What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square meter ?

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 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{We know area of square =}a^2 \\
\text{Area of triangle =}\frac{1}{2}*a*h \\
=> \frac{1}{2}*a*h = a^2 \\
=> h = 2a

\end{aligned}

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:

Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
\begin{aligned}
\text{Area of triangle =}\frac{1}{2}*b*h1\\
\text{Area of rectangle =}b*h2\\
\text{As per question }\\
\frac{1}{2}*b*h1 = b*h2 \\

\frac{1}{2}*b*h1 = b*100 \\

h1 = 100*2 = 200 m \\

\end{aligned}