Question

Find the area of a triangle whose sides have the measures: 6in, 9in, and 11 in.

Answer 1

Heron's formula s=(1/2)(a+b+c)

Answer 2

Im not following you. So the semipermimeter is 13 so what do you do from there?

Answer 3

herons formula can be tedious; but so can the law of cosines...

Answer 4

11^2 -6^2 -9^2
--------------- = cos(t)
-2(6)(9)

9.6.sin(t)
--------- = area
2

Answer 5

18sqrt(110)
---------- = abt. 26.97
7

Answer 6

Im lost. Law of cosines is c^2 = a^2 + b^2 -2ab cos y right? And I know it can be manipulated but Im still lost

Answer 7

that is the law of cosines yes; and i used it to find the angle between the legs 6 and 9 so that i could use the sin of that angle to find the area

Answer 8

6sin(t) = h

9 = base

hb/2 = area

Answer 9

This is confusing...
Im trying to work it

Answer 10

Can you make a picture? I think it would help alot.

Answer 11

what does 6sin(t) mean?

Answer 12

it means 6 times the sin(angle)

sin(angle) = h/6 ; so h = 6 sin(angle)

Answer 13

Oka I got that. So how are you supposed to find t then? Can you work out the theorom because I think that's what I'm having trouble with

Answer 14

i used the law of cosine and modified it to find t

c^2 = a^2 + b^2 -2ab cos(C)

c^2 -a^2-b^2
------------ = cos(C)
-2ab

the cosine inverse of that left side = t

Answer 15

but why isn't it c^2 -a^2-b^2+2
---------------- = cos(C)

Answer 16

ab

Answer 17

becasue "-2ab cosC" is not addition; to solve for cosC you have to divide both sides by "-2ab"