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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 

 #1
avatar+1794 
+1

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 \(\frac{8}{3}\)

 

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

 

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

 

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

 

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

 

 

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

 

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

\(A=\sqrt{\frac{5}{3}}\)

gibsonj338  Jul 1, 2017
Sort: 

3+0 Answers

 #1
avatar+1794 
+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 \(\frac{8}{3}\)

 

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

 

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

 

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

 

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

 

 

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

 

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

\(A=\sqrt{\frac{5}{3}}\)

gibsonj338  Jul 1, 2017
 #3
avatar+1364 
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, \(\sqrt{a}*\sqrt{b}=\sqrt{a*b}\)
\(\frac{\sqrt{15}}{3}\) The square root of 15 has no perfect-square factors and is therefore irreducible. No more simplification is possible. 
   
TheXSquaredFactor  Jul 2, 2017
 #2
avatar+78620 
+1

 

                 

Here's # 2.......we need to manipulate the Law of Cosines twice.....

 

We can first  find  angle RST...so we have

 

[ RS^2  +ST^2 - RT^2] / [ 2(RS)(ST)] = cos RST

 

So

 

[13^2 + 14^2  - 15^2] / [ 2(13) (14) ]  =  cos RST

 

So  

 

arcos  ( [13^2 + 14^2  - 15^2] / [ 2(13) (14) ] )  =  RST  ≈  67.38°

 

Since M is the mid-point of ST, then SM  = 7

 

 

Now manipulating the Law of Cosines again to find RM, we have that

 

RM  =  sqrt ( SM^2  + RS^2  - 2(SM)(RS) cos  67.38° )  

 

RM =  sqrt ( 7^2  + 13^2 - 2 (7)(13) cos  67.38° ) ≈ 12.166

 

Here's an approximate pic :

 

 

 

 

 

 

 

 

 

cool cool cool

CPhill  Jul 1, 2017

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