Practice MCQ Questions and Answer on Area

11.

How many metres of carpet 63 cm wide will be required to cover the floor of a room 14 metres by 9 metres ?

• (A) 185 metres
• (B) 200 metres
• (C) 210 metres
• (D) 220 metres

Show Answer

 Area of the floor = (14 × 9) m2 = 126 m2 ∴ Length of the carpet : $$\eqalign{ & = \left( {\frac{{126}}{{63}} \times 100} \right){\text{m}} \cr & = 200{\text{ metres}} \cr} $$

12.

The ratio of the areas of the in-circle and the circum-circle of a square is :

• (A) 1 : 2
• (B) $$\sqrt{2}:1$$
• (C) $$1:\sqrt{2}$$
• (D) 2 : 1

Show Answer

 Let r1 and r2 be the radii of the in-circle and circum-circle of a square respectively and let each side of the square be a. Then, $$\eqalign{ & {r_1} = \frac{a}{2} \cr & {r_2} = \frac{1}{2} \times {\text{diagonal of the sequence}} \cr & {r_2} = \frac{1}{2} \times \sqrt 2 a \cr & {r_2} = \frac{{\sqrt 2 a}}{2}cm \cr} $$ ∴ Required ratio : $$\eqalign{ & = \frac{{\pi \times {{\left( {\frac{a}{2}} \right)}^2}}}{{\pi \times {{\left( {\frac{{\sqrt 2 a}}{2}} \right)}^2}}} \cr & = 1:2 \cr} $$

13.

A diagonal of a rhombus is 6 cm. If its area is 24 cm² then the length of each side of the rhombus is :

• (A) 5 cm
• (B) 6 cm
• (C) 7 cm
• (D) 8 cm

Show Answer

 $$\eqalign{ & \frac{1}{2} \times 6 \times d = 24 \cr & \Rightarrow d = \frac{{24}}{3} \cr & \Rightarrow d = 8\,cm \cr & OA = 4\,cm\,{\text{and}}\,OB = 3\,cm \cr & \therefore AB = \sqrt {{{\left( {OA} \right)}^2} + {{\left( {OB} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{4^2} + {3^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5\,cm \cr} $$

14.

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $$12\sqrt{2}$$ cm, then the area of the triangle is :

• (A) $$24\sqrt{2}$$ cm2
• (B) $$24\sqrt{3}$$ cm2
• (C) $$48\sqrt{3}$$ cm2
• (D) $$64\sqrt{3}$$ cm2

Show Answer

 Let the side of the square be a cm Then, its diagonal = $$\sqrt 2 $$ a cm Now, $$\sqrt 2 $$ a = $$12\sqrt 2 $$ ⇒ a = 12 cm Perimeter of the square = 4a = 48 cm Perimeter of the equilateral triangle = 48 cm Each side of the triangle = 16 cm Area of the triangle : $$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times 16 \times 16} \right)c{m^2} \cr & = \left( {64\sqrt 3 } \right)c{m^2} \cr} $$

15.

The diagonal of a rectangle is $$\sqrt{41}$$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be :

• (A) 9 cm
• (B) 18 cm
• (C) 20 cm
• (D) 41 cm

Show Answer

 $$\eqalign{ & \sqrt {{l^2} + {b^2}} = \sqrt {41} \cr & \Rightarrow {l^2} + {b^2} = 41 \cr & Also,\,\,lb = 20 \cr & {\left( {l + b} \right)^2} = \left( {{l^2} + {b^2}} \right) + 2lb \cr & {\left( {l + b} \right)^2} = 41 + 40 \cr & {\left( {l + b} \right)^2} = 81 \cr & \left( {l + b} \right) = 9 \cr & \therefore {\text{ Perimeter}} = 2\left( {l + b} \right) = 18\,cm \cr} $$

16.

An order was placed for supply of carpet of breadth 3 metres, the length of carpet was 1.44 times of breadth. Subsequently the breadth and length were increased by 25 and 40 percent respectively. At the rate of 45 per square metre, what would be the increase in the cost of the carpet ?

• (A) Rs. 398.80
• (B) Rs. 437.40
• (C) Rs. 583.20
• (D) Rs. 1020.60

Show Answer

 Original breadth = 3 m Original length : $$\eqalign{ & = \left( {1.44 \times 3} \right)m \cr & = 4.32\,m \cr} $$ New breadth : $$\eqalign{ & = \left( {125\% {\text{ of 3}}} \right)m \cr & = \left( {\frac{{125}}{{100}} \times 3} \right)m \cr & = 3.75\,m \cr} $$ New length : $$\eqalign{ & = \left( {140\% {\text{ of 4}}{\text{.32}}} \right)m \cr & = \left( {\frac{{140}}{{100}} \times 4.32} \right)m \cr & = 6.048\,m \cr} $$ Original area : $$\eqalign{ & = \left( {4.32 \times 3} \right){m^2} \cr & = 12.96\,{m^2} \cr} $$ New area : $$\eqalign{ & = \left( {6.048 \times 3.75} \right){m^2} \cr & = 22.68\,{m^2} \cr} $$ Increase in area : $$\eqalign{ & = \left( {22.68 - 12.96} \right){m^2} \cr & = 9.72\,{m^2} \cr} $$ ∴ Increase in cost : $$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {9.72 \times 45} \right) \cr & = {\text{Rs}}{\text{.}}\,{\text{437}}{\text{.40}} \cr} $$

17.

Four circles having equal radii are drawn with centres at the four corners of a square. Each circle touches the other two adjacent circles. If the remaining area of the square is 168 cm², what is the size of the radius of the circle ? (in centimetres)

• (A) 14 centimetres
• (B) 1.4 centimetres
• (C) 35 centimetres
• (D) 21 centimetres

Show Answer

 Let the radius of each circle be r cm Then the side of the square will be 2r cm Area covered by the four circle in the square $$\eqalign{ & = 4 \times \frac{1}{4} \times \pi {r^2} \cr & = \pi {r^2}c{m^2} \cr} $$ Area of the square : $$\eqalign{ & = {\left( {2r} \right)^2} \cr & = 4{r^2}c{m^2} \cr} $$ Now, according to the question, Remaining area of the square $$\eqalign{ & 4{r^2} - \pi {r^2} = 168 \cr & \Rightarrow {r^2}\left( {4 - \frac{{22}}{7}} \right) = 168 \cr & \Rightarrow {r^2} \times \left( {28 - 22} \right) = 168 \times 7 \cr & \Rightarrow {r^2} = \frac{{168 \times 7}}{6} \cr & \Rightarrow {r^2} = 28 \times 7 = 7 \times 4 \times 7 \cr & \Rightarrow r = \sqrt {7 \times 7 \times 2 \times 2} \cr & \Rightarrow r = 7 \times 2 \cr & \Rightarrow r = 14\,cm \cr} $$

18.

If the length and breadth of a rectangular field are increased, the area increases by 50%. If the increase in length was 20 %, by what percentage was the breadth increased ?

• (A) 20%
• (B) 25%
• (C) 30%
• (D) Data inadequate

Show Answer

 Let the original length and breadth of the rectangle be l and b respectively Then, original area = lb New length = 120% of l = $$\frac{6l}{5}$$ New area = 150% of lb = $$\frac{3lb}{2}$$ New breadth :$$\eqalign{ & = \left( {\frac{{3lb}}{2} \times \frac{5}{{6l}}} \right) \cr & = \frac{{5b}}{4} \cr} $$ Increase in breadth :$$\eqalign{ & = \left( {\frac{{5b}}{4} - b} \right) \cr & = \frac{b}{4} \cr} $$ ∴ Increase % : $$\eqalign{ & = \left( {\frac{b}{4} \times \frac{1}{b} \times 100} \right)\% \cr & = 25\% \cr} $$

19.

Two boys are running on two different circular paths with same centre. If their radii are 5 m and 10 m, the maximum possible distance between them is :

• (A) 5 m
• (B) 10 m
• (C) 15 m
• (D) 20 m

Show Answer

 Since the diameter is the longest chord of a circle So, maximum possible distance = (10 + 5) m = 15 m

20.

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 = $$5\sqrt 3 \,m$$