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# helphhelp

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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?

Aug 16, 2018
edited by Guest  Aug 16, 2018

#1
+972
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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.

Aug 16, 2018
#2
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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

Aug 16, 2018