Question 61

# 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?

Solution

There are 3 parallel lines AB, CD, and EF, in that order.Â
Let the distance between CD and AB be 2x.
It has been given that the distance between CD and EF is x.

A quadrilateral PQRS is formed such that P is onÂ AB, Q and S are onÂ CD, and R is on EF.Â
Also, the length of SR is the least possible value it can take. Therefore, SR must be perpendicular to the parallel lines.Â

Area of quadrilateral PQRS = Area of triangle PQS + Area of triangle SRQ = 30 square cm.Â

Area of triangle PQS = 2*area of triangle SRQ (Since they rest on the same base and height of SRQ is half the height of PQS)
=> Area of triangle SRQ = 10 square cm.Â
Let the length of SQ be $$b$$. We know that SR= $$x$$
$$0.5*x*s = 10$$
=> $$xs=20$$
$$s=20/x$$
We do not have any other detail to evaluate the value of the expression.Â But, we have been given that d1 and d2 are integers. Therefore, the least value that 'x' can take is 1.Â
The least value that S can take is 1.Â
By Pythagoras theorem, $$QR =\sqrt{s^2 + x^2}$$
$$QR=\sqrt{20^2+1}$$
$$QR=\sqrt{401}$$
Therefore, the value of QR will be slightly greater than 20. Therefore, optionÂ E is the right answer.Â

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