Practice MCQ Questions and Answer on Area

41.

A courtyard is 25 m long and 16 m broad is to be paved with bricks of dimensions 20 cm by 10 cm. What is the total number of bricks required ?

• (A) 16000 $${\text{ bricks}}$$
• (B) 18000 $${\text{ bricks}}$$
• (C) 20000 $${\text{ bricks}}$$
• (D) 22000 $${\text{ bricks}}$$

Show Answer

 Given length and breadth of courtyard is 25 m long and 16 m respectively Area of courtyard : $$\eqalign{ & = \left( {25 \times 16} \right)sq.{m^2} \cr & = 400\,sq.{m^2} \cr} $$ Dimensions of bricks 20 cm by 10 cm Area of the surface of brick $$ = \left( {\frac{{20 \times 10}}{{10000}}} \right){m^2}$$ ∴ Number of bricks : $$\eqalign{ & = \frac{{400}}{{\frac{{20 \times 10}}{{10000}}}} \cr & = \frac{{400 \times 10000}}{{20 \times 10}} \cr & = 20000\,{\text{ bricks}} \cr} $$

42.

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 :

• (A) 7.5 m
• (B) 8.5 m
• (C) 10.5 m
• (D) 12.5 m

Show Answer

 Area of the floor : $$\eqalign{ & = \left( {\frac{{510}}{{8.50}}} \right){m^2} \cr & = 60\,{m^2} \cr} $$ ∴ Breadth of the room : $$\eqalign{ & = \left( {\frac{{60}}{8}} \right)m \cr & = 7.5\,m \cr} $$

43.

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

• (A) 16 cm
• (B) 18 cm
• (C) 24 cm
• (D) Data inadequate

Show Answer

 $$\eqalign{ & \frac{{2\left( {l + b} \right)}}{b} = \frac{5}{1} \cr & \Rightarrow 2l + 2b = 5b \cr & \Rightarrow 3b = 2l \cr & b = \frac{2}{3}l \cr & {\text{Then,}} \cr & {\text{Area = }}216\,c{m^2} \cr & \Rightarrow l \times b = 216 \cr & \Rightarrow l \times \frac{2}{3}l = 216 \cr & \Rightarrow {l^2} = 324 \cr & l = 18\,cm \cr} $$

44.

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

• (A) 34
• (B) 40
• (C) 68
• (D) 88

Show Answer

 We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

45.

The base of an isosceles is 14 cm and its perimeter is 36 cm. Find its area.

• (A) 42$$\sqrt 2 $$ sq. cm
• (B) 42 sq. cm
• (C) 84 sq. cm
• (D) 48 sq. cm

Show Answer

 Let each equal side of isosceles triangle be x cm Perimeter of an isosceles triangle = 36 cm $$\eqalign{ & \therefore x + x + 14 = 36 \cr & \Rightarrow 2x = 36 - 14 \cr & \Rightarrow x = \frac{{22}}{2} \cr & \Rightarrow x = 11\,cm \cr} $$ 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 $$\eqalign{ & = \frac{1}{2} \times BC \times AD \cr & = \frac{1}{2} \times 14 \times 6\sqrt 2 \cr & = 42\sqrt 2 {\text{ sq}}{\text{. cm}} \cr} $$

46.

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

• (A) 814
• (B) 820
• (C) 840
• (D) 844

Show Answer

 $$\eqalign{ & {\text{Length}}\,{\text{of}}\,{\text{largest}}\,{\text{tile}} = \cr & {\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\,{\text{of}}\,1517\,cm\,{\text{and}}\,902\,cm = 41\,cm \cr & {\text{Area}}\,{\text{of}}\,{\text{each}}\,{\text{tile}} = \left( {41 \times 41} \right)c{m^2} \cr & \therefore {\text{Required}}\,{\text{number}}\,{\text{of}}\,{\text{tiles}} \cr & = {\frac{{1517 \times 902}}{{41 \times 41}}} \cr & = 814 \cr} $$

47.

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 :

• (A) Rs. 60000
• (B) Rs. 65000
• (C) Rs. 70000
• (D) Rs. 75000

Show Answer

 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

48.

Three boys are standing on a circular boundary of a fountain. They are at equal distance from each other. If the radius of the boundary is 5 m, the shortest distance between any two boys is :

• (A) $$\frac{{5\sqrt 3 }}{2}m$$
• (B) $$5\sqrt 3 \,m$$
• (C) $$\frac{{15\sqrt 3 }}{2}m$$
• (D) $$\frac{{10\pi }}{3}m$$

Show Answer

 Let A, B and C denote the portions of the three boys Then, AB = BC = AC So, ΔABC is equilateral Let the side of ΔABC be a Then, $$\eqalign{ & \frac{a}{{\sqrt 3 }} = 5 \cr & \Rightarrow a = 5\sqrt 3 \cr} $$ ∴ Required shortest distance = $$5\sqrt 3 \,m$$

49.

An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square ?

• (A) 2 : $$\sqrt 3 $$
• (B) 4 : $$\sqrt 3 $$
• (C) $$\sqrt 3 $$ : 2
• (D) $$\sqrt 3 $$ : 4

Show Answer

 Let the side of the square be a cm Then, the length of its diagonal = $$\sqrt 2 $$ a cm Area of equilateral triangle with side : $$\eqalign{ & = \sqrt 2 a \cr & = \frac{{\sqrt 3 }}{4} \times {\left( {\sqrt 2 a} \right)^2} \cr & = \frac{{\sqrt 3 {a^2}}}{2} \cr} $$ ∴ Required ratio : $$\eqalign{ & = \frac{{\sqrt 3 {a^2}}}{2}:{a^2} \cr & = \sqrt 3 :2 \cr} $$

50.

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 ?

• (A) 51 m
• (B) 88 m
• (C) 104 m
• (D) 153 m

Show Answer

 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