What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle ?

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 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}

Answer: Option C

Explanation:

Let each side be a cm, then
\begin{aligned}
\left(\frac{a}{2}\right)^2+{10}^2 = a^2 \\
<=>\left(a^2-\frac{a^2}{4}\right) = 100 \\
<=> \frac{3a^2}{4} = 100 \\
a^2 = \frac{400}{3} \\
Area = \frac{\sqrt{3}}{4}*a^2 \\
= \left(\frac{\sqrt{3}}{4}*\frac{400}{3}\right)cm^2 \\
= \frac{100}{\sqrt{3}}cm^2
\end{aligned}

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}

Answer: Option C

Explanation:

Let Breadth = x cm,
then, Length = 3x cm
\begin{aligned}
x^2+{(3x)}^2 = {(8\sqrt{10})}^2 \\
=> 10x^2 = 640 \\
=> x = 8 \\
\end{aligned}
So, length = 24 cm and breadth = 8 cm
Perimeter = 2(l+b)
= 2(24+8) = 64 cm