+0

geometry

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86
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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.

May 15, 2020

#1
+2422
+3

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.

$$P=EC+EF+CF\\ P=(10-x)+(8-\frac{4}{5}x)+(6-\frac{3}{5}x)\\ P=24-\frac{12}{5}x$$

May 15, 2020

#1
+2422
+3

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.

$$P=EC+EF+CF\\ P=(10-x)+(8-\frac{4}{5}x)+(6-\frac{3}{5}x)\\ P=24-\frac{12}{5}x$$

TheXSquaredFactor May 15, 2020
#2
+111456
0

Very nice, X^2    !!!!!!

May 15, 2020