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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

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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\$?

Dec 6, 2017

#1
+7354
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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

Dec 6, 2017

#1
+7354
+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