Equilateral Triangle: find area, given distances to point P

[melai]

The problem: Located inside an equilateral triangle ABC is a point P such that PA = 6, PB = 8, and PC = 10. Find an exact value for the area of triangle ABC.

My thoughts:

Seeing as how PA, PB, and PC are of unequal lengths, P is not in the dead center of the triangle. Since P is inside triangle ABC, there would be three triangles embedded in it: triangle APB, triangle APC, and Triangle BPC.

I figure that I could solve for the area of each triangle and add them together to give me the exact are for ABC. The only problem that I think this method seems to pose are the values I would use. For instance, for triangle APB, I used PB for the height and AP for the base. Will this work?

Thanks for any feedback.

 

[Denis]

Re: Equilateral Triangle: find area, given distances to poin

For instance, for triangle APB, I used PB for the height and AP for the base. Will this work?

NO. PB is the length of a side, NOT the height.
The height of triangle APB is the length of a line from P to AB, perpendicular to AB.

Did you notice that 6^2 + 8^2 = 10^2 ?

 

[pka]

For any equilateral triangle with side length S then its area is given by
\(\displaystyle \L A = \frac{{\sqrt 3 S^2 }}{4}\).
Using Dennis’ hint you can find S.

 

[TchrWill]

Re: Equilateral Triangle: find area, given distances to poin

The problem: Located inside an equilateral triangle ABC is a point P such that PA = 6, PB = 8, and PC = 10. Find an exact value for the area of triangle ABC.

My thoughts:

Seeing as how PA, PB, and PC are of unequal lengths, P is not in the dead center of the triangle. Since P is inside triangle ABC, there would be three triangles embedded in it: triangle APB, triangle APC, and Triangle BPC.

I figure that I could solve for the area of each triangle and add them together to give me the exact are for ABC. The only problem that I think this method seems to pose are the values I would use. For instance, for triangle APB, I used PB for the height and AP for the base. Will this work?

1--Construct equilateral triangle ABC, A at lower left, B at top and C at lower right.
2--Locate a point P just above AC and closer to C.
3--Draw lines PA, PB and PC.
4--Label the lines PC = 6, PA = 8 and PB = 10.
5--Using PC as a radius, swing two arcs with P and C as centers, meeting at F, below line AC.
6--Triangle PCF is an equilateral triangle with sides of 6
7--Extend CP to the left to point E such that AE is perpendicular to CE.
8--Angle PCB = 60º - angle PCA = angle FCA
9--Triangle PCB is congruent to triangle FCA.
10 Therefore, AF= BP = 10
11--Triangle APF is a right triangle.
12--Angle APE = 180 - 60 - 90 - = 30º
13--Therefore, AE = 4 and EP = 4sqrt(3).
13--Therefore, AC = sqrt[4^2 +[(6 + (4sqrt3)^2]
14--Having the triangle side, the area easily follows.

Compliments of Charles W. Trigg, Mathematical Quickies, McGraw Hill, 1967, Problem 201.

I'll let you work it out with your numbers.