Question

Given: ABCD is a trapezoid, AB =13 CD = 14, BC = 5, and AD = 20 Find: AABCD

Answer 1

140

Explanation:

RSM

Answer 2

`\text{The area is }140units^2`

Explanation:

Given trapezoid ABCD, AB =13, CD = 14, BC = 5 and AD = 20 units.

we have to find the area of trapezoid ABCD

Quadrilateral BEFC is a rectangle, then

EF = BC = 5 units and BE = CF = y units

Let AE = x, then FD = AD - AE - EF = 20 - x - 5 = 15 - x.

By the Pythagorean theorem,

In ΔDFC

`CD^2=CF^2+FD^2`

`14^2=y^2+(15-x)^2` → (1)

In ΔAEB,

`AB^2=BE^2+AE^2`

`13^2=y^2+x^2` → (2)

Subtract equation 2 from 1

`14^2-13^2=(15-x)^2-x^2`

`196-169=225+x^2-30x-x^2`

`30x=225-196+169=198`

`x=6.6 units`

(2) ⇒
`13^2=y^2+x^2`

`169-43.56=y^2`

`y=11.2units`

Hence, the height is 11.2 units

`\text{Area of trapezium ABCD=}(1)/(2)* (BC+AD)* height`

`=(1)/(2)* (5+20)* 11.2=140 units^2`

`\text{Hence, the area is }140units^2`