Side Section Lengths in a Right Triangle

Question

Right triangle ABC has its right angle at C. Let M and N be the midpoints of AC and BC, respectively, with AN=19 and BM=22. What is AB?

Answer 1

we have $\frac{b^2}{4}+a^2=22^2$
$b^2+\frac{a^2}{4}=19^2$ summing up we obatain $c^2+\frac{c^2}{4}=19^2+22^2$ the rest is easy

Answer 2

Consider the figure below.

Since $M$ is the midpoint of $\overline{AC}$,

$$AM = CM = \frac{1}{2}AC$$

Since $N$ is the midpoint of $\overline{BC}$,

$$BN = CN = \frac{1}{2}BC$$

Since $\triangle ABC$ is a right triangle, the Pythagorean Theorem yields

$$AC^2 + BC^2 = AB^2$$

Since $\triangle ANC$ is a right triangle, the Pythagorean Theorem yields

$$AC^2 + CN^2 = AN^2$$

Substituting $\dfrac{1}{2}BC$ for $CN$ and $19$ for $AN$ yields

$$AC^2 + \frac{1}{4}BC^2 = 19^2$$

Multiplying both sides of the equation by $4$ yields

$$4AC^2 + BC^2 = 4 \cdot 19^2$$

Since $\triangle BMC$ is a right triangle, the Pythagorean Theorem yields

$$CM^2 + BC^2 = BM^2$$

Substituting $\dfrac{1}{2}AC$ for $CM$ and $22$ for $BM$ yields

$$\frac{1}{4}AC^2 + BC^2 = 22^2$$

Multiplying both sides of the equation by $4$ yields

$$AC^2 + 4BC^2 = 4 \cdot 22^2$$

This gives us the system of equations

\begin{alignat*}{5} 4AC^2 & + & BC^2 & = & 4 \cdot 19^2\\ AC^2 & + & 4BC^2 & = & 4 \cdot 22^2 \end{alignat*}

Summing yields

\begin{align*} 5AC^2 + 5BC^2 & = 4(19^2 + 22^2)\\ 5(AC^2 + BC^2) & = 4(361 + 484)\\ 5AB^2 & = 4(845)\\ AB^2 & = 4(169)\\ AB^2 & = 2^2 \cdot 13^2\\ AB & = 2 \cdot 13\\ AB & = 26 \end{align*}