SSC GD 2024 PYQ — Question 42

The correct answer is option 4 i.e. 25 metre.

The perimeter ( P ) of a rectangle is given by:

⇒ P = 2l + 2w

⇒ 2l + 2w = 70

⇒ l + w = 35  (Equation 1)

The area ( A ) of a rectangle :

⇒ A = l × w

⇒ l × w = 300   (Equation 2)

 

From equation  1:

⇒ l = 35 - w

 into Equation 2:

⇒ (35 - w) × w = 300

⇒ 35w - w² = 300

⇒ w² - 35w + 300 = 0

⇒ $w = \frac{-(-35) \pm \sqrt{(-35)^2 - 4(1)(300)}}{2(1)}$

⇒ $w = \frac{35 \pm \sqrt{1225 - 1200}}{2}$

⇒ $w = \frac{35 \pm \sqrt{25}}{2}$

⇒ $w = \frac{35 \pm 5}{2}$

⇒ $w = \frac{35 + 5}{2} = 20 \quad \text{or} \quad w = \frac{35 - 5}{2} = 15$

 

If ( w = 20 ), then from Equation 1:

⇒ l = 35 - 20 = 15

If ( w = 15 ), then from Equation 1:

⇒ l = 35 - 15 = 20

 

The length of the diagonal ( d) of a rectangle can be found using the Pythagorean theorem:

⇒ $d = \sqrt{l^2 + w^2}$

⇒ $d = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25$