MCQ Questions on Mensuration 2D for Competitive Exams

Diagonal of square ABCD = 8√2
Diagonal of square EBGF = 6√2
Height of triangle EDG = 8√2 - 3√2 = 5√2
Area of triangle EDG = $$\frac{1}{2}$$ × 6√2 × 5√2 = 30
Area of triangle EFG = (3√2)² = 18
Area of shaded region = 30 - 18 = 12

Remember: area of isosceles triangle
$$ = \frac{1}{2}{a^2}\sin \theta $$
(θ is angle between equal sides)
$$\eqalign{ & = \frac{1}{2}{\left( {10} \right)^2}\sin {45^ \circ } \cr & = \frac{{100}}{2} \times \frac{1}{{\sqrt 2 }} \cr & = \frac{{50}}{{\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }} \cr & = 25\sqrt 2 {\text{ c}}{{\text{m}}^2} \cr} $$

So, If
OF = OE = ED
So, FC = 24 cm
So, RQ $$ = \frac{{{\text{FC}} + {\text{ED}}}}{2} = \frac{{36}}{2}$$
RQ = 18 cm
Area of $$\Delta {\text{PQR}} = \frac{{\sqrt 3 }}{4} \times 18 \times 18 = 81\sqrt 3 {\text{ c}}{{\text{m}}^2}$$

$$\eqalign{ & {\text{Length of arc}} = \frac{\theta }{{360}} \times 2\pi r \cr & \frac{{72}}{{360}} \times 2 \times \frac{{22}}{7} \times r = 88 \cr & r = \frac{{88 \times 7 \times 360}}{{72 \times 2 \times 22}} \cr & r = 70{\text{ m}} \cr} $$

Area of portion enclosed by the three circles = Area of Equilateral triangle - 3 × Area of sector
$$\eqalign{ & = \frac{{\sqrt 3 }}{4} \times {\left( {{\text{side}}} \right)^2} - \left( {\pi {r^2}\left( {\frac{\theta }{{{{360}^ \circ }}}} \right) \times 3} \right) \cr & = \frac{1}{2}\left( {\frac{{\sqrt 3 }}{4} \times 42 \times 42 - \frac{{22}}{7} \times 21 \times 21 \times \frac{1}{3} \times 3} \right) \cr & = 441\sqrt 3 - 693 \cr} $$

Side of square PQRS
$$\eqalign{ & = \frac{{{\text{Diameter of circle}}}}{{\sqrt 2 }} \cr & = \frac{{14\sqrt 2 \times 2}}{{\sqrt 2 }} \cr & = 28{\text{ cm}} \cr} $$

In ΔABO
AB = a
OB = 14√2
AO = 14 + 2a
OB² = AB² + OA²
(14√2)² = a² + (14 + 2a)²
196 × 2 = a² + 196 + 4a² + 56a
5a² + 56a - 196 = 0
5a² + 70a - 14a - 196 = 0
5a(a + 14) - 14(a + 14) = 0
(a + 14)(5a - 14) = 0
a = -14, a = $$\frac{{14}}{5}$$
Side of small square = 2a = $$\frac{{28}}{5}$$
Required Area $$ = 4 \times {\left( {\frac{{28}}{5}} \right)^2} = 125.44{\text{ c}}{{\text{m}}^2}$$

$$\eqalign{ & {\text{Diagonal of square}} = \sqrt 2 a = 2r \cr & \therefore a = \sqrt 2 r \cr & {\text{Area of square}} = {a^2} = {\left( {\sqrt 2 r} \right)^2} = 2{r^2} \cr} $$

$$\eqalign{ & {\text{Side of hexagon}} = a = r \cr & {\text{Area of hexagon}} = 6\frac{{\sqrt 3 }}{4}{a^2} = 3\frac{{\sqrt 3 }}{2} \times {r^2} \cr & {\text{Required ratio}} = 2{r^2}:3\frac{{\sqrt 3 }}{2}{r^2} = 4:3\sqrt 3 \cr} $$

According to the question,

$$\eqalign{ & \sin {15^ \circ } = \frac{P}{H} = \frac{{AB}}{1} \cr & AB = \sin {15^ \circ } \cr & \cos {15^ \circ } = \frac{B}{H} = \frac{{BC}}{1} \cr & BC = \cos {15^ \circ } \cr & {\text{Area of }}\Delta ABC = \frac{1}{2} \times AB \times BC \cr & = \frac{1}{2} \times \sin {15^ \circ } \times \cos {15^ \circ } \cr & = \frac{1}{{2 \times 2}} \times 2\sin {15^ \circ } \times \cos {15^ \circ } \cr & = \frac{1}{4} \times \sin 2 \times {15^ \circ }\,\,\,\,\left[ {\because \sin 2\theta = 2\sin \theta \cos \theta } \right] \cr & = \frac{1}{4} \times \sin {30^ \circ } \cr & = \frac{1}{4} \times \frac{1}{2} \cr & = \frac{1}{8}{\text{ }}{{\text{m}}^2} \cr & = \frac{1}{8} \times 100 \times 100 = 1250{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & {\text{Let }}\angle ACB = \alpha \cr & \angle DAC = {90^ \circ } - \alpha \cr & \cos \alpha = \frac{{{{\left( {\sqrt {41} } \right)}^2} + {{\left( 8 \right)}^2} - {{\left( 5 \right)}^2}}}{{2\sqrt {41} \times 8}} = \frac{{DC}}{{\sqrt {41} }} \cr & \frac{{41 + 64 - 25}}{{16\sqrt {41} }} = \frac{{DC}}{{\sqrt {41} }} \cr & \frac{{80}}{{16}} = DC \cr & DC = 5{\text{ cm}} \cr & BD = 8 - 5 = 3{\text{ cm}} \cr} $$

$${\text{Area of }}\Delta ABD = \frac{1}{2} \times 4 \times 3 = 6{\text{ c}}{{\text{m}}^2}$$