In triangle ABC, angle B is right. E is on AC and F is on BC such that EF is parallel to AB. If AB = 8, BC = 6, and AE = x, then find the perimeter of triangle EFC in terms of x.
+2441 It is vitally important to use a diagram, I would say. Luckily, I have taken the time to create one for you so that you can follow along. I have put all the information we know into the diagram below:
In order to find the perimeter, we need to express the length of each side of \(\triangle EFC\) in terms of x. \(\angle B\) is a right triangle, so by Pythagorean's theorem, \(AC=10\) and \(EC = 10-x\).
\(\triangle ACB \cong \triangle ECF \) by Angle-Angle Similarity Postulate, and we will use this similarity in order for us to express the remaining side lengths in terms of x like so:
\(\frac{10-x}{10}=\frac{EF}{8}\quad\quad\quad\frac{10-x}{x}=\frac{CF}{6}\\ EF = 8-\frac{4}{5}x\quad\quad CF=6-\frac{3}{5}x\)
Now, find the perimeter by taking the sum of all the sidelngths of the desired triangle.
\(P=EC+EF+CF\\ P=(10-x)+(8-\frac{4}{5}x)+(6-\frac{3}{5}x)\\ P=24-\frac{12}{5}x\)
+2441 It is vitally important to use a diagram, I would say. Luckily, I have taken the time to create one for you so that you can follow along. I have put all the information we know into the diagram below:
In order to find the perimeter, we need to express the length of each side of \(\triangle EFC\) in terms of x. \(\angle B\) is a right triangle, so by Pythagorean's theorem, \(AC=10\) and \(EC = 10-x\).
\(\triangle ACB \cong \triangle ECF \) by Angle-Angle Similarity Postulate, and we will use this similarity in order for us to express the remaining side lengths in terms of x like so:
\(\frac{10-x}{10}=\frac{EF}{8}\quad\quad\quad\frac{10-x}{x}=\frac{CF}{6}\\ EF = 8-\frac{4}{5}x\quad\quad CF=6-\frac{3}{5}x\)
Now, find the perimeter by taking the sum of all the sidelngths of the desired triangle.
\(P=EC+EF+CF\\ P=(10-x)+(8-\frac{4}{5}x)+(6-\frac{3}{5}x)\\ P=24-\frac{12}{5}x\)
