The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares .

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 B

Explanation:

Let breadth =x metres.
Then length =(115x/100)metres.
\begin{aligned}
=x*\frac{115x}{100}= 460\\
x^2=(460 x 100/115) \\
x^2=400 \\
x= 20 \\
\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}

Answer: Option D

Explanation:

Let original length = x
and original width = y
Decrease in area will be
\begin{aligned}
= xy-\left( \frac{80x}{100}\times\frac{90y}{100}\right) \\
= \left(xy- \frac{18}{25}xy\right) \\
= \frac{7}{25}xy \\
\text{Decrease = }\left(\frac{7xy}{25xy} \times100\right) \% \\
= 28\%
\end{aligned}

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}