Discussion

Explanation:

​​​​​​​

Let d1 = x and d2 = 2x.

Now, SR would be shortest when SR = d1 = x

Area of PQRS = 30 square units.

∴ Area of PQRS = ½ × QS × 2x + ½ × QS × x

= (3/2) × QS × x = 30 

⇒ QS × x = 20

Smallest value of x = 1 units.

⇒ QS = 20

∴ ∆QSR is right angled at S

⇒ QR2 = QS2 + SR2

⇒ QR2 = 400 + 1

⇒ QR = slightly more than 20.

Hence, option (e).

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