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Home > Practice > Arithmetic Aptitude > Area > Miscellaneous

61. The area of the largest triangle that can be inscribed in a semi-circle of radius r, is :

A. r²

B. 2r²

C. r³

D. 2r³

Discuss

Solution:

Required area :

\begin{aligned} = \frac{1}{2} \times Base \times Height \\ = (\frac{1}{2} \times 2 r \times r) \\ = r^{2} \end{aligned}

62. If radius of a circle is 3 cm, what is the area of the circle in sq. cm ?

A. $6 π$ sq. cm

B. $9 π$ sq. cm

C. $\frac{3 π}{2}$ sq. cm

D. $9 π^{2}$ sq. cm

Discuss

Solution:

Given: Radius of a circle = 3 cm
Area of circle :

\begin{aligned} = π r^{2} \\ = π \times 3^{2} \\ = 9 π sq . cm \end{aligned}

63. 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

Discuss

Solution:

Let each equal side of isosceles triangle be x cm
Perimeter of an isosceles triangle = 36 cm

\begin{aligned} ∴ x + x + 14 = 36 \\ \Rightarrow 2 x = 36 - 14 \\ \Rightarrow x = \frac{22}{2} \\ \Rightarrow x = 11 c m \end{aligned}

BD = DC = 7cm
From ΔABD
By using Pythagoras theorem :

\begin{aligned} A D = \sqrt{A B^{2} - B D^{2}} \\ = \sqrt{11^{2} - 7^{2}} \\ = \sqrt{121 - 49} \\ = \sqrt{72} \\ = 3 \times 2 \sqrt{2} \\ = 6 \sqrt{2} c m \end{aligned}

∴ Area of ΔABC

\begin{aligned} = \frac{1}{2} \times B C \times A D \\ = \frac{1}{2} \times 14 \times 6 \sqrt{2} \\ = 42 \sqrt{2} sq . cm \end{aligned}

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 :

A. 7.5 m

B. 8.5 m

C. 10.5 m

D. 12.5 m

Discuss

Solution:

Area of the floor :

\begin{aligned} = (\frac{510}{8.50}) m^{2} \\ = 60 m^{2} \end{aligned}

∴ Breadth of the room :

\begin{aligned} = (\frac{60}{8}) m \\ = 7.5 m \end{aligned}

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 :

A. $9 m^{2}$

B. $12 m^{2}$

C. $21 m^{2}$

D. $108 m^{2}$

Discuss

Solution:

Let l and b be the length and breadth of the rectangle respectively.
Then,

\begin{aligned} \Rightarrow \sqrt{l^{2} + b^{2}} = 15 \\ \Rightarrow (l^{2} + b^{2}) = {(15)}^{2} \\ \Rightarrow l^{2} + b^{2} = 225 \end{aligned}

And,

\begin{aligned} \Rightarrow l + b = 3 \\ \Rightarrow {(l - b)}^{2} = 9 \\ \Rightarrow l^{2} + b^{2} - 2 l b = 9 \\ \Rightarrow 225 - 2 l b = 9 \\ \Rightarrow 2 l b = 216 \\ \Rightarrow l b = 108 \end{aligned}

Hence, area of the field

= l b = 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 :

A. Rs. 673.92

B. Rs. 682.46

C. Rs. 691.80

D. Rs. 702.60

Discuss

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 ?

A. 51 m

B. 88 m

C. 104 m

D. 153 m

Discuss

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 :

A. Rs. 60000

B. Rs. 65000

C. Rs. 70000

D. Rs. 75000

Discuss

Solution:

Number of tiles required :

\begin{aligned} = \frac{Area of hall}{Area of each tile} \\ = (\frac{60 \times 40}{0.4 \times 0.4}) \\ = 15000 \end{aligned}

∴ 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 :

A. 1 : 1

B. 1 : $\sqrt{2}$

C. 1 : 2

D. 1 : 4

Discuss

Solution:

\begin{aligned} Required ratio \\ = \frac{a^{2}}{{(\sqrt{2} a)}^{2}} \\ = \frac{a^{2}}{2 a^{2}} \\ = \frac{1}{2} \\ = 1 : 2 \end{aligned}

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 :

A. 48 m²

B. 49 m²

C. 50 m²

D. 52 m²

Discuss

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

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