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We have a triangle $\triangle ABC$ and a point $K$ on $BC$ such that $AK$ is an altitude to $\triangle ABC$. If $AC = 10,$ $BK = 7$, and $BC=13$ then what is the area of $\triangle ABC$?

RektTheNoob  Dec 6, 2017

Best Answer 

 #1
avatar+7324 
+2

If   BC = 13   and   BK = 7  ,  then   KC  =  13 - 7  =  6

 

 

And we can use the Pythagorean theorem to find  AK .

 

CK2 + AK2  =  AC2

 

62 + AK2  =  102          Subtract  62  from both sides of this equation.

 

AK2  =  102 - 62

 

AK2  =  64                 Take the positive square root of both sides.

 

AK  =  8

 

And let  BC  be the triangle's base, so  AK  is the triangle's height.

 

area of triangle ABC  =  (1/2) * BC * AK

 

area of triangle ABC  =  (1/2) * 13 * 8

 

area of triangle ABC  =  52   sq units

hectictar  Dec 6, 2017
 #1
avatar+7324 
+2
Best Answer

If   BC = 13   and   BK = 7  ,  then   KC  =  13 - 7  =  6

 

 

And we can use the Pythagorean theorem to find  AK .

 

CK2 + AK2  =  AC2

 

62 + AK2  =  102          Subtract  62  from both sides of this equation.

 

AK2  =  102 - 62

 

AK2  =  64                 Take the positive square root of both sides.

 

AK  =  8

 

And let  BC  be the triangle's base, so  AK  is the triangle's height.

 

area of triangle ABC  =  (1/2) * BC * AK

 

area of triangle ABC  =  (1/2) * 13 * 8

 

area of triangle ABC  =  52   sq units

hectictar  Dec 6, 2017

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