We have a triangle $\triangle ABC$ and a point $K$ on $BC$ such that $AK$ is an altitude of $\triangle ABC$. If $AC = 10,$ $BK = 7$, and $BC = 13,$ then what is the area of $\triangle ABC$?
I know that the area of a triangle is given by the formula $\frac{1}{2} bh,$ where $b$ is the base and $h$ is the height. In this case, $AK$ is the height of $\triangle ABC,$ and $BC$ is the base. So, the area of $\triangle ABC$ is $\frac{1}{2} BC \cdot AK.$ I have $BC = 13,$ but I don't have $AK.$ However, I can use the Pythagorean theorem to find $AK,$ since $\triangle AKC$ is a right triangle. The Pythagorean theorem says that $AC^2 = AK^2 + CK^2.$ I have $AC = 10,$ and I can find $CK$ by subtracting $BK$ from $BC.$ That is, $CK = BC - BK = 13 - 7 = 6.$ So, I can plug these values into the Pythagorean theorem and get $10^2 = AK^2 + 6^2.$ Simplifying, I get $AK^2 = 10^2 - 6^2 = 100 - 36 = 64.$ Taking the square root of both sides, I get $AK = \sqrt{64} = 8.$ Now, I have everything I need to find the area of $\triangle ABC.$ Plugging in the values of $BC$ and $AK,$ I get $\frac{1}{2} BC \cdot AK = \frac{1}{2} \cdot 13 \cdot 8 = 52.$

