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.
$\endgroup$ 42 Answers
$\begingroup$Hint: let $AC$ and $BM$ intersect at $P$, then $\Delta BCP$ is equal to $\Delta AMP$.
$\endgroup$ 8 $\begingroup$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 $$Area (BMD) ={8\cdot 12\over 2}=48$$
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