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​​​​​​​​​​​​​​​​​​​​​​​​​​​​The sum of the perimeters of an equilateral triangle and a rectangle is $90 \; \text{cm}.$ The area, $\text{T},$ of the triangle and the area, $\text{R},$ of the rectangle, both in $\text{sq cm},$ satisfy the relationship $\text{R = T}^{2}.$ 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

1. $24$
2. $27$
3. $21$
4. $18$

Given that, the sum of the perimeter of an equilateral triangle and a rectangle is $90 \; \text{cm}.$

Let the side of an equilateral triangle be $x\text{’} \; \text{cm}.$

Perimeter of equilateral triangle $= x + x + x = 3x \; \text{cm}.$

Let the length and width of a rectangle be $a\text{’} \; \text{cm}$ respectively. $(b>a)$

We have, $a : b = 1 : 3$

$\Rightarrow \frac{a}{b} = \frac{1}{3} = k \; (\text{let})$

$\Rightarrow \boxed {a = k, b = 3k}$

Perimeter of rectangle $= 2 (a+b) = 2 (k+3k)= 2 (4k) = 8k$

So, $3x + 8k = 90 \quad \longrightarrow (1)$

The area of equilateral triangle $\text{T} = \frac{\sqrt{3}}{4} x^{2} \; \text{sq. cm}$

The area of rectangle $\text{R} = ab = (k) (3k) = 3k^{2} \; \text{sq. cm}$

We have, $\text{R} = \text{T}^{2}$

$\Rightarrow 3k^{2} = \left( \frac{\sqrt{3}}{4} x^{2} \right)^{2}$

$\Rightarrow 3k^{2} = \frac{(\sqrt{3})^{2}}{16} x^{4}$

$\Rightarrow k^{2} = \frac{1}{16} x^{4}$

$\Rightarrow \boxed{ k = \frac{1}{4} x^{2}}$

Put the value of $`k\text{’}$ in the equation $(1),$ we get.

$3x + 8k = 90$

$\Rightarrow 3x + 8 \left( \frac{1}{4} x^{2} \right) = 90$

$\Rightarrow 2x^{2} + 3x – 90 = 0$

$\Rightarrow 2x^{2} + 15x – 12x – 90 = 0$

$\Rightarrow x (2x+15) – 6(2x+15) = 0$

$\Rightarrow (2x+15) (x-6) = 0$

$\Rightarrow 2x + 15 = 0 (\text{or}) x – 6 = 0$

$\Rightarrow \boxed{x = \frac{-15}{2} ; x = 6}$

The side of the equilateral triangle can’t be negative.

So, $\boxed{x = 6}$

From the equation $(1),$

$3x + 8k = 90$

$\Rightarrow 3(6) + 8k = 90$

$\Rightarrow 8k = 90 – 18$

$\Rightarrow 8k = 72$

$\Rightarrow \boxed {k = 9}$

Thus, the longer side of the rectangle $= 3k = 3(9) = 27 \; \text{cm.}$

$\therefore$ The longer side of the rectangle is $27 \; \text{cm}.$

Correct Answer $: \text{B}$

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