Question

ABCD is a rectangle. P, Q and R are the midpoint of BC, CD and DA. The point S lies on the line QR in such a way that SR: QS = 1:3. The ratio of the area of triangle APS to area of rectangle ABCD is

Accepted Answer

A. Let $AB = 4, BC = 2$. $R = (0, 1)$, $Q = (2, 0)$, $P = (4, 1)$, $A = (0, 2)$, $D = (0, 0)$. $S$ divides $RQ$ in ratio $RS:SQ = 1:3$. Coordinates of $S = \left(\frac{3(0) + 1(2)}{4}, \frac{3(1) + 1(0)}{4}ight) = (0.5, 0.75)$. Area of $	riangle APS$ via coordinates: $A = (0, 2), P = (4, 1), S = (0.5, 0.75)$. Area $= \frac{1}{2} | 0(1 - 0.75) + 4(0.75 - 2) + 0.5(2 - 1) | = \frac{1}{2} | -5 + 0.5 | = \frac{4.5}{2} = 2.25 = \frac{9}{4}$. Area of ABCD $= 4 	imes 2 = 8$. Ratio $= \frac{2.25}{8} = \frac{9}{32} = \frac{36}{128}$.

Area Ratio in Rectangle

Explanation

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