In triangle ABC, AB = 17, AC = 8, and BC = 20. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.
+14995 Find the area of triangle ACD.
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\(f(x)=\sqrt{8^2-x^2}\\ g(x)=\sqrt{20^2-(x-17)^2}\\ \sqrt{8^2-x^2}=\sqrt{20^2-(x-17)^2}\\64-x^2=400-x^2+34x-289\\ 34x=-47\\ x=-\dfrac{47}{34}\\ y=\sqrt{8^2-(-\frac{47}{34})^2}\)
\(A_{ACD}=\dfrac{xy}{2}\\ A_{ACD}=\dfrac{1.382\cdot 7.8985}{2}\\ \color{blue}A_{ACD}=5.469 \)
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+14995 Find the area of triangle ACD.
Hello Guest!
\(f(x)=\sqrt{8^2-x^2}\\ g(x)=\sqrt{20^2-(x-17)^2}\\ \sqrt{8^2-x^2}=\sqrt{20^2-(x-17)^2}\\64-x^2=400-x^2+34x-289\\ 34x=-47\\ x=-\dfrac{47}{34}\\ y=\sqrt{8^2-(-\frac{47}{34})^2}\)
\(A_{ACD}=\dfrac{xy}{2}\\ A_{ACD}=\dfrac{1.382\cdot 7.8985}{2}\\ \color{blue}A_{ACD}=5.469 \)
!
+129852 This triangle is obtuse.. altitude CD falls outside triangle ABC as shown
Let x be the distance from A to D
We have that
DC^2 = BC^2 + (AB + x)^2 = 20^2 - (17+x)^2
DC^2 = (AC)^2 - x^2 = 8^2 - x^2
So
20^2 - (17+x)^2 = 8^2 - x^2
400 - x^2 - 34x -289 = 64 - x^2
-34x = 289 + 64 - 400
-34x = -47
x = 47/34
DC = sqrt [8^2 - (47/34)^2 ]
Area of ACD = (1/2) (AD) (CD) = (1/2)(47/34)(sqrt [ 8^2 - (47/34)^2 ] = 5.446
