Halp!

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Halp!

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hi, Im new to this site, and I need help with this problem. Please respond ASAP!

Find the area of a triangle wih sides of 4/3, 2, and 8/3.

Put in sqrt pls! Dont round, just exact form using sqrtt

Please help with this problem too!

In triangle RST,RS=13 ST=14, and RT=15

Let M be the midpoint of ST. Find RM

MeepMeep Jun 30, 2017

edited by MeepMeep Jun 30, 2017

Best Answer

3⁺² Answers

#1

+1904

+1

Best Answer

I do not know how to solve the second problem; however, I can help solve the first problem. To solve the first problem, use Heron's Formula.

$A=\sqrt{\frac{a+b+c}{2}\times(\frac{a+b+c}{2}-a)\times(\frac{a+b+c}{2}-b)\times(\frac{a+b+c}{2}-c)}$

A = Area

a = side a or $\frac{4}{3}$

b = side b or $2$

c = side c or $\frac{8}{3}$

$A=\sqrt{\frac{\frac{4}{3}+2+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{\frac{\frac{10}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{\frac{\frac{18}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times(\frac{\frac{18}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times(\frac{6}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times(\frac{18}{6}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times(\frac{18}{6}-\frac{8}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times(\frac{10}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times(\frac{5}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{3\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{\frac{3}{1}\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{\frac{15}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{6}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(3-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(1)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times1\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{6}{2}-\frac{8}{3})}$

$A=\sqrt{5\times(3-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{9}{3}-\frac{8}{3})}$

$A=\sqrt{5\times(\frac{1}{3})}$

$A=\sqrt{5\times\frac{1}{3}}$

$A=\sqrt{\frac{5}{1}\times\frac{1}{3}}$

$A=\sqrt{\frac{5}{3}}$

gibsonj338 Jul 1, 2017

#3

+2446

0

Technically, $\sqrt{\frac{5}{3}}$ isn't completely simplified, so I'll simplify it; you did a lot of work for that problem...

$\sqrt{\frac{5}{3}}$	"Distribute" the square root to the numerator and denominator. It follows the rule that $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
$\frac{\sqrt{5}}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}$	Because there is a radical in the denominator, you must rationalize it by multiplying the denominator by itself. Doing this gets rid of the radical in the denominator.
$\frac{\sqrt{5}\sqrt{3}}{\sqrt{3}\sqrt{3}}$	In the denominator, multiplying by itself is the same as squaring it, which undoes the square root.
$\frac{\sqrt{5}*\sqrt{3}}{3}$	Multiplying radicals is quite simple, actually. Just multiply the radicands (the number inside the radical) together. In general, $\sqrt{a}\sqrt{b}=\sqrt{ab}$
$\frac{\sqrt{15}}{3}$	The square root of 15 has no perfect-square factors and is therefore irreducible. No more simplification is possible.

TheXSquaredFactor Jul 2, 2017

2 Online Users

gibsonj338 · Answer 1 · 2017-07-01T07:02:59