Pythagorean Theorem: Intergrated Math

[Cutiepie]

I am just starting to learn how to use the pythagorean theorem. I have encountered a problem that asks me to apply it to an isosceles triangle. I can see that each side forms a right triangle and both legs are equal. The problem: An isosceles triangle has a height of 12in and two sides of 18in. How long is the base?Iam not sure how to begin. I do know that I should double the answer for the measurement of the entire length. Please help me. Thanks in advance.

 

[pka]

The summit angle if an isosceles triangle is the angle formed by the congruent sides.
The median from the summit angle is perpendicular bisector of the base of a isosceles triangle.
Given isosceles ΔABC, AB=AC then if M is the midpoint of BC then ΔAMB is a right triangle. Apply the he Pythagorean theorem: AB<SUP>2</SUP>=BM<SUP>2</SUP>+MA<SUP>2</SUP>.

You are given AB & AM. You are to find 2BM.

 

[Cutiepie]

thanks for the reply but I am only in the 7th grade do not under stand half of what u wrote. Please send the simple version. Thanks. CP

 

[stapel]

Hint: The height line divided the isosceles triangle into two of what sort of triangle? (Review the definition of "altitude", if you're not sure.)

Eliz.

 

[Cutiepie]

a2+b2=c2 324+_=144, 324-144=180 square root of 180= 13.4 the double measurement = 26.8in? is this correct? problem is isosceles triangle has a height of 12in and two sides of 18in. How long is the base?

 

[Denis]

a2+b2=c2 324+_=144, 324-144=180 square root of 180= 13.4 the double measurement = 26.8in? is this correct? problem is isosceles triangle has a height of 12in and two sides of 18in. How long is the base?

Correct...but by luck! Next time, type this way:
a^2 + b^2 = c^2
12^2 + b^2 = 18^2
b^2 = 324 - 144
b^2 = 180
b = sqrt(180) = ~13.4

 

[soroban]

Hello, Cutiepie!

I'll give it a try . . .

An isosceles triangle has a height of 12in and two sides of 18in.
How long is the base?

            A
            *
           /:\
          / : \
      18 /  :  \ 18
        / 12:   \
       /    :    \
      * - - + - - *
      B     M     C

We have isosceles triangle \(\displaystyle ABC\) with \(\displaystyle AB\,=\,AC\,=\,18\)
\(\displaystyle \;\;\) and its height is \(\displaystyle AM\,=\,12\)

Using Pythagorus on right triangle \(\displaystyle AMC\) we have:
\(\displaystyle \;\;MC^2\,+\,12^2\:=\:18^2\;\;\Rightarrow\;\;MC^2\,+\,144\:=\:324\;\;\Rightarrow\;\;MC^2\:=\:180\)

Hence: \(\displaystyle MC\:=\:\sqrt{180}\:=\:6\sqrt{5}\)

Therefore: \(\displaystyle \;\)base \(\displaystyle BC\:=\:2\,\times\,MC\:=\:2(6\sqrt{5})\:=\:12\sqrt{5}\)

 

[Ti-Pro.doc.]

stapel is saying that the altitude splits the triangle into two right triangles

cutiepie u know that a^2+b^2=c^2
right now you know a and c.

\(\displaystyle sqrt{18^2-12^2}=B\) B=13.42 (which is rounded to 2 decimal places)

(when denis says "b = sqrt(180)", he means \(\displaystyle b\,=\,\sqrt{180}\))

right now "B" is only half of the entire base, B*2=26.84 (which is rounded to 2 decimal places)

hope this helps.