Question 16.

Let the sides of the rectangle be "a" and "3a" m. Hence the perimeter of the rectangle is 8a.

Let the side of the equilateral triangle be "m" cm. Hence the perimeter of the equilateral triangle is "3m" cm. Now we know that 8a+3m=90......(1)

Moreover area of the equilateral triangle is  $\frac{\sqrt3}4m^2$ and area of the rectangle is $3a^2$

According to the relation given $(\frac{\sqrt3}4m^2)^2$= $3a^2$

$\frac3{16}m^4$= $3a^2$ or $a^2 = \frac{m^4}{16}$

a = $\frac{m^2}4$

Substituting this in (1) we get $2m^2+3m−90$ = 0

solving this we get m=6 (ignoring the negative value since side can't be negative)

Hence a=9 and the longer side of the rectangle will be 3a=27cm