Answer :

kate200468
[tex]A_1=6\cdot5=30\ [in^2]\\\\A_2= \frac{1}{2} \cdot[15+(7+5)]\cdot4=2\cdot(15+12)=2\cdot27=54\ [in^2]\\\\A=A_1+A_2=30+54=84\ [in^2][/tex]
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AL2006
-- The little block on top is 6-high by 5-wide = 30 square inches.

-- The dotted line across the middle is 5 inches long.

-- The long skinny block on the bottom is (10-6)=4-high by 12-wide = 48 inches² .

-- The height of the triangle at the end is (10-6) = 4 inches.
The base of the triangle is (15 - 7 - 5) = 3 inches.
The area of the triangle is (1/2)(base)(height) = 6 inches².

-- Sum of the the 3 sections = (20 + 48 + 6) = 74 square inches.

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