Question Detail

The area of incircle of an equilateral triangle of side 42 cm is :

$\begin{aligned} 462 cm^2 \end{aligned}$
$\begin{aligned} 452 cm^2 \end{aligned}$
$\begin{aligned} 442 cm^2 \end{aligned}$
$\begin{aligned} 432 cm^2 \end{aligned}$

Answer: Option A

Explanation:

$\begin{aligned} \text{Radius of incircle} = \frac{a}{2\sqrt3} \\ = \frac{42}{2\sqrt3} \\ = 7\sqrt{3} \\ \text{Area of incircle =} \\ \frac{22}{7}*49*3 = 462 cm^2 \end{aligned}$

1. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :

22 cm
20 cm
18 cm
10 cm

Answer: Option C

Explanation:

$\begin{aligned} \text{Area of triangle, A1 = }\frac{1}{2}*base*height \\ = \frac{1}{2}*15*12 = 90 cm^2 \\ \text{Area of second triangle =} 2*A1 \\ = 180 cm^2 \\ \frac{1}{2}*20*height = 180 \\ => height = 18 cm \end{aligned}$

2. 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.ft, how many feet of fencing will be required ?

88 feet
86 feet
84 feet
82 feet

Answer: Option A

Explanation:

We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 680
=> Breadth = 34 feet

Area to be fenced = 2B + L = 2*34 + 20
= 88 feet

3. A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.

10 m * 5 m
15 m * 5 m
20 m * 5 m
25 m * 5 m

Answer: Option C

Explanation:

From the question, 2b+l = 30
=> l = 30-2b

$\begin{aligned} Area = 100m^2 \\ => l \times b = 100 \\ => b(30-2b) = 100 \\ b^2 - 15b + 50 = 0 \\ =>(b-10)(b-5)=0 \\ \end{aligned}$
b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m

4. The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:

15 cm
20 cm
25 cm
30 cm

Answer: Option D

Explanation:

We know the product of diagonals is 1/2*(product of diagonals)

Let one diagonal be d1 and d2
So as per question

$\begin{aligned} \frac{1}{2}*d1*d2 = 150 \\ \frac{1}{2}*10*d2 = 150 \\ d2 = \frac{150}{5} = 30 \\ \end{aligned}$

5. The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.

4:25
4:15
3:25
3:15

Answer: Option A

Explanation:

Let the diagonals of the squares be 2x and 5x.
Then ratio of their areas will be

$\begin{aligned} \text{Area of square} = \frac{1}{2}*{Diagonal}^2 \\ \frac{1}{2}*{2x}^2:\frac{1}{2}*{5x}^2 \\ 4x^2:25x^2 = 4:25 \end{aligned}$

Question Detail

The area of incircle of an equilateral triangle of side 42 cm is :

1. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :

2. 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.ft, how many feet of fencing will be required ?

3. A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.

4. The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:

5. The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.

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Question Detail

The area of incircle of an equilateral triangle of side 42 cm is :

1. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :

2. 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.ft, how many feet of fencing will be required ?

3. A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.

4. The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:

5. The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.

Thanks ! Your comment will be approved shortly !

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