Question Detail

If the area of a square with the side a is equal to the area of a triangle with base a, then the altitude of the triangle is.

a
a/2
2a
None of above

Answer: Option C

Explanation:

\begin{aligned}
\text{We know area of square =}a^2 \\
\text{Area of triangle =}\frac{1}{2}*a*h \\
=> \frac{1}{2}*a*h = a^2 \\
=> h = 2a

\end{aligned}

1. 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}

2. 50 square stone slabs of equal size were needed to cover a floor area of 72 sq.m. Find the length of each stone slab.

110 cm
116 cm
118 cm
120 cm

Answer: Option D

Explanation:

Area of each slab =
\begin{aligned}
\frac{72}{50}m^2 = 1.44 m^2\\
\text{Length of each slab =}\sqrt{1.44} \\
= 1.2m = 120 cm

\end{aligned}

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. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

200 m
150 m
148 m
140 m

Answer: Option A

Explanation:

Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
\begin{aligned}
\text{Area of triangle =}\frac{1}{2}*b*h1\\
\text{Area of rectangle =}b*h2\\
\text{As per question }\\
\frac{1}{2}*b*h1 = b*h2 \\

\frac{1}{2}*b*h1 = b*100 \\

h1 = 100*2 = 200 m \\

\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