The perimeters of 5 squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to the sum of the area of these square is:

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 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 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 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 A

Explanation:

Clearly first we need to find the areas of the given squares, for that we need its side.

Side of sqaure = Perimeter/4

So sides are,

\begin{aligned}
\left(\frac{24}{4}\right),\left(\frac{32}{4}\right),\left(\frac{40}{4}\right),\left(\frac{76}{4}\right),\left(\frac{80}{4}\right) \\
= 6,8,10,19,20 \\
\text{Area of new square will be }\\
= [6^2+8^2+10^2+19^2+20^2] \\
= 36+64+100+361+400 \\
= 961 cm^2 \\
\text{Side of new Sqaure =}\sqrt{961} \\
= 31 cm \\
\text{Required perimeter =}(4\times31) \\
= 124 cm

\end{aligned}