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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?

Guest Aug 16, 2018
edited by Guest  Aug 16, 2018
 #1
avatar+963 
+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
avatar+91019 
+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

 

 

cool cool cool

CPhill  Aug 16, 2018

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