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

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}

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}

Answer: Option C

Explanation:

Let the original radius be R cm. New radius = 2R

\begin{aligned}
Area = \pi R^2 \\
\text{New Area =} \pi {2R}^2 \\
= 4\pi R^2 \\
\text{Increase in area =}(4\pi R^2 - \pi R^2) \\
= 3\pi R^2 \\
\text{Increase percent =} \frac{3\pi R^2}{\pi R^2}*100 \\
= 300 \%
\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}

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}