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Area - Aptitude MCQ Questions and Solutions with Explanations

Area MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Area with easy and logical explanations. Area MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-7 section-1

Each section contains 100 MCQs on Area .
61
The area of the largest triangle that can be inscribed in a semi-circle of radius r, is :
Discuss
Answer & Solution
Answer: Option A
Solution:
Required area :
$$\eqalign{ & = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr & = \left( {\frac{1}{2} \times 2r \times r} \right) \cr & = {r^2} \cr} $$
Area mcq solution image
62
If radius of a circle is 3 cm, what is the area of the circle in sq. cm ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Given: Radius of a circle = 3 cm
Area of circle :
$$\eqalign{ & = \pi {r^2} \cr & = \pi \times {3^2} \cr & = 9\pi {\text{ sq}}{\text{.}}\,{\text{cm}} \cr} $$
63
The base of an isosceles is 14 cm and its perimeter is 36 cm. Find its area.
Discuss
Answer & Solution
Answer: Option A
Solution:
Let each equal side of isosceles triangle be x cm
Perimeter of an isosceles triangle = 36 cm
$$\eqalign{ & \therefore x + x + 14 = 36 \cr & \Rightarrow 2x = 36 - 14 \cr & \Rightarrow x = \frac{{22}}{2} \cr & \Rightarrow x = 11\,cm \cr} $$
Area mcq solution image
BD = DC = 7cm
From ΔABD
By using Pythagoras theorem :
$$\eqalign{ & AD = \sqrt {A{B^2} - B{D^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{11}^2} - {7^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {121 - 49} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {72} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 3 \times 2\sqrt 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 6\sqrt 2 \,cm \cr} $$
∴ Area of ΔABC
$$\eqalign{ & = \frac{1}{2} \times BC \times AD \cr & = \frac{1}{2} \times 14 \times 6\sqrt 2 \cr & = 42\sqrt 2 {\text{ sq}}{\text{. cm}} \cr} $$
64
The total cost of flooring a room at Rs. 8.50 per square metre is Rs. 510. If the length of the room is 8 m, its breadth is :
Discuss
Answer & Solution
Answer: Option A
Solution:
Area of the floor :
$$\eqalign{ & = \left( {\frac{{510}}{{8.50}}} \right){m^2} \cr & = 60\,{m^2} \cr} $$
∴ Breadth of the room :
$$\eqalign{ & = \left( {\frac{{60}}{8}} \right)m \cr & = 7.5\,m \cr} $$
65
The diagonal of a rectangular field is 15 metres and the difference between its length its length and width is 3 metres. The area of the rectangular field is :
Discuss
Answer & Solution
Answer: Option D
Solution:
Let l and b be the length and breadth of the rectangle respectively.
Then,
$$\eqalign{ & \Rightarrow \sqrt {{l^2} + {b^2}} = 15 \cr & \Rightarrow \left( {{l^2} + {b^2}} \right) = {\left( {15} \right)^2} \cr & \Rightarrow {l^2} + {b^2} = 225 \cr} $$
And,
$$\eqalign{ & \Rightarrow l + b = 3 \cr & \Rightarrow {\left( {l - b} \right)^2} = 9 \cr & \Rightarrow {l^2} + {b^2} - 2lb = 9 \cr & \Rightarrow 225 - 2lb = 9 \cr & \Rightarrow 2lb = 216 \cr & \Rightarrow lb = 108 \cr} $$
Hence, area of the field $$ = lb = 108\,{m^2}$$
66
A room 5m × 8 m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is Rs. 18 per sq. meter, the cost of carpeting the room will be :
Discuss
Answer & Solution
Answer: Option A
Solution:
Area of the carpet :
= [(5 - 0.20) × (8 - 0.20)] m2
= (4.8 × 7.8) m2
= 37.44 m2
∴ Cost of carpeting :
= Rs. (37.44 × 18)
= Rs. 673.92
67
The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 7803 sq. mtr, what is the breadth of the rectangular plot ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the breadth of the plot be x metres
Then, length of the plot = (3x) metres
x × 3x = 7803
⇒ 3x2 = 7803
⇒ x2 = 2601
⇒ x = $$\sqrt {2601} $$
⇒ x = 51 m
68
A coaching institute wants to execute tiling work for one of its teaching halls 60 m long and 40 m wide with a square tile of 0.4 m side. If each tile costs Rs. 5, the total cost of tiles would be :
Discuss
Answer & Solution
Answer: Option D
Solution:
Number of tiles required :
$$\eqalign{ & = \frac{{{\text{Area of hall}}}}{{{\text{Area of each tile}}}} \cr & = \left( {\frac{{60 \times 40}}{{0.4 \times 0.4}}} \right) \cr & = 15000 \cr} $$
∴ Total cost of tiles :
= Rs. (15000 × 5)
= Rs. 75000
69
The ratio of the area of a square to that of the square drawn on diagonal is :
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Required ratio}} \cr & = \frac{{{a^2}}}{{{{\left( {\sqrt 2 a} \right)}^2}}} \cr & = \frac{{{a^2}}}{{2{a^2}}} \cr & = \frac{1}{2} \cr & = 1:2 \cr} $$
70
The area of the four walls of a room is 120 m2 and the length is twice the breadth. If the height of the room is 4 m, then the area of the floor is :
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the breadth = x metres and length = (2x) metres
Area of 4 walls = [2(2x + x)× 4] m2 = (24x) m2
∴ 24x = 120
⇒ x = 5
So, length = 10 m, and breadth = 5 m
Area of the floor = (10 × 5) m2 = 50 m2