# Area of triangle within a triangle

• Jan 9th 2013, 12:25 PM
TimsBobby2
Area of triangle within a triangle
Given a triangle ABC. On the sides: Let R lie one third of the way from A to B. Let S lie one third of the way from B to C. Let T lie one third of the way from C to A. Now connect points RST to make a triangle RST within triangle ABC. Prove that the area of triangle RST is equal to one third the area of triangle ABC.

Any help please?? I've tried law of sines and such to no avail.
• Jan 9th 2013, 12:41 PM
MarkFL
Re: Area of triangle within a triangle
The area A of triangle ABC is:

$A=\frac{1}{6}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)$

The area $A_O$ within ABC but outside of RST is:

$A_O=\frac{1}{9}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)$

Hence the area within RST is:

$A-A_O=\frac{1}{18}\left(bc\sin(A)+ac\sin(B)+ab\sin(C ) \right)=\frac{1}{3}A$
• Jan 9th 2013, 01:03 PM
TimsBobby2
Re: Area of triangle within a triangle
Where did the 1/6 and 1/9 come from?
• Jan 9th 2013, 03:19 PM
MarkFL
Re: Area of triangle within a triangle
I used the formula for the area of a triangle:

$A=\frac{1}{2}ab\sin(C)$

Now, for triangle ABC, I included all three angles so that each term is 1/3 of the entire area, or equivalently the area of ABC can be found from:

$3A=\frac{1}{2}bc\sin(A)+\frac{1}{2}ac\sin(B)+\frac {1}{2}ab\sin(C)$ and so:

$A=\frac{1}{6}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)$

For the 3 triangles outside of RST, I used the same formula:

$A_O=\frac{1}{2}\cdot\frac{2}{3}b\cdot\frac{1}{3}c \sin(A)+\frac{1}{2}\cdot\frac{1}{3}a\cdot\frac{2}{ 3}c\sin(B)+\frac{1}{2}\cdot\frac{2}{3}a\cdot\frac{ 1}{3}b\sin(C)$

$A_O=\frac{1}{9}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)$