2 cm

Area\ of\ hexagon\ = \frac{3\sqrt{3}}{2} side^{2} \\\ \\ = \frac{3\sqrt{3}}{2} \times 2 \times 2 \\\ \\ = 6\sqrt{3} cm ^{2}

Option (C) is the right answer

(09) The length of each side of an equilateral triangle is 14 \sqrt{3} cm. The area of the incircle (in cm) is

A) 450
B) 308
C) 154
D) 77