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 $\triangle ABC$?
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
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