Finding the area of a triangle in a trapezoid and the area of the trapezoid based on the given information.

Question

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 $ABCD$?

Image 1: (For the area of ΔBDM) Image 2 : (For the area of the whole figure)

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.

Answer 1

Hint: let $AC$ and $BM$ intersect at $P$, then $\Delta BCP$ is equal to $\Delta AMP$.

Answer 2

Put whole thing in coordinate system so that $A(-{17\over 2},0)$, $D({17\over 2},0)$, $B(-{5\over 2},8)$ and $C({5\over 2},8)$. Then midpoint of $AC$ is $R(-3,4)$ and a line $BR$ has equation $$ y= 8x+28$$

This line cuts x-axis at $M(-{7\over 2},0)$, so $DM= 12$ and $$Area (BMD) ={8\cdot 12\over 2}=48$$