There is a plot which is rectangular in shape and it area is 460 square metres. What will be breadth of the plot if length of plot is 15% more than the breadth of plot ?

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:

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 B

Explanation:

In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344

So required cost will be = 344 * 10 = 3440

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 C

Explanation:

$$\begin{aligned} \frac{a^2}{b^2} = \frac{225}{256} \\ \frac{15}{16} \\ <=> \frac{4a}{4b} = \frac{4*15}{4*16} \\ = \frac{15}{16} = 15:16 \end{aligned}$$