A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.

Answer: Option D

Explanation:

\begin{aligned} l*b = 460 m^2 \end{aligned} .. (i)
Lets suppose breadth of plot is = b rnLength will be,\begin{aligned}l=b*\frac{100+15}{100}\\=\frac{115b}{100}\end{aligned} .. (ii)

From (i) and (ii), we get :
\begin{aligned}
\frac{115b}{100}*b = 460 \\
b^2 = \frac{46000}{115} = 400 \\
=> b = \sqrt{400} = 20 meter
\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 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{Ratio of sides =}\frac{1}{2}:\frac{1}{3}:\frac{1}{4} \\
=6:4:3\\
Perimeter = 52 cm \\
\text{So sides are =} \\
\left( 52*\frac{6}{13}\right)cm,\left( 52*\frac{4}{13}\right)cm, \left( 52*\frac{3}{13}\right)cm
\end{aligned}
a = 24 cm, b = 16 cm and c = 12 cm
Length of the smallest side = 12 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