Question

Surya has a land in the shape of a rhombus. He wants to cultivate rice and wheat. he divided the land in two equal parts. The perimeter of the land is $400m$ and one of the diagonals is $160m$, how much area is available for rice ?

• A. $4600m^2$
• B. $4500m^2$
• C. $4700m^2$
• D. $4800m^2$

Answer 1

Answer: (D)

$4800m^2$

Let $ABCD$ be the land in which $△ADB$ represents the area of rice and $△BDC$ represents the area of wheat.

Since it is given that the perimeter of the land is $400$ m therefore, each side $=\frac{400}{4}=100$ m and the given diagonal $BD=160$ m.

To find the area available for rice we need to find the area of $△ADB$.

Using herons formula:

$S=\frac{100+100+160}{2}=\frac{360}{2}=180$ m

Area of $△ADB$:

$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{180(180-100)(180-100)(180-160)}=\sqrt{180\times 80\times 80\times 20}=\sqrt{23040000}$

$=4800$ m${}^2$

Hence, the area available for rice is $4800$m${}^2$.