Re: If ABCD is a square and the area of ∆AFG is 10, then what is
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20 Mar 2019, 10:02
We have to be able to identify proper bases and heights.
In triangle AFG, the base, FG, is 4, and the height is AB.
The area, given as 10, is equal to half of the base times the height, so:
\(\frac{4AB}{2}=10\)
Multiply both sides by 2:
\(4AB=20\)
Divide by 4:
\(AB=5\)
Now that is one side of the square, so all sides have a length of 5.
The triangle AEC has a base of 2 and a height of 5 (the height is AD, another side of the square). Its area, then, is:
\(\frac{2*5}{2} = 5\)