The bases of trapezoid $ABCD$ are $\overline{AB}$ and $\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid.

Question

[asy]
unitsize(1 cm);

pair A, B, C, D;

A = (1,2);
B = (3,2);
C = (4,0);
D = (0,0);

draw(A--B--C--D--cycle);
draw(B--D);

label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, SW);
label("$7$", (A + D)/2, W);
label("$7$", (B + C)/2, E);
label("$9$", (B + D)/2, SE);
label("$8$", (C + D)/2, S);
[/asy]
I know I can use Heron's Formula, but I still need side AB.

Answer 1

huh? An easy isosceles trapezoid? Mark off P and Q on AB so you have a rectangle PQCD.

That means that AP=QB=(9-8)/2 = 1/2

Now the height h of the trapezoid is

h^2 = 7^2 - (1/2)^2

And the area is h(8+9)/2

Answer 2

You even copied the asymptote? How pathetic. And Steve, don't you realize that the answer is a number? You have to find the height. That's wrong anyway.

Answer 3

"You even copied the asymptote? How pathetic. And Steve, don't you realize that the answer is a number? You have to find the height. That's wrong anyway." -very funny (aug. 1 2016)