+136 In triangle $ABC$, $AB = 10$ and $AC = 15$. Let $D$ be the foot of the perpendicular from $A$ to $BC$. If $BD:CD = 1:3$, then find $AD$.
+975 First off, let's let \(DC = 3BD\)
We can write two equations off the problem
\(AD = \sqrt{AC^2 - (3BD)^2 } = \sqrt{15^2 - 9BD^2 } \\ AD = \sqrt{AB^2 - BD^2} = \sqrt{10^2 - BD^2 }\)
Notice that we have AD on both equations, so we can equate the two to get
\( \sqrt{15^2 - 9BD^2} = \sqrt {10^2 -BD^2 }\)
Now, we just have to solve for BD^2.
\(15^2 - 9BD^2 = 10^2 - BD^2 \\ 225 - 9BD^2 = 100 - BD^2 \\ 125 = 8BD^2 \\ 125/8 = BD^2\)
We know that
\(AD =\sqrt{10^2 - BD^2}\) so we can just plug in BD^2 to find AD.
\(AD = \sqrt{100 - 125/8} = \sqrt { 675/8} = \frac{15\sqrt3 }{2 \sqrt2}\)
This rounds to about 9.19.
Thanks! :)
+975 First off, let's let \(DC = 3BD\)
We can write two equations off the problem
\(AD = \sqrt{AC^2 - (3BD)^2 } = \sqrt{15^2 - 9BD^2 } \\ AD = \sqrt{AB^2 - BD^2} = \sqrt{10^2 - BD^2 }\)
Notice that we have AD on both equations, so we can equate the two to get
\( \sqrt{15^2 - 9BD^2} = \sqrt {10^2 -BD^2 }\)
Now, we just have to solve for BD^2.
\(15^2 - 9BD^2 = 10^2 - BD^2 \\ 225 - 9BD^2 = 100 - BD^2 \\ 125 = 8BD^2 \\ 125/8 = BD^2\)
We know that
\(AD =\sqrt{10^2 - BD^2}\) so we can just plug in BD^2 to find AD.
\(AD = \sqrt{100 - 125/8} = \sqrt { 675/8} = \frac{15\sqrt3 }{2 \sqrt2}\)
This rounds to about 9.19.
Thanks! :)
