MCQ Questions on Mensuration 3D for Competitive Exams

$$\eqalign{ & {l^2} = {r^2} + {h^2} \cr & {l^2} = {\left( {7.5} \right)^2} + {\left( {18} \right)^2} \cr & {l^2} = \frac{{225}}{4} + 324 \cr & {l^2} = \frac{{1521}}{4} \cr & l = \frac{{39}}{2} \cr & {\text{Total surface area}} = \pi rl + 2\pi rh + \pi {r^2} \cr & = \pi r\left[ {l + 2h + r} \right] \cr & = \pi \times \frac{{15}}{2}\left[ {\frac{{39}}{2} + 36 + \frac{{15}}{2}} \right] \cr & = \pi \times \frac{{15}}{2}\left[ {\frac{{126 \times 15}}{2}} \right] \cr & = \left[ {\frac{{945}}{2}} \right]\pi \cr & = 472.5\pi \cr} $$

Curved surface area = $$\frac{1}{3}$$ × total surface area
2πrh = $$\frac{1}{3}$$ × [2πrh + 2πr²]
6πrh = 2πrh + 2πr²
4πrh = 2πr²
4h = 2r
h : r = 1x : 2x
Volume of cylinder = πr²h = 4312
$$\frac{{22}}{7}$$ × (2x)² × x = 4312
$$\frac{{22}}{7}$$ × 4x³ = 4312
x³ = 49 × 7
x = 7
Curved surface area = 2πrh
= 2 × $$\frac{{22}}{7}$$ × (2 × 7) × 7
= 22 × 28
= 616 cm²

BX = BY = 8 cm
∴ XY = YZ = XZ = 8√2

$$l$$² = 8² - (4√2)²
$$l$$² = 32
$$l$$ = 4√2
Total surface area = $$\frac{1}{2}$$ × Perimeter of base × $$l$$ + Area of base
= $$\frac{1}{2} \times 3 \times 8\sqrt 2 \times 4\sqrt 2 + \frac{{\sqrt 3 }}{4}{\left( {8\sqrt 2 } \right)^2}$$
= 96 + 32√3
= 32(3 + √3) cm²

$$\eqalign{ & \Delta ABC = {\text{equilateral }}\Delta \cr & \therefore \angle ACB = {60^ \circ } \cr & \& \angle BCP = {30^ \circ } \cr & \Delta CDO,\,\angle CDO = {90^ \circ } \cr & \left( {{\text{Angle between radius and tangent is }}{{90}^ \circ }} \right) \cr} $$

$$\eqalign{ & OD = 1P = 1{\text{ cm}} \cr & OC = 2P = 2\left( 1 \right) = 2{\text{ cm}} \cr & {\text{Then, }}CZ = OC + OZ = 2 + 1 = 3{\text{ cm}} \cr & \Delta CZY,\,\angle CZY = {90^ \circ } \cr & CZ = \sqrt 3 P = 3{\text{ cm}} \cr & YZ = 1P = \sqrt 3 {\text{ cm}} \cr & {\text{Now, in cone }}XYC \cr & r = ZY = \sqrt 3 {\text{ cm}} \cr & h = CZ = 3{\text{ cm}} \cr & {\text{Volume of cone}} = \frac{1}{3}\pi {r^2}h \cr & = \frac{1}{3}\pi {\left( {\sqrt 3 } \right)^2}\left( 3 \right) \cr & = 3\pi {\text{ c}}{{\text{m}}^2} \cr & {\text{Volume of sphere}} = \frac{4}{3}\pi r_s^3 \cr & \left( {\therefore {r_s} = 1{\text{ cm}}} \right) \cr & = \frac{4}{3}\pi {\text{ c}}{{\text{m}}^3} \cr & {\text{Volume of water that can immerse the ball}} \cr & = \left( {3\pi - \frac{{4\pi }}{3}} \right){\text{c}}{{\text{m}}^3} \cr & = \frac{{5\pi }}{3}{\text{ c}}{{\text{m}}^3} \cr} $$

$$\eqalign{ & R \to 1:2 \cr & V \to 1:3 \cr & {\text{Height}} \to h:H \cr & \frac{1}{3} = \frac{{1 \times h}}{{4 \times H}} \cr & \left\{ {\frac{1}{3},\,\pi {\text{ is constant}}} \right. \cr & \frac{h}{H} = \frac{4}{3} \cr} $$

$$\eqalign{ & \frac{2}{3}\pi {r^3} = 2425\frac{1}{2} \cr & \frac{2}{3} \times \frac{{22}}{7} \times {r^3} = \frac{{4851}}{2} \cr & {r^3} = \frac{{3 \times 7 \times 441}}{8} \cr & r = \frac{{21}}{2} \cr & r = 10.5{\text{ cm}} \cr} $$

$$l$$ + b + h = 20
$$\sqrt {{l^2} + {{\text{b}}^2} + {{\text{h}}^2}} = 12$$
$$l$$² + b² + h² = 144
2($$l$$b + bh + h$$l$$) = ?
20² = 144 + 2($$l$$b + bh + h$$l$$)
400 - 144 = 2($$l$$b + bh + h$$l$$)
256 = 2($$l$$b + bh + h$$l$$)

$$\eqalign{ & {\text{Volume of bucket}} \cr & = \frac{1}{3}\pi h\left( {{R^2} + {r^2} + Rr} \right) \cr & = \frac{1}{3} \times \frac{{22}}{7} \times 45\left( {{{28}^2} + {7^2} + 28 \times 7} \right) \cr & = \frac{{22}}{7} \times 15 \times 1029 \cr & = 48510{\text{ c}}{{\text{m}}^3} \cr} $$

Side of cube = 2.8 cm
Then radius of sphere $$ = \frac{{2.8}}{2} = 1.4{\text{ cm}}$$
Because, sphere be taken out of the cube.
⇒ Volume of sphere $$ = \frac{4}{3}\pi {r^3}$$
$$\eqalign{ & = \frac{4}{3} \times \frac{{22}}{7} \times 1.4 \times 1.4 \times 1.4 \cr & = \frac{{88 \times 0.392}}{3} \cr & = \frac{{34.496}}{3} \cr & = 11.498 \cr & = 11.50{\text{ c}}{{\text{m}}^3} \cr} $$

Length of smaller cuboid = $$\frac{{30}}{2}$$ = 15 cm
Breadth = $$\frac{{40}}{2}$$ = 20 cm
Height = $$\frac{{50}}{2}$$ = 25 cm
Total surface area of cuboid
= 8[2($$l$$b + bh + h$$l$$)]
= 8[2(15 × 20 + 20 × 25 + 25 × 15)]
= 16(300 + 500 + 375)
= 16 × 1175
= 18800