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MCQ Questions on Mensuration 3D for Competitive Exams

Ace your competitive exams with our MCQ questions on Mensuration 3D. Learn to solve volume and surface area problems with questions designed to test your knowledge effectively. Page-9 section-1

Each section contains 100 MCQs on Mensuration 3D .
81
Height of a prism-shaped part of a machine is 8 cm and its base is an isosceles triangle, whose each of the equal sides is 5 cm and remaining side is 6 cm. The volume of the part is
Discuss
Answer & Solution
Answer: Option B
Solution:
Volume of the part (prism) = Area of base × height
Area of base (Isosceles Δ)
$$\eqalign{ & = \frac{b}{4}\sqrt {4{a^2} - {b^2}} \cr & = \frac{6}{4}\sqrt {4{{\left( 5 \right)}^2} - {{\left( 6 \right)}^2}} \cr & = 12{\text{ c}}{{\text{m}}^2} \cr} $$
Volume of prism = 12 × 8 = 96 cm3
82
The base of a right pyramid is an equilateral triangle with area 16√3 cm2. If the area of one of its lateral faces is 30 cm2, then its height (in cm) is:
Discuss
Answer & Solution
Answer: Option C
Solution:
Mensuration 3D mcq question image
$$\eqalign{ & {\text{Area of }}\Delta ABC = \frac{{\sqrt 3 }}{4}{a^2} \cr & 16\sqrt 3 = \frac{{\sqrt 3 }}{4}{a^2} \cr & a = 8 \cr & {\text{In }}\Delta ABD, \cr & \frac{1}{2} \times 8 \times MD = 30 \cr & MD = \frac{{15}}{2} \cr & {\text{Now, in }}\Delta MOD, \cr & MO = \frac{a}{{2\sqrt 3 }} = \frac{4}{{\sqrt 3 }} \cr & M{D^2} = M{O^2} + O{D^2} \cr & \frac{{225}}{4} = {\left( {\frac{4}{{\sqrt 3 }}} \right)^2} + O{D^2} \cr & OD = \sqrt {\frac{{225}}{4} + \frac{{16}}{3}} \cr & OD = \sqrt {\frac{{611}}{{12}}} \cr} $$
83
A metallic solid spherical ball of radius 3 cm is melted recast into three spherical balls. The radii of two of these balls are 2 cm and 1.5 cm. What is the surface area (in cm2) of the third ball?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\frac{4}{3}$$π × 3 × 3 × 3 = $$\frac{4}{3}$$π × 23 + $$\frac{4}{3}$$π × (1.5)3 + $$\frac{4}{3}$$π × r3
27 = 8 + 3.375 + r3
15.625 = r3
r = 2.5 cm
∴ The surface area of the third ball = 4π × 2.5 × 2.5 = 25π cm2
84
The circumference of the base of a cylindrical vessel is 158.4 cm and height is 1 m. How many litres of water can it hold (correct to one decimal place)?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & h = 1\,{\text{m}} \cr & 2\pi r = 158.4 \cr & 2 \times \frac{{22}}{7} \times r = 158.4 \cr & r = 3.6 \times 7 \cr & V = \pi {r^2}h \cr & = \frac{{\frac{{22}}{7} \times 3.6 \times 7 \times 3.6 \times 7 \times 100}}{{1000}} \cr & = \frac{{199584}}{{1000}} \cr & = 199.6 \cr} $$
85
The height of a right prism with a square base is 15 cm. If the area of the total surface of the prism is 608 sq. cm, its volume is
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the side of square base = a cm
⇒ 2a2 + 4a × h = 608
⇒ 2a2 + 4a × 15 = 608
⇒ a2 + 30a = 304
⇒ a2 + 38a - 8a - 304 = 0
⇒ a(a + 38) - 8(a + 38) = 0
⇒ a = -38, 8
⇒ a = 8 cm
∴ Volume of prism = 8 × 8 × 15 = 960 cm3
86
The area of the base of a right circular cone is 81π cm2 its height is 12 cm. What is curved surface area (in cm2) of the cone?
Discuss
Answer & Solution
Answer: Option B
Solution:
πr2 = 81π
r = 9
Curved surface area of cone = πr$$l$$
$$l$$ = $$\sqrt {{{12}^2} + {9^2}} $$
$$l$$ = 15
Curved surface area = π × 9 × 15 = 135π
87
A rectangular sheet of metal is 40 cm by 15 cm. Equal squares of side 4 cm are cut off at the corners and the remaining is folded up to form an open rectangular box. The volume of the box is
Discuss
Answer & Solution
Answer: Option A
Solution:
Volume of the box = $$l$$ × b × h
= (40 - 8) × (15 - 8) × 4
= 32 × 7 × 4
= 896 cm3
88
If the sum of three dimensions and the total surface area of a rectangular box are 12 cm and 94 cm2 respectively, then the maximum length of a stick that can be placed inside the box is
Discuss
Answer & Solution
Answer: Option A
Solution:
Let length = $$l$$, breadth = b, height = h
Given that
($$l$$ + b + h) = 12 cm
Total surface area of box
= 2($$l$$b + bh + h$$l$$)
= 94 m2 (Given)
⇒ ($$l$$ + b + h)2 = $$l$$2 + b2 + h2 + 2($$l$$b + bh + h$$l$$)
⇒ (12)2 = $$l$$2 + b2 + h2 + 94
⇒ 144 - 94 = $$l$$2 + b2 + h2
⇒ 50 = $$l$$2 + b2 + h2
Diagonal of box $$ = \sqrt {{l^2} + {b^2} + {h^2}} $$
∴ Length of longest rod that can be put inside the box
$$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {50} \cr & = 5\sqrt 2 {\text{ cm}} \cr} $$
89
If the length of each side of a regular tetrahedron is 12 cm, then the volume of the tetrahedron is
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Volume of tetrahedron}} = \frac{{{a^3}}}{{6\sqrt 2 }} \cr & = \frac{{{{12}^3}}}{{6\sqrt 2 }} \cr & = \frac{{1728}}{{6\sqrt 2 }} \cr & = 144\sqrt 2 {\text{ c}}{{\text{m}}^3} \cr} $$
90
A right circular cylinder is partially filled with water. Two iron spherical balls are completely immersed in the water so that the height of the water in the cylinder rises by 4 cm. If the radius of one ball is half of the other and the diameter of the cylinder is 18 cm, then the radii of the spherical balls are
Discuss
Answer & Solution
Answer: Option C
Solution:
Mensuration 3D mcq question image
Total volume of 2 balls = volume of cylinder of height = 4 cm
$$\eqalign{ & \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 = \pi {9^2} \times 4 \cr & {r_1} = 2{r_2}{\text{ }}\left( {{\text{given}}} \right) \cr & \frac{4}{3}\pi \left( {8r_2^3 + r_2^3} \right) = \pi \times 9 \times 9 \times 4 \cr & 9r_2^3 = 9 \times 9 \times 3 \cr & {r_2} = 3 \cr & {r_1} = 2{r_2} = 2 \times 3 = 6 \cr} $$