Q 3

Geometry Level 2

ABCD is a rectangle. P is the mid-point of DC and Q is a point on AB such that AQ = 1 / 3 × A B 1/3 \times AB . What fraction of the area of ABCD is AQPD ?

Details and Assumptions

The answer is in the form p/q.

Enter your answer as p+q


The answer is 17.

Tushar Malik by Tushar Malik , 20, India

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1 solution

Erza Scarlet
Apr 22, 2015

Area of ABCD = AB x BC

let Q' be the foot of a straight line segment from Q on CD

AQPD = rectangle AQQ'C + triangle QQ'P

hence Q'P = A B 2 − A B 3 \frac{AB}{2} - \frac{AB}{3}

Q'P= A B 6 \frac{AB}{6}

Area of AQPD = A B 3 × B C + 1 2 × B C × A B 6 \frac{AB}{3} \times BC + \frac{1}{2} \times BC \times \frac{AB}{6}

A Q P D A B C D = ( A B × B C × 1 3 ) ( 1 + 1 4 ) A B × B C \frac{AQPD}{ABCD} = \frac{(AB\times BC\times\frac{1}{3})(1+\frac{1}{4})}{AB\times BC}

cancelling out

5 12 \frac{5}{12}

hence answer is 17 \boxed{17}

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