MCQ Questions on Mensuration 2D for Competitive Exams

$$\eqalign{ & {\text{Side of equilateral }}\Delta = 8{\text{ cm}} \cr & {\text{In circle radius}} = \frac{a}{{2\sqrt 3 }} = \frac{8}{{2\sqrt 3 }} = \frac{4}{{\sqrt 3 }} \cr & {\text{Circum radius}} = \frac{a}{{\sqrt 3 }} = \frac{8}{{\sqrt 3 }} \cr & {\text{Area bounded by both circle is}} \cr & = \pi \left\{ {{{\left( {\frac{8}{{\sqrt 3 }}} \right)}^2} - {{\left( {\frac{4}{{\sqrt 3 }}} \right)}^2}} \right\} \cr & = \frac{{22}}{7}\left( {\frac{{64}}{3} - \frac{{16}}{3}} \right) \cr & = \frac{{22}}{7}\left( {\frac{{48}}{3}} \right) \cr & = \frac{{22 \times 16}}{7} \cr & = 50\frac{2}{7}{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & {\text{Area of trapezium}} = \frac{1}{2}\left( {AB + CD} \right) \times h \cr & = \frac{1}{2} \times 6 \times 1.44 \cr & = 4.32{\text{ sq}}{\text{. yards}} \cr & \left( {{\text{1 sq}}{\text{. yards}} = 1800} \right) \cr & \therefore 4.32 = 1800 \times \frac{{432}}{{100}} = {\text{Rs}}{\text{. }}7776 \cr} $$

$$\eqalign{ & = 3 \times 2\sqrt {Rr} \cr & = 3 \times 2\sqrt {4 \times 4} \cr & = 24 \cr & = 24 + 2\pi r \cr & = 24 + 2\pi \times 4 \cr & = 24 + 8\pi {\text{ cm}} \cr} $$

Distance covered in 30 seconds = 30 m/min × $$\frac{{30}}{{60}}$$ = 15 m
This is the difference of distance of the boundary and the diameter.
Let 'R' be the radius

$$\eqalign{ & 2\pi R - 2R = 15 \cr & 2R\left( {\pi - 1} \right) = 15 \cr & 2R = \frac{{15}}{{\pi - 1}} \cr & 2R = \frac{{15}}{{\frac{{22}}{7} - 1}} \cr & 2R = \frac{{15 \times 7}}{{15}} \cr & 2R = 7 \cr & R = \frac{7}{2} \cr & R = 3.5{\text{ m}} \cr} $$

Perimeter of equilateral triangle = 100 + 100 + 100 = 300
Perimeter of New triangle
= 120 + 150 + 130
= 400
% increase $$ = \frac{{100}}{{300}} \times 100 = 33\frac{1}{3}\% $$

In ΔABC & ΔDCQ
∠ABC = ∠DCQ
∠ACB = ∠DQC
BC = CQ
ΔABC ≅ ΔDCQ
ar ΔABC = ar ΔDCQ

According to the figure,
⇒ Perimeter = r + r + $$l$$
⇒ 75 cm = 2r + length of arc
⇒ 75 cm = 2r + $$\frac{{2\pi r}}{4}$$
⇒ 75 cm = 2r + $$\frac{{22 \times r}}{{7 \times 2}}$$
⇒ r = 21 cm
⇒ Its area $$ = \frac{1}{4}\left[ {\frac{{22}}{7} \times 21 \times 21} \right] = 346.5{\text{ c}}{{\text{m}}^2}$$

Area of parallelogram
= BC × FC
= 15 × 12
= 180 cm²
Area of parallelogram
DC × AE = 180
18 × AE = 180
AE = 10 cm
∴ Distance between bigger sides = 10 cm

Let P be the point inside the equilateral ΔABC
Let, PD = √3, PE = 2√3, PF = 5√3 and AB = BC = AC = $$x$$
$$\eqalign{ & {\text{ar}}{\text{.}}\,\Delta {\text{ABC}} = {\text{ar}}{\text{.}}\,\Delta {\text{ABP}} + {\text{ar}}{\text{.}}\,\Delta {\text{ACP}} + {\text{ar}}{\text{.}}\,\Delta {\text{BCP}} \cr & \frac{{\sqrt 3 }}{4}{x^2} = \frac{1}{2} \times x \times \sqrt 3 + \frac{1}{2} \times x \times 2\sqrt 3 + \frac{1}{2} \times x \times 5\sqrt 3 \cr & \sqrt 3 x = 2\sqrt 3 + 4\sqrt 3 + 10\sqrt 3 \cr & x = 16 \cr} $$
∴ Perimeter of triangle = 3$$x$$ = 3 × 16 = 48 cm

$$\eqalign{ & {\text{Area Grazed by the cow}} \cr & = \frac{{{A^ \circ }}}{{360}}\pi {\left( 7 \right)^2} + \frac{{{B^ \circ }}}{{360}}\pi {\left( 7 \right)^2} + \frac{{{C^ \circ }}}{{360}}\pi {\left( 7 \right)^2} \cr & = \pi {\left( 7 \right)^2}\left[ {\frac{{{A^ \circ } + {B^ \circ } + {C^ \circ }}}{{360}}} \right] \cr & = \pi {\left( 7 \right)^2} \times \frac{{180}}{{360}} \cr & = \frac{1}{2}\pi {\left( 7 \right)^2} \cr & = 77{\text{ }}{{\text{m}}^2} \cr & s = \frac{{26 + 30 + 28}}{2} = 42 \cr & {\text{Area of field}} = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr & = \sqrt {42 \times 16 \times 14 \times 12} \cr & = 336{\text{ }}{{\text{m}}^2} \cr & \Rightarrow {\text{Remaining area}} = 336 - 77 = 259{\text{ }}{{\text{m}}^2} \cr} $$