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P be any point on the parallelogram ABCD and the area of the triangle ΔAPD and triangle ΔBCP are 32 cm2 and 37 cm2 respectively. If the length of the side AB is 9 cm and the perpendicular distance from P to AB is 6 cm then what is the area of the triangle ΔPCD?

P be any point on the parallelogram ABCD and the area of the triangle ΔAPD and triangle ΔBCP are 32 cm2 and 37 cm2 respectively. If the length of the side AB is 9 cm and the perpendicular distance from P to AB is 6 cm then what is the area of the triangle ΔPCD? Correct Answer 42 cm<sup>2</sup>

Given:

P be any point on the parallelogram ABCD and the area of the triangle ΔAPD and triangle ΔBCP are 32 cm2 and 37 cm2 respectively. The length of the side AB is 9 cm and the perpendicular distance from P to AB is 6 cm

Concept Used:

If P be any point on the parallelogram then (Sum of the area of the pair of opposite triangles) = (Sum of the area of the other pair of the opposite triangle) = ½ × Area of the parallelogram

Calculation:

The length of the side AB is 9 cm and the perpendicular distance from P to AB is 6 cm

Area of the triangle ΔAPB = ½ × 9 × 6

⇒ 27 cm2

The area of the triangle ΔAPD and triangle ΔBCP are 32 cm2 and 37 cm2 respectively.

Area of ΔAPD + Area of ΔBPC = Area of ΔAPB + Area of ΔPCD

⇒ 32 cm2 + 37 cm2 = 27 cm2 + Area of ΔPCD

⇒ Area of ΔPCD = (32 + 37 – 27) cm2

⇒ Area of ΔPCD = 42 cm2

∴ The area of the triangle ΔPCD is 42 cm2.

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