AB, CD and EF are three parallel lines, in that order. Let d1 and d2 be the distances from CD to AB and EF respectively. d1 and d2 are integers, where d1 : d2 = 2 : 1. P is a point on AB, Q and S are points on CD and R is a point on EF. If the area of the quadrilateral PQRS is 30 square units, what is the value of QR when value of SR is the least?
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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