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In triangle ABC, AB = 17, AC = 8, and BC = 16. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.

 Nov 9, 2021
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 Find the area of triangle ACD.

 

Hello Guest!

 

\(\overline{AD} =x\)

\(h^2=8^2-x^2=16^2-(17-x)^2\\ 8^2-x^2=16^2-(17^2-34x+x^2)\\ 8^2-x^2=16^2-17^2+34x-x^2=0\\ 8^2-16^2+17^2=34x\\ x=\dfrac{8^2-16^2+17^2}{34}\)

\(x=\overline{AD} =2.853\)

\(h=\sqrt{8^2-x^2}=\sqrt{64-2.853^2}\)

\(h=7.474\)

\(A_{ACD}=\dfrac{h\cdot x}{2}=\dfrac{7.474\cdot 2.853}{2}\)

\(A_{ACD}=10.661\)

laugh  !

 Nov 9, 2021

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