Question

In a trapezium ABCDwith AB parallel to CD the diagonals intersect at P the area of triangle ABP is 72 square cm and of triangle CDP is 50 square centimetre find the area of trapezium

Answer 1

Answer

The area of trapezium = 122 square centimeter

Solution

In the figure attached with this answer, a trapezium ABCD with AB parallel to CD

The diagonals intersect at P

The area of triangle ABP is 72 square cm

Let h1 be the height of  triangle ABP and h2 be the height of triangle CDP

Area of ABP = 1/2x AB xh1 = 72

        1/2x b xh1 = 72

        b xh1 = 72 x 2 =144

and of triangle CDP is 50 square centimeter

Area of CDP = 1/2x CD xh2

        1/2x a xh1 = 50

        a xh1 = 50 x 2 =100

Tthe area of trapezium

1/2(ah + bh) = 1/2(100 + 144) = 122 square centimeter

Answer 2

Triangle ABP:

Let the height be H1

Area = 1/2 x base x height

72 = 1/2 x AB x H1

AB x H1 = 144 cm

Triangle CDP:

Let the height be H1

Area = 1/2 x base x height

50 = 1/2 x CD x H2

CD x H2 = 100 cm

Find the area of the trapezium:

Area of trapezium = 1/2 ( a + b ) x h

Height of the trapezium = H1 + H2

Area of trapezium = 1/2 ( AB x H1  + CD x H2)

Area of trapezium = 1/2 ( 144 + 100)

Area of trapezium = 1/2 (244) = 122 cm²

Answer:  122 cm²