MCQ Questions on Mensuration 2D for Competitive Exams

Given, AB = 2
CD = 1
Let BC = x
∵ Area of middle circle = Average of areas of other two circle
$$\eqalign{ & \frac{\pi }{4}{\left( {2 + x} \right)^2} = \frac{{\frac{\pi }{4}{{\left( {2 + x + 1} \right)}^2} + \frac{\pi }{4}{{\left( 2 \right)}^2}}}{2} \cr & 2\left[ {\frac{\pi }{4}{{\left( {2 + x} \right)}^2}} \right] = \frac{\pi }{4}{\left( {3 + x} \right)^2} + \frac{\pi }{4} \times 4 \cr & 2{\left( {2 + x} \right)^2} = {\left( {3 + x} \right)^2} + 4 \cr & 2\left( {4 + {x^2} + 4x} \right) = 9 + {x^2} + 6x + 4 \cr & 8 + 2{x^2} + 8x = 9 + {x^2} + 6x + 4 \cr & {x^2} + 2x - 5 = 0 \cr & x = \frac{{ - 2 \pm \sqrt {4 + 20} }}{2} \cr & x = \frac{{ - 2 \pm 2\sqrt 6 }}{2} \cr & x = - 1 \pm \sqrt 6 \cr & \therefore x = \sqrt 6 - 1 = {\text{BC}} \cr} $$

$$\eqalign{ & 2S = 50 + 70 + 80 \cr & S = 100 \cr & A = \sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)} \cr & A = \sqrt {100 \times 50 \times 30 \times 20} \cr & A = 1000\sqrt 3 \cr & \frac{{\sqrt 3 }}{4}{x^2} = 1000\sqrt 3 \cr & {x^2} = 4000 \cr & x = 63.2 \cr} $$

$$\eqalign{ & {\text{Length of side}} = \frac{2}{{\sqrt 3 }}\left( {{P_1} + {P_2} + {P_3}} \right) \cr & = \frac{2}{{\sqrt 3 }}\left( {6 + 7 + 8} \right) \cr & = \frac{2}{{\sqrt 3 }} \times 21 \cr & = \frac{{42}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{42\sqrt 3 }}{3} \cr & = 14\sqrt 3 {\text{ cm}} \cr} $$

AB = 56 cm
AE = EF = FG = GB = 14 cm
∴ EM = MF = 7 cm
PF = $$\frac{{56}}{2}$$ = 28 cm
OF = 28 - r
OM = 7 + r
MF = 7
In ΔOMF,
OM² = OF² + MF²
(7 + r)² = (28 - r)² + 7²
49 + r² + 14r = 784 + r² - 56r + 49
70r = 784
r = $$\frac{{784}}{{70}}$$ = 11.2
Area of circle = πr²
= $$\frac{{22}}{7} \times \frac{{112}}{{10}} \times \frac{{112}}{{10}}$$
= 394.24 cm²

Circumference of paper sheet = 352
$$\eqalign{ & 2\pi R = 352 \cr & R = \frac{{352}}{{2\pi }} = \frac{{352 \times 7}}{{2 \times 22}} = 56{\text{ cm}} \cr & r = \frac{R}{2} = \frac{{56}}{2} = 28{\text{ cm}} \cr} $$
∴ Circumference of circular plate
$$\eqalign{ & = 2\pi r \cr & = 2 \times \frac{{22}}{7} \times 28 \cr & = 176{\text{ cm}} \cr} $$

Let the length of square park = L
Area = L²
L² = 16x² + 8x + 1
Length of the park
$$\eqalign{ & = \sqrt {16{x^2} + 8x + 1} \cr & = \sqrt {{{\left( {4x} \right)}^2} + 2 \times 4x + {1^2}} \cr & = \sqrt {{{\left( {4x + 1} \right)}^2}} \cr & = \left( {4x + 1} \right){\text{ units}} \cr} $$

$$\eqalign{ & \sqrt {{a^2} + {b^2}} ,\,\sqrt {{{\left( {2a} \right)}^2} + {b^2}} ,\,\sqrt {{a^2} + {{\left( {2b} \right)}^2}} \cr & {\text{Let }}a = b \cr & \sqrt 2 a,\,\sqrt 5 a,\,\sqrt 5 a \cr} $$

$$\eqalign{ & {\text{Height}}\left( h \right) = \sqrt {{l^2} - {b^2}} \cr & = \sqrt {{{\left( {\sqrt 5 a} \right)}^2} - {{\left( {\frac{{\sqrt 2 }}{2}a} \right)}^2}} \cr & = \sqrt {\frac{{9{a^2}}}{2}} \cr & = \frac{{3a}}{{\sqrt 2 }} \cr & {\text{Area}} = \frac{1}{2} \times \frac{{3a}}{{\sqrt 2 }} \times \sqrt 2 a \cr & = \frac{3}{2}{a^2} \cr & = \frac{3}{2}ab \cr} $$

Let the radius of swimming pool = R

Outer radius of pool with concrete wall = (R + 4)
According to question
πR² × $$\frac{{11}}{{25}}$$ = π(R + 4)² - πR²
R² × $$\frac{{11}}{{25}}$$ = R² + 16 + 8R - R²
$$\frac{{11}}{{25}}$$ R² = 16 + 8R
11R² - 200R - 400 = 0
By option (D), (In such type of equation go through the option to save your valuable time)
R = 20
11 × (20)² - 200 × 20 - 400 = 0
4400 - 4000 - 400 = 0
0 = 0 (satisfy)
Therefore, radius of pool R = 20 m

$$\eqalign{ & {\text{ar }}\Delta EDC = \frac{1}{2} \times ED \times DC \times \sin \angle EDC \cr & = \frac{1}{2} \times 12 \times 12 \times \sin {120^ \circ } \cr & = 36\sqrt 3 {\text{ c}}{{\text{m}}^2} \cr} $$

AO = OC = 12
AB = 13
BO = 5
Area of rhombus 4 × Area of ΔAOB
= 4 × $$\frac{1}{2}$$ × 12 × 5
= 120 cm²