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. Find the circumference of a circle, whose area is 24.64 meter sqaure

17.90m
17.80m
17.60m
17.40m

Answer: Option C

Explanation:

\begin{aligned}
\text{Area of Square =} \pi*r^2\\
=> \pi*r^2 = 24.64 \\
=> r^2 = \frac{24.64}{22}*7 \\
=> r^2 = 7.84 \\
=> r = \sqrt{7.84} \\
=> r = 2.8 \\
\text{Circumference =}2\pi*r\\
= 2*\frac{22}{7}*2.8 \\
= 17.60m

\end{aligned}

2. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.

\begin{aligned} 120m^2 \end{aligned}
\begin{aligned} 130m^2 \end{aligned}
\begin{aligned} 140m^2 \end{aligned}
\begin{aligned} 150m^2 \end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2

\end{aligned}

3. If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?

Area is quadrupled
Area is tripled
Area is doubles
Area become half

Answer: Option A

Explanation:

\begin{aligned}
2\pi R1 = 4 \pi \\
=> R1 = 2 \\
2\pi R2 = 8 \pi \\
=> R2 = 4 \\
\text{Original Area =} 4\pi * 2^2 \\
= 16 \pi \\
\text{New Area =} 4\pi * 4^2 \\
= 64 \pi
\end{aligned}

So the area quadruples.

4. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

Answer: Option D

Explanation:

Let original length = x metres and original breadth = y metres.

\begin{aligned}
\text{Original area } = \text{xy } m^2 \\
\text{New Length }= \frac{120}{100}x = \frac{6}{5}x \\
\text{New Breadth }= \frac{120}{100}y = \frac{6}{5}y \\
=>\text{New Area }= \frac{6}{5}x * \frac{6}{5}y \\
=>\text{New Area }= \frac{36}{25}xy \\

\text{Area Difference} = \frac{36}{25}xy - xy \\
= \frac{11}{25}xy \\

Increase \% = \frac{Differnce}{Actual}*100 \\
= \frac{11xy}{25}*\frac{1}{xy}*100 = 44\%

\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}