Question 1166927: In the triangle to the right, segments AB and AC are trisected, and D is the midpoint of BC. If the area of triangle ABC is 630 cm*2, then the area of the section marked x, in cm*2 is:
A) 105
B) 175
C) 150
D) 185
E) 170
https://imageshack.com/i/pnLdNNl3j
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The two horizontal segments divide triangle ABC into a small triangle, a small trapezoid, and a larger trapezoid. The heights of all three figures are the same.
Segment BC and the two horizontal segments can also be viewed as forming three similar triangles all with vertex A; the ratio of the heights of those three triangles is 1:2:3.
That means the height of the smallest triangle is 1/3 the height of triangle ABC, so the area of the smallest triangle is 1/9 of the area of triangle ABC.
Similarly, the height of the middle sized triangle is 2/3 the height of triangle ABC, so the area of the middle sized triangle is 4/9 of the area of triangle ABC.
That means the area of the larger trapezoid is 5/9 of the area of triangle ABC.
Finally, since D is the midpoint of BC, segment AD bisects that larger trapezoid, so the area of the figure marked x is 5/18 of the area of triangle ABC.
ANSWER: (5/18)*630 = 5*35 = 175 square cm
|
|
|
