Question 58

The sum of the perimeters of an equilateral triangle and a rectangle is 90cm. The area, T, of the triangle and the area, R, of the rectangle, both in sq cm, satisfying the relationship $$R=T^{2}$$. If the sides of the rectangle are in the ratio 1:3, then the length, in cm, of the longer side of the rectangle, is


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 $$\frac{\sqrt{\ 3}}{4}m^2$$ and area of the rectangle is $$3a^2$$

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

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


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

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