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Geometry

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135
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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

#1
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Find the area of triangle ACD.

Hello Guest!

$$\overline{AD} =x$$

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

$$x=\overline{AD} =2.853$$

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

$$h=7.474$$

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

$$A_{ACD}=10.661$$

!

Nov 9, 2021