Answer:
(81√15)/4 ≈ 78.43 cm²
Step-by-step explanation:
It can help to draw a diagram.
The base angle at the side of length 6 can be found using the Law of Cosines. Its opposite side is of length 8, and the two adjacent sides are of length 6 and 7. So, we have ...
8² = 6² +7² -2·6·7·cos(α)
cos(α) = (36+49-64)/(84) = 1/4
α = arccos(1/4) ≈ 75.522°
Then the height of the trapezium is ...
6·sin(75.522°) ≈ 5.80948 . . . . cm
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The area of the trapezium is given by the formula ...
A = 1/2(b1 +b2)h
A = (1/2)(17 +10)(5.80948) . . . . cm²
A ≈ 78.43 cm²
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Of course, sin(arccos(1/4)) = (√15)/4, so the exact formulation of the area is ...
A = (81√15)/4 cm²
