renegade05
Full Member
- Joined
- Sep 10, 2010
- Messages
- 260
I think i got this one, i just wanted to confirm that i didnt screw something up.
The tirangle shown is isosceles with base x.
Find the function, P, that gives the perimeter of the triangle as a function of the base, x, if its area is 10m².
So...this is what i did.
variables:
\(\displaystyle x = base\)
\(\displaystyle h = height\)
\(\displaystyle a = \frac{1}{2} of base = \frac{x}{2}\)
\(\displaystyle c = hypotonus\)
work:
\(\displaystyle \frac{xh}{2} = 10m^2\)
\(\displaystyle h = \frac{20m^2}{x}\)
then:
\(\displaystyle a^2+h^2=c^2\)
\(\displaystyle (\frac{x}{2})^2+(\frac{20}{x})^2=c^2\)
\(\displaystyle \sqrt{(\frac{x}{2})^2+(\frac{20}{x})^2}=c\)
ANSWER:
\(\displaystyle P(x)=x+2\sqrt{(\frac{x}{2})^2+(\frac{20}{x})^2}\)
The tirangle shown is isosceles with base x.
Find the function, P, that gives the perimeter of the triangle as a function of the base, x, if its area is 10m².
So...this is what i did.
variables:
\(\displaystyle x = base\)
\(\displaystyle h = height\)
\(\displaystyle a = \frac{1}{2} of base = \frac{x}{2}\)
\(\displaystyle c = hypotonus\)
work:
\(\displaystyle \frac{xh}{2} = 10m^2\)
\(\displaystyle h = \frac{20m^2}{x}\)
then:
\(\displaystyle a^2+h^2=c^2\)
\(\displaystyle (\frac{x}{2})^2+(\frac{20}{x})^2=c^2\)
\(\displaystyle \sqrt{(\frac{x}{2})^2+(\frac{20}{x})^2}=c\)
ANSWER:
\(\displaystyle P(x)=x+2\sqrt{(\frac{x}{2})^2+(\frac{20}{x})^2}\)
