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

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.

• You draw AC as it is angle bisector. Is it?
– Aqua
Apr 14 '18 at 18:33
• Sorry I assumed that when I wrote the question. I will remove that. Apr 14 '18 at 18:37
• If you re-drew your drawings on gridded paper, whenever you determine each new detail, you would have ended up with this image, and found the problem easy to solve. Apr 14 '18 at 19:03
• Yes, that drawing on the grid paper is much accurate. My trapezoid is not accurately drawn. Apr 15 '18 at 16:31

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

• Ok, even though there are two congruent right triangles, how would this help because we are looking for the the area of ΔBDM not the area of ΔABM? Apr 14 '18 at 18:33
• ΔBDM only has a small part of ΔBPC in it. Apr 14 '18 at 18:34
• Further hints: Can you find $AM$? Can you find $MD$? How do you find the area of $\Delta BMD$? Apr 14 '18 at 18:37
• Ah, ok I can see the length of MD is 17-5=12 because the two right triangles are congruent. Because we already know that the height is 8, we can find the area of ΔBMD. Apr 14 '18 at 18:49
• ΔBMD is 0.5*(height*base) = 0.5*(8*12) = 0.5*(96) = 48 Apr 14 '18 at 18:50

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 $$Area (BMD) ={8\cdot 12\over 2}=48$$

• That another great way to find the answer to the question too. Apr 15 '18 at 1:48