The difference of the areas of two squares drawn on two line segments in 32 sq. cm. Find the length of the greater line segment if one is longer than the other by 2 cm.

Answer: Option A

Explanation:

We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 680
=> Breadth = 34 feet

Area to be fenced = 2B + L = 2*34 + 20
= 88 feet

Answer: Option C

Explanation:

Let the lengths of the line segments be x and x+2 cm
then,

$$\begin{aligned} {(x+2)}^2 - x^2 = 32 \\ x^2 + 4x + 4 - x^2 = 32 \\ 4x = 28 \\ x = 7 cm \end{aligned}$$

Answer: Option D

Explanation:

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

$$\begin{aligned} \text{Original area } = \text{xy } m^2 \\ \text{New Length }= \frac{120}{100}x = \frac{6}{5}x \\ \text{New Breadth }= \frac{120}{100}y = \frac{6}{5}y \\ =>\text{New Area }= \frac{6}{5}x * \frac{6}{5}y \\ =>\text{New Area }= \frac{36}{25}xy \\ \text{Area Difference} = \frac{36}{25}xy - xy \\ = \frac{11}{25}xy \\ Increase \% = \frac{Differnce}{Actual}*100 \\ = \frac{11xy}{25}*\frac{1}{xy}*100 = 44\% \end{aligned}$$

Answer: Option B

Explanation:

We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm

Area of third Square = 10*10 - 8*8
= 36 cm

So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm

Answer: Option A

Explanation:

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