Trapezoid problem

The bases of trapezoid ABCD are line AB and line CD. We are given that AB < CD, CD = 8, AD = BC = 7, and BD = 9. Find the area of the trapezoid.

Thaks Omi......here's another approach....

By the Law of Cosines, we have

9^2  = 7^2 + 8^2  -  (2 * 8 * 7)cos(DCB)

-32   / -112  =  cos(DCB)

2/7  = cos(DCB)      (1)

And since angle DAB is supplemental to DCB, cos DAB  = -cos(DCB)

So....by the Law of Cosines again, we have

9^2 =  7^2  + AB^2  - (2 * 7 * AB) (-cosDCB)

[32- AB^2] = (14AB) cos(DCB)

[32 - AB^2] / [14AB ]   = cos(DCB)   (2)

Set  (1)  = (2)

2/7  =  [32 - AB^2] / [ 14AB ]

2 = [32 - AB^2] / [ 2AB]

4AB = 32 - AB^2

AB^2 + 4AB - 32  =  0      factor

(AB + 8) (AB - 4)  = 0

Setting both factors to 0 and solving for AB produces  AB  = -8  or AB = 4

Reject the first answer, accept the second

Dropping a perpendicular line from B to intersect DC at E.....then, by symmetry,  EC  =  2 

The height of the trapezoid  is given by √[BC^2 - EC^2 ] =  √[7^2 - 2^2]  = √45 = 3√5

So....the area is

height / 2   *  (sum of the bases )  =

[ 3√5  ( 8 + 4) ]  / 2  =    18√5  units^2  ≈  40.25 units^2

Thanks Omi and CPhill, you guys are really helpful!

Here's yet another way.

Using the formula for the area of a triangle \(\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\text{ , }\)

where s is the semi perimeter, the area of the larger triangle BCD is \(\displaystyle \sqrt{12.3.4.5}=12\sqrt{5}.\)

That means that the height of this triangle h, will be given by  \(\displaystyle 12\sqrt{5}=8h/2,\text{ so }h=3\sqrt{5}\)

Now, borrowing the notation from Omi67's diagram below, \(\displaystyle x^{2}+(3\sqrt{5})^{2}=7^{2}, \text{ so } x = 2.\)

Given the symmetrical nature of the figure it will follow that AB = 4, so the area of the whole figure wiil be

\(\displaystyle (4 + 8).3\sqrt{5}/2=18\sqrt{5}.\)

Tiggsy