The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

Answer: Option A

Explanation:

$$\begin{aligned} 2\pi R1 = 4 \pi \\ => R1 = 2 \\ 2\pi R2 = 8 \pi \\ => R2 = 4 \\ \text{Original Area =} 4\pi * 2^2 \\ = 16 \pi \\ \text{New Area =} 4\pi * 4^2 \\ = 64 \pi \end{aligned}$$

So the area quadruples.

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 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:

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,

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

Answer: Option A

Explanation:

$$\begin{aligned} \text{Radius of incircle} = \frac{a}{2\sqrt3} \\ = \frac{42}{2\sqrt3} \\ = 7\sqrt{3} \\ \text{Area of incircle =} \\ \frac{22}{7}*49*3 = 462 cm^2 \end{aligned}$$