Question

A rectangular park is to be designed whose breadth is $3m$ less than its length. Its area is to be $4$ square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude $12m$. Find length and breadth of rectangular park.

Answer 1

Here, Length of Rectangular Park $=l$

Breadth of Rectangular Park $=l-3$

Altitude of Triangular Park $=12$

Breadth of Triangular Park $=l-3$

Now,

$A_{Rectangle}=l(l-3)$

$A_{Triangle}=\frac{1}{2}\times 12(l-3)$

$A_{Rectangle}-A_{Triangle}=4$

$\Rightarrow l(l-3)-\frac{1}{2}\times 12(l-3)=4$
$\Rightarrow l^2-9l+14=0$
$\Rightarrow (l-2)(l-7)=0\Rightarrow l=2,7$

Since, breadth is $l-3$ , $l\ne 2$

$\therefore l=7m$ and breadth $=7-3=4m$