MCQ Questions on Mensuration 2D for Competitive Exams

(∴ Rhombus is a parallelogram ∴ area of Rhombus = base × height)
Area of Rhombus
= base × height
= 6.5 × 10
= 65 cm²
Also area of Rhombus = $$\frac{1}{2}$$ × d₁ × d₂
⇒ $$\frac{1}{2}$$ × 26 × d₂ = 65
⇒ 13 × d₂ = 65
⇒ d₂ = 5 cm

Breadth = 12 m
Length = 24 m
Cost = 100 Rs./m²
r = 6
R = 12

π × 144 + π × 36 + 288
= 3.14 × 144 + 3.14 × 36 + 288
= 853.2
∴ Total cost = 853.2 × 100 = Rs. 85320

$$\eqalign{ & {\text{Let EN }} \bot \,{\text{AD}} \cr & {\text{Area of }}\Delta {\text{AED}} = \frac{1}{2} \times {\text{EN}} \times {\text{AD}} \cr & {\text{Area of trapezium ABCD}} \cr & = \frac{1}{2}\left( {{\text{AD}} + {\text{BC}}} \right) \times {\text{EN}} \cr & \frac{{{\text{ar}}\left( {{\text{ABCD}}} \right)}}{{{\text{ar}}\left( {{\text{AED}}} \right)}} \cr & = \frac{{\frac{1}{2}\left( {{\text{AD}} + {\text{BC}}} \right) \times {\text{EN}}}}{{\frac{1}{2} \times {\text{EN}} \times {\text{AD}}}} \cr & = \frac{{{\text{AD}} + {\text{BC}}}}{{{\text{AD}}}} \cr} $$

Length of perpendicular drawn from the right angle to hypotenuse,
$$\eqalign{ & {\text{P}} = \frac{{{\text{a}} \times {\text{b}}}}{{\text{H}}} \cr & {{\text{P}}^2} = \frac{{{{\text{a}}^2}{{\text{b}}^2}}}{{{{\text{H}}^2}}} \cr & {{\text{P}}^2} = \frac{{{{\text{a}}^2}{{\text{b}}^2}}}{{{{\text{a}}^2} + {{\text{b}}^2}}}\,\,\,\,\left[ {\because {{\text{H}}^2} = {{\text{a}}^2} + {{\text{b}}^2}} \right] \cr} $$

BD = length of diagonal = speed × time
$$\eqalign{ & = \frac{{52}}{{60}} \times 15 \cr & = 13{\text{ m}} \cr & {\text{BD}} = \sqrt {{l^2} + {b^2}} \cr & \Rightarrow {l^2} + {b^2} = {13^2} \cr & \Rightarrow {l^2} + {b^2} = 169 \cr & {\text{Again, }}l + b = \frac{{68}}{{60}} \times 15 = 17 \cr & {\left( {l + b} \right)^2} = {l^2} + {b^2} + 2lb \cr & {17^2} = 169 + 2lb \cr & lb = \frac{{120}}{2} = 60{\text{ }}{{\text{m}}^2} \cr} $$

Area of shaded region = Area of semicircle - Area of triangle
$$\eqalign{ & = \frac{{\pi {{\left( a \right)}^2}}}{2} - \frac{1}{2} \times a \times 2a \cr & = \frac{{\pi {a^2}}}{2} - {a^2} \cr & = {a^2}\left( {\frac{\pi }{2} - 1} \right){\text{ sq}}{\text{. units}} \cr} $$

Let the length of interior square = $$x$$
The length of exterior square = ($$x$$ + 9)
According to question
$$\eqalign{ & {\left( {x + 9} \right)^2} - {x^2} = 105.75 \cr & \left( {x + 9 + x} \right)\left( {x + 9 - x} \right) = \frac{{10575}}{{100}} \cr & \left( {2x + 9} \right) \times 9 = \frac{{423}}{4} \cr & 2x + 9 = \frac{{47}}{4} \cr & 2x = \frac{{47}}{4} - 9 \cr & 2x = \frac{{11}}{4} \cr & x = \frac{{11}}{8} \cr & {\text{Total cost of fencing}} \cr & = \frac{{11}}{8} \times 4 \times 100 \cr & = {\text{Rs}}{\text{. }}550 \cr} $$

Area of Hexagon $$ = 6 \times \frac{{\sqrt 3 }}{4}{\left( 6 \right)^2} = 54\sqrt 3 {\text{ c}}{{\text{m}}^2}$$
Area of one part of Hexagon $$ = \frac{{54}}{6}\sqrt 3 = 9\sqrt 3 $$
Required area $$ = 54\sqrt 3 + 2 \times 9\sqrt 3 = 72\sqrt 3 {\text{ c}}{{\text{m}}^2}$$

Alternate Solution:
Counting the equilateral triangle figure = 8
Area of equilateral triangle is equal to $$ = 8 \times \frac{{\sqrt 3 }}{4} \times 6 \times 6 = 72\sqrt 3 {\text{ c}}{{\text{m}}^2}$$

Each interior angle of polygon is given by
$$\eqalign{ & = \frac{{\left( {x - 2} \right)}}{x} \times 180 \cr & {\text{Sides is }}a,\,2a \cr & \frac{{\left( {\frac{{a - 2}}{a}} \right) \times 180}}{{\left( {\frac{{2a - 2}}{{2a}}} \right) \times 180}} = \frac{3}{4} \cr & \frac{{\left( {a - 2} \right)}}{a} \times \frac{{2a}}{{2\left( {a - 1} \right)}} = \frac{3}{4} \cr & \frac{{\left( {a - 2} \right)}}{{\left( {a - 1} \right)}} = \frac{3}{4} \cr & 4a - 8 = 3a - 3 \cr & a = 5 \cr & {\text{So, sides }}5,\,10 \cr} $$