What is the area of this trapezoid?

Question

http://img.photobucket.com/albums/v371/Raushan/trapeziod.jpg

I've tried all I can and applied everything I know! (I'm in 7th grade)

How do you solve this kind of problem without the height?

Answer 1

If you look at the left end of the figure, and drop a line from point D perpendicular to line AB, you will have a small right triangle with hypotenuse 5. This would be a 3-4-5 right triangle (have you studied those in 7th grade yet?).

Start with that fact. You can drop another perpendicular line from point C to line AB to check on the triangle on the right. You can't assume the height is 4 over there - use the Pythagorean theorem to figure out the height there based on the length you determine for the base of that triangle given that you know AB = 52 and CD is 39 and the base of the left triangle must be 3.

Check back if this doesn't start you in the right direction.

Answer 2

Drop two perpendiculars DP and CQ from points D & C respectively. So the area of the trapezoid is the area of the triangles ADP and CQB plus the rectangle DPQC.
Now in triangle DPA,
ADsq = DPsq + APsq or 5sq = DPsq + APsq
or, DPsq + APsq = 25 [1]
In triangle BCQ,
BCsq = CQsq + QBsq
= DPsq + QBsq (CQ = DP)
or DPsq + QBsq = 144 [2]
Substituting [1] from [2],
QBsq-APsq = 144-25 = 119
or, (QB + AP)(QB - AP) = 119
But from the drawing, QB + AP = 52 - 39 = 13
Therefore, 13 X (QB - AP) = 119
QB - AP = 119/13 [3]
But QB + AP = 13
So 2QB = 119/13 + 13 = (119 + 169)/13 = 288/13
QB = 288/26 = 11.08
Hence AP = 13 - 11.08 = 1.92
Now determine the value of DP, using [1] or [2]
You can calculate the areas of the triangles (1/2 base X height) and the area of the rectangle, sum them and get the area of the trapezoid.

Answer 3

it is the square root of 9 , so the answer is 3