+12 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
+1904 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 \(\frac{8}{3}\)
\(A=\sqrt{\frac{\frac{4}{3}+2+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{\frac{10}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{\frac{18}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{\frac{18}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{\frac{18}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{6}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{18}{6}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{18}{6}-\frac{8}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{10}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{5}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{3}{1}\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{15}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{6}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(3-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(1)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times1\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{6}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(3-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{9}{3}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{1}{3})}\)
\(A=\sqrt{5\times\frac{1}{3}}\)
\(A=\sqrt{\frac{5}{1}\times\frac{1}{3}}\)
\(A=\sqrt{\frac{5}{3}}\)
+1904 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 \(\frac{8}{3}\)
\(A=\sqrt{\frac{\frac{4}{3}+2+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{\frac{10}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{\frac{18}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{\frac{18}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{\frac{18}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{6}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{18}{6}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{18}{6}-\frac{8}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{10}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times(\frac{5}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{3\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{3}{1}\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{\frac{15}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{6}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(3-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(1)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times1\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{6}{2}-\frac{8}{3})}\)
\(A=\sqrt{5\times(3-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{9}{3}-\frac{8}{3})}\)
\(A=\sqrt{5\times(\frac{1}{3})}\)
\(A=\sqrt{5\times\frac{1}{3}}\)
\(A=\sqrt{\frac{5}{1}\times\frac{1}{3}}\)
\(A=\sqrt{\frac{5}{3}}\)
+2441 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. |
+128460
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 :
