Question Detail

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}

1. 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

2. 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}

3. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is ?

Answer: Option D

Explanation:

Let original length = x
and original width = y
Decrease in area will be
\begin{aligned}
= xy-\left( \frac{80x}{100}\times\frac{90y}{100}\right) \\
= \left(xy- \frac{18}{25}xy\right) \\
= \frac{7}{25}xy \\
\text{Decrease = }\left(\frac{7xy}{25xy} \times100\right) \% \\
= 28\%
\end{aligned}

4. 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}

5. What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square meter ?

Rs. 3430
Rs. 3440
Rs. 3450
Rs. 3460

Answer: Option B

Explanation:

In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344

So required cost will be = 344 * 10 = 3440