Question

ABCD is a trapezium in which AB is parallel to CD, CD=30 cm and AB = 50 cm. if x and y are the respectively the mid points of AD and BC prove that ar (DCXY) =7/9(XYBA)

Answer 1

Conditions

ABCD is a trapezium in which AB is parallel to CD, CD=30 cm and AB = 50 cm. if x and y are the respectively the mid points of AD and BC prove that ar (DCXY) =7/9(XYBA)

Solution

Consider a graph:

As X and Y are respectively the mid points of AD and BC, so XY is the midline, and:

$$X Y = \frac {A B + C D}{2} = \frac {3 0 + 5 0}{2} = 4 0$$

DCXY and XYBA are also trapeziums.

The area of DCXY is:

$$S _ {D C X Y} = \frac {D C + X Y}{2} h _ {D C X Y}$$

The area of XYBA is:

$$S _ {X Y B A} = \frac {A B + X Y}{2} h _ {X Y B A}$$

Note, that:

$$h _ {D C X Y} = h _ {X Y B A} = \frac {1}{2} h _ {A B C D}$$

The rate of areas of DCXY and XYBA is:

$$\frac {\frac {D C + X Y}{2}}{\frac {A B + X Y}{2}} = \frac {D C + X Y}{A B + X Y} = \frac {3 0 + 4 0}{5 0 + 4 0} = \frac {7 0}{9 0} = \frac {7}{9}$$

Q.E.D.