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 $ABC$?

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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?

Guest Aug 16, 2018

edited by
Guest
Aug 16, 2018

#1**+2 **

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.

GYanggg Aug 16, 2018

#2**+1 **

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

CPhill Aug 16, 2018