Started "doing" Pythagorean theorem problems with isosceles triangles today ...

Question

Yesterday, I solved my very first Pythagorean theorem problem! Everything was going good so far, I was solving harder problems very easily. However, today's lesson is a little bit different. I am working with isosceles triangles, and I have the following:

The two equal sides of the isosceles triangle are 25 cm long. The base, on the other hand, is 40 cm. I have to find the height of the triangle (CH).

So, I did the following:

(CH)2 + (AC)2 = (AB)2

(CH)2 + 625 = 1600

(CH)2 = 975

CH = 5 √ 39

Am I correct? Any kind of help is appreciated!

Answer 1

As you noticed in your comment, "I should be using half of AC, instead of the whole side". Indeed. Then you have a right triangle with one leg $20$, and the hypotenuse, say $AB$ of length $25$.

So we get the equation, to determine height, $$20^2 + \text{height}^2 = 25^2$$

Now you just need to solve for height: $$\text{height}^2 = 25^2 - 20^2.$$

Answer 2

What you have done wrong is that you messed up with your notations! Let's keep them organized!

Denote your isosceles triangle as $\triangle ABC$ where

• $AB=40$ is the base,
• $CB=CA=25$ are the two sides of equal length,
• and $CH$ is the height that you want to find.

Notice that the point $H$ is the midpoint of the base $AB$. Now let's apply the Pythagorean theorem to the right triangle $\triangle CHA$: $$ CH^2+AH^2=AC^2\;. $$ Now substitution: $AH=\frac12 AB = 20$, $AC=25$. So $$ CH^2 + 20^2 = 25^2 $$ which is $CH^2 +400=625$. Hence $$ CH = \sqrt{225} = 15. $$ Don't you get excited that the length is an integer?! :)