Question

Find the leg of each isosceles right triangle when the hypotenuse is of the given measure.

Given = 5sqrt6

Answer 1

hmmm is it a right triangle?

Answer 2

\[2x ^{2}=(5\sqrt{6})2\]

Answer 3

solve for x

Answer 4

hmm

Answer 5

* \[(5\sqrt{6})^{2}\]

Answer 6

I got that from Pythagorean theorem. a^2 + b^2 = c^2

Answer 7

since a and b have to be equal, i just made it 2x^2=c^2

Answer 8

where c is 5sqrt6

Answer 9

Ok, so solving for x?

Answer 10

When i solved for x i got 125?

Answer 11

did you take the square root?

Answer 12

I got 5sqrt3 or 8.66

Answer 13

https://www.wolframalpha.com/input/?i=+2x^2%3D%285sqrt%286%29%29^2

Answer 14

so... hmmm is it a right triangle?

Answer 15

http://www.mathsisfun.com/definitions/hypotenuse.html <--- notice that

so... what do you think, is it a right triangle or maybe not?

Answer 16

ohhh wait a sec... seems to have been added now... ohhh alrite

Answer 17

and "isosceles" triangle
means

two sides are equal

a hypotenuse is the "longest" side, so it cannot by definition have another equal

so the other two sides should be equal then

Answer 18

Ok wow
i feel dumb
i was using the wrong formula
the entire time

Answer 19

that's why my answer was far off

Answer 20

Did you get it though?

Answer 21

Yea i got it finally.

Answer 22

Another question though, Find the diagonal of a square whose sides are of the given measure. Given = 3^2?

Answer 23

i cant seem to plug this one in right

Answer 24

given = 3sqrt2*

Answer 25

so

you could use the pythagorean theorem for that

since \(\bf c^2=a^2+b^2\implies \qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite
\end{cases}
\\ \quad \\
c^2=b^2+b^2\qquad
\begin{cases}
\textit{recall, sides are equal, thus}\\
a=b
\end{cases}
\\ \quad \\
c^2=2b^2\implies \cfrac{c^2}{2}=b^2\implies \sqrt{\cfrac{c^2}{2}}=b=a
\\ \quad \\
\sqrt{\cfrac{{\color{brown}{ (5\sqrt{6}}})^2}{2}}=b=a\)