MCQ Questions on Mensuration 3D for Competitive Exams

$$\eqalign{ & {\text{In }}\Delta ABD, \cr & BD = \sqrt {A{B^2} + A{D^2}} \cr & = \sqrt {{9^2} + {{12}^2}} \cr & = \sqrt {81 + 144} \cr & = \sqrt {225} \cr & = 15\,{\text{cm}} \cr & {\text{Area of }}\Delta ABD = \frac{1}{2} \times AB \times AD \cr & = \frac{1}{2} \times 9 \times 12 \cr & = 54{\text{ c}}{{\text{m}}^2} \cr & {\text{In }}\Delta BCD \cr & {\text{Semiperimeter}} = \frac{{13 + 14 + 15}}{2} = \frac{{42}}{2} = 21 \cr & {\text{Area of }}\Delta BCD = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr & = \sqrt {21\left( {21 - 13} \right)\left( {21 - 14} \right)\left( {21 - 15} \right)} \cr & = \sqrt {21 \times 8 \times 7 \times 6} \cr & = 21 \times 4 \cr & = 84{\text{ c}}{{\text{m}}^2} \cr & {\text{Area }}ABCD = 84 + 54 = 138{\text{ c}}{{\text{m}}^2} \cr & {\text{Height of prism}} = \frac{{{\text{Volume}}}}{{{\text{Area of base}}}} \cr & = \frac{{2070}}{{138}} \cr & = 15{\text{ cm}} \cr & {\text{Perimeter of base}} = 9 + 14 + 13 + 12 = 48{\text{ cm}} \cr & {\text{Area of lateral surface}} = {\text{Perimeter}} \times {\text{Height}} \cr & = 48 \times 15 \cr & = 720{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & 2\pi r = 44 \cr & 2 \times \frac{{22}}{7} \times r = 44 \cr & r = 7 \cr & {\text{Curved surface area}} = \pi rl \cr & = \frac{{22}}{7} \times 7 \times 25 \cr & = 550{\text{ c}}{{\text{m}}^2} \cr} $$

Decrease in base radius = (Decrease in base area)$$^{\frac{1}{2}} = {\left( {\frac{1}{9}} \right)^{\frac{1}{2}}} = \frac{1}{3}$$
Let initial radius and height be 3r and h
∴ New radius and height are r and 6h
Old lateral surface area = 2 × π × 3r × h = 6πrh
New lateral surface area = 2 × π × r × 6h = 12πrh
Required factor $$ = \frac{{12\pi rh}}{{6\pi rh}} = 2$$

Volume of cone = $$\frac{1}{3}\pi {r^2}h$$
Volume of cylinder = $$\pi {r^2}h$$
Volume of water = Volume of cylinder - volume of cone
$$\eqalign{ & = \pi {r^2}h - \frac{1}{3}\pi {r^2}h \cr & = \frac{2}{3}\pi {r^2}h \cr & = 2\left( {\frac{1}{3}\pi {r^2}h} \right) \cr & = 2 \times 27\pi \cr & = 54\pi {\text{ c}}{{\text{m}}^3} \cr} $$

$$\eqalign{ & \pi {r^2} = 154 \cr & {r^2} = \frac{{154 \times 7}}{{22}} \cr & {r^2} = 49 \cr & r = \sqrt {49} \cr & r = 7{\text{ m}} \cr & {\text{Also volume}} = 1232 \cr & \frac{1}{3}\pi {r^2}h = 1232 \cr & h = \frac{{1232 \times 3}}{{\pi {r^2}}} \cr & h = \frac{{1232 \times 3}}{{154}} \cr & h = 24{\text{ m}} \cr & {\text{Area of canvas required}} = \pi rl \cr & = \pi r\sqrt {{r^2} + {h^2}} \cr & = \frac{{22}}{7} \times 7 \times \sqrt {{{24}^2} + {7^2}} \cr & = \frac{{22}}{7} \times 7 \times 25 \cr & = 550{\text{ }}{{\text{m}}^2} \cr & {\text{Length}} \times {\text{2}} = 550{\text{ }}{{\text{m}}^2} \cr & {\text{length}}\left( l \right) = \frac{{550}}{2} = 275{\text{ m}} \cr} $$

$$\eqalign{ & {\text{Volume of Metallic shell}} \cr & = V = \frac{4}{3}\pi \left( {{R^3} - {r^3}} \right) \cr & {\text{Total weight of Metallic shell}} \cr & \Rightarrow 10.5 \times V = 6688 \cr & \Rightarrow 10.5 \times \frac{4}{3} \times \frac{{22}}{7}\left( {{6^3} - {r^3}} \right) = 6688 \cr & \Rightarrow r = 4 \cr & {\text{Thickness}} = 6 - 4 = 2 \cr} $$

$$\eqalign{ & 396 = \frac{{22}}{7} \times \frac{{3 \times 3}}{{10000}} \times h \cr & 140000 = h \cr & 1\,{\text{m}} \to 1000\,{\text{dm}} \cr & h = 140\,{\text{m}} \cr} $$

$$\eqalign{ & {V_1}{\text{ of cube}} = {a^3} \cr & {V_2}{\text{ of cylinder}} = \pi {\left( {\frac{a}{2}} \right)^2} \times a \cr & {V_1}:{V_2} = {a^3}:\frac{{22}}{7} \times \frac{{{a^3}}}{4} = 14:11 \cr & {\text{Remaining part}}:{V_1} = 3:14 \cr & \% = \frac{3}{{14}} \times 100 = \frac{{150}}{7} = 21\% \cr} $$

$$ - 40\% \to \frac{{ - 2}}{5}$$

$$\eqalign{ & {\text{Decrease }}\% = \frac{{98}}{{125}} \times 100 \cr & = \frac{{98 \times 4}}{5} \cr & = \frac{{392}}{5} \cr & = 78.4\% \cr} $$

$$\eqalign{ & 20\% \to \frac{{ + 1}}{5} \cr & V = \pi {R^2}h \cr & R \to 5:6 \cr & {R^2} \to 25:36 \cr & \boxed{{R^2} \propto \frac{1}{h}}{\text{ If volume same}} \cr} $$

$$\frac{{11}}{{36}} \times 100 = 30\frac{5}{9}\% $$