Area of trapezium ABCD

CAT 2021 Slot 2 · QA · Easy · Geometry

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The sides ABAB and CDCD of a trapezium ABCDABCD are parallel, with ABAB being the smaller side. PP is the midpoint of CDCD and ABPDABPD is a parallelogram. If the difference between the areas of the parallelogram ABPDABPD and the triangle BPCBPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCDABCD is

  1. A.

    20

  2. B.

    25

  3. C.

    40

  4. D.

    30

Answer

D

Explanation

Since ABPDABPD is a parallelogram, AB=DPAB = DP. Since PP is the midpoint of CDCD, DP=PC=ABDP = PC = AB.

Let the height of trapezium be hh. Area of parallelogram ABPD=DP×h=AB×hABPD = DP \times h = AB \times h. Area of triangle BPC=12×PC×h=12AB×hBPC = \frac{1}{2} \times PC \times h = \frac{1}{2} AB \times h.

Given difference = 10: AB×h−12AB×h=10AB \times h - \frac{1}{2} AB \times h = 10 12AB×h=10  ⟹  AB×h=20\frac{1}{2} AB \times h = 10 \implies AB \times h = 20

Area of trapezium $ABCD = \text{Area}(ABPD) + \text{Area}(\triangle BPC) = 20 + 10 = 30\text{ sq cm.}$$$$$

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