Question

The perimeter of a square field is $40$ m more than another square field. Three times the area of the smaller field is $50$ $m^2$ more than the area of the larger field. Find the sides of both the fields.

Answer 1

Let the side of small square be $xm.$

so , the area of small square will be $x^2{m}^2$

Thus the side of big square will be $x+40$ $m.$

so , the area of big square will be $(x+40{)}^2{m}^2$

According to question,

$3x^2=(x+40{)}^2+50$

$3x^2=x^2+1600+80x+50$

$2x^2=1650+80x$

$x^2=825+40x$

$x^2-40x-825$

by quadratic formula,

$x=$ $\frac{40\pm \sqrt{{\left(40\right)}^2-4\left(1\right)(-825)}}{2\left(1\right)}$

$x=$ $\frac{40\pm \sqrt{1600+3300}}{2\left(1\right)}$

$x=$ $\frac{40\pm \sqrt{4900}}{2}$

$x=$ $\frac{40\pm \sqrt{{70}^2}}{2}$

$x=$ $\frac{40\pm 70}{2}$

$x=$ $20\pm 35$

Length can not be negative so,

$x=$ $20+35m$

$x=$ $55m$

Thus length of side of a larger square will be $95m$