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) enter image description here Image 2 : (For the area of the whole figure) enter image description here

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.

  • $\begingroup$ You draw AC as it is angle bisector. Is it? $\endgroup$
    – Aqua
    Apr 14 '18 at 18:33
  • $\begingroup$ Sorry I assumed that when I wrote the question. I will remove that. $\endgroup$ Apr 14 '18 at 18:37
  • $\begingroup$ 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. $\endgroup$ Apr 14 '18 at 19:03
  • $\begingroup$ Yes, that drawing on the grid paper is much accurate. My trapezoid is not accurately drawn. $\endgroup$ Apr 15 '18 at 16:31

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

  • $\begingroup$ 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? $\endgroup$ Apr 14 '18 at 18:33
  • $\begingroup$ ΔBDM only has a small part of ΔBPC in it. $\endgroup$ Apr 14 '18 at 18:34
  • $\begingroup$ Further hints: Can you find $AM$? Can you find $MD$? How do you find the area of $\Delta BMD$? $\endgroup$
    – farruhota
    Apr 14 '18 at 18:37
  • $\begingroup$ 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. $\endgroup$ Apr 14 '18 at 18:49
  • $\begingroup$ ΔBMD is 0.5*(height*base) = 0.5*(8*12) = 0.5*(96) = 48 $\endgroup$ 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 $$Area (BMD) ={8\cdot 12\over 2}=48$$

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

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