An isosceles trapezoid $ABCD$ has bases $AD = 17$cm, $BC = 5$cm, and leg $AB = 10 $cm. A line is drawn through vertex $B$ so that it bisects diagonal $AC$ and intersect $AD$ at point $M$.
1) Find the area of $ΔBDM$.
2) What is the area of $ABCD$?
What I did:
So I didn't know what to do for the first question so I skipped it. Any help would be appreciated.
The second one was easy because the height makes a right triangle and we can see that the right triangle has a $6-8-10 (3-4-5)$ Pythagorean triple. So from there, the height is 8. I can plug all of this information into the formula for the area of a trapezoid: $0.5*(b1+b2)*h$. I finally get the area of the trapezoid as 88 cm^2.
So I don't know how to do the first question (the area of the triangle) so any help would be appreciated.
Sorry for the crude drawings.