Point $P$ is inside equilateral triangle $ABC$ such that the altitudes from $P$ to $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$ have lengths 5, 6, and 7 respectively. What is the area of triangle $ABC$?
nvm
need help with this one
We have a triangle $\triangle ABC$ and a point $K$ on segment $\overline{BC}$ such that $AK$ is an altitude to$\triangle ABC$. If $AK = 6,$ $BK = 8$, and $CK = 6,$ then what is the perimeter of the triangle?
+983 
Point \(P\) is inside equilateral triangle \(ABC\) such that the altitudes from \(P\) to \(\overline{AB}, \overline{BC}, \text{and} \ \overline{CA}\) have lengths 5, 6, and 7 respectively. What is the area of triangle \(ABC\).
Connect \(P\) to \(A,B, \text{and } C\), to form \(\triangle ABP, \triangle ACP, \triangle BCP.\)
\([ABC]=[ABP]+[ACP]+[BCP]=\frac12x(5+6+7).\)
The area of \(\triangle ABC\) can also be represented as \(\frac{\sqrt3}{4}x^2\).
Therefore: \(\frac{\sqrt3}{4}x^2=\frac12x(5+6+7)\Rightarrow x=12\sqrt3\).
Plugging \(x\) into one of our area formulas, \(\frac{\sqrt3}{4}x^2=\frac12x(5+6+7)=108\sqrt3\).
I hope this helped,
Gavin.
+129852 If AK is an altitude drawn to base BC.....then angles AKB and AKC form right angles
So......using the Pythagorean Theorem......
√[AK^2 + BK^2] = AB = √ [ 6^2 + 8^2 ] = √100 = 10 units
Ans
√] AK^2 + CK^2 ] = AC = √ [ 6^2 + 6^2 ] = √72 = 6√2 units
And CK + BK = BC = 6 + 8 = 14
So...the perimeter is [ 10 + 6√2 + 14 ] = 24 + 6√2 units
