MCQ Questions on Mensuration 2D for Competitive Exams

$$\eqalign{ & {\text{Area of half }}\Delta = \frac{1}{2}\left( {\frac{{\sqrt 3 }}{4}{a^2}} \right) \cr & = \frac{1}{2} \times \frac{{\sqrt 3 }}{4} \times 6 \times 6 \cr & = \frac{{9\sqrt 3 }}{2} \cr & {\text{Area of shaded region}} \cr & \Rightarrow \frac{{9\sqrt 3 }}{2} - \frac{{90}}{{360}}\pi {r^2} \cr & = \frac{{9\sqrt 3 }}{2} - \frac{1}{4} \times \frac{{22}}{7} \times 1 \times 1 \cr & = \frac{{9\sqrt 3 }}{2} - \frac{{11}}{{2 \times 7}} \cr & = \frac{1}{2}\left( {9\sqrt 3 - \frac{{11}}{7}} \right){\text{c}}{{\text{m}}^2} \cr} $$

In this condition always square area is greater than any other quadrilateral
So, option (C) is correct.

Alternate Solution:-
Let perimeter = 16 cm
Perimeter of square = 4a = 16
                                      a = 4
Area (A) of square = 16 m²
Perimeter of rectangle = 2($$l$$ + b)
16 = 2($$l$$ + b)
$$l$$ + b = 8
Area (B) of rectangle = $$\left( {\frac{{4,\,4}}{{6,\,2}}} \right)$$
= 6 × 2
= 12 cm²

a√2 = 9√2
a = 9

a + b + c = 36
Circle of maximum radius is only exist in equilateral Δ
3a = 36
a = 12
$$\eqalign{ & S = \frac{{a + b + c}}{2}, \cr & r = \frac{{\frac{{\sqrt 3 }}{4} \times {a^2}}}{{\frac{{12 + 12 + 12}}{2}}} \cr & \Rightarrow r = \frac{{\sqrt 3 }}{4} \times \frac{{12 \times 12 \times 2}}{{36}} \cr & \Rightarrow r = 2\sqrt 3 \cr} $$
∴ Area of circle = πr² = 12π

Perimeter of rectangle
2(5x + 7x) = 168
5x + 7x = 84
12x = 84
x = 7
Sides of rectangle = 35 cm, 49 cm

$$\eqalign{ & {\text{AC}} = \sqrt {{{35}^2} + {{49}^2}} \cr & {\text{AC}} = \sqrt {1225 + 2401} \cr & {\text{AC}} = \sqrt {3626} \cr} $$

$$\eqalign{ & {\text{Area of square}} = {\left( {14} \right)^2} = 196 \cr & {\text{Area of semicircle}} = \frac{\pi }{2}{\left( 7 \right)^2} = \frac{{49}}{2}\pi \cr & {\text{Side of square LMNO}} = \frac{{{\text{diagonal}}}}{{\sqrt 2 }} = \frac{7}{{\sqrt 2 }} \cr & {\text{Area of square LMNO}} = {\left( {\frac{7}{{\sqrt 2 }}} \right)^2} = \frac{{49}}{2} \cr & {\text{Shaded portion}} = 196 - \frac{{49}}{2}\pi - \frac{{49}}{2} \cr & = 196 - \frac{{49}}{2} \times \frac{{22}}{7} - \frac{{49}}{2} \cr & = 119 - \frac{{49}}{2} \cr & = \frac{{189}}{2} \cr & = 94.5 \cr} $$

$$\eqalign{ & {\text{Area}} = \frac{{270}}{{360}}\pi {r^2} \cr & = \frac{3}{4} \times \frac{{22}}{7} \times 20 \times 20 \cr & = \frac{{6600}}{7}\,{\text{c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & {\text{The area of largest }}\Delta = \frac{1}{2} \times b \times h \cr & = \frac{1}{2} \times 12 \times 6 \cr & = 36{\text{ }}{{\text{m}}^2} \cr} $$

$$\eqalign{ & {\text{Area of }}\Delta {\text{SQU}} = \frac{1}{2}{\text{ area of hexagon}} \cr & = \frac{1}{2} \times \left[ {6 \times \frac{{\sqrt 3 }}{4}{{\left( {12} \right)}^2}} \right] \cr & = \frac{1}{2} \times 216\sqrt 3 \cr & = 108\sqrt 3 {\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & \Delta DHC \sim \Delta DFG \cr & {\text{So, }}\frac{{DC}}{{DG}} = \frac{{CH}}{{FG}} \cr & \Rightarrow \frac{8}{{28}} = \frac{{CH}}{{20}} \cr & \Rightarrow CH = \frac{{40}}{7} \cr & {\text{So, area of }}\Delta DCH \cr & = \frac{1}{2} \times DC \times CH \cr & = \frac{1}{2} \times 8 \times \frac{{40}}{7} \cr & = \frac{{160}}{7}{\text{ c}}{{\text{m}}^2} \cr} $$

Distance between two parallel line = $$\frac{2}{{11}}$$ (sum of both parallel line)
= $$\frac{2}{{11}}$$ × (7x + 4x)
= 2x
Area = $$\frac{1}{2}$$ (sum of parallel sides) × (distance between them)
⇒ $$\frac{1}{2}$$ (7x + 4x) × 2x = 176
⇒ 11x² = 176
⇒ x² = 16
⇒ x = 4
AB = 7 × 4 = 28 cm
CD = 4 × 4 = 16 cm
CM = 2 × 4 = 8 cm
AM = AN + NM
⇒ AM = AN + 16
⇒ AM = 6 + 16
⇒ AM = 22
$$\eqalign{ & \left( {{\text{AN}} = {\text{BM}} = \frac{{12}}{2} = 6} \right) \cr & {\text{A}}{{\text{C}}^2} = {\text{C}}{{\text{M}}^2} + {\text{A}}{{\text{M}}^2} \cr & \Rightarrow {\text{A}}{{\text{C}}^2} = {8^2} + {22^2} \cr & \Rightarrow {\text{AC}} = \sqrt {64 + 484} \cr & \Rightarrow {\text{AC}} = \sqrt {548} \cr & \Rightarrow {\text{AC}} = 2\sqrt {137} \cr} $$