Question 1112160: Suppose that AD, BC, AC, and BD are line segments with line AD parallel to line BC. If AD=3, BC=1, and the distance from AD to BC is 5, then what is the sum of the areas of the two triangles formed?
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
The two triangles are similar, because BC and AD are parallel.
The ratio of similarity is BC:AD = 1:3.
So the ratio of the altitudes of the two triangles is 1:3.
The (perpendicular) distance between BC and AD is 5.
Therefore the altitude of the small triangle is 1/4 of 5, or 5/4.
Then the area of the small triangle is one-half base times height = (1/2)(1)(5/4) = 5/8.
Since the ratio of similarity between the two triangles is 1:3, the ratio of their areas is 1^2:3^2, or 1:9.
So the area of the large triangle is 9*(5/8) = 45/8.
And so the sum of the areas of the two triangles is 5/8+45/8 = 50/8 = 25/4.
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