Question

AB is the hypotenuse of a right triangle ABC. Median AD = 7 and median $\overline{BE}$ = 4. The length of AB is:

• A. $10$
• B. $5\sqrt{3}$
• C. $5\sqrt{2}$
• D. $2\sqrt{13}$

Answer 1

The correct option is D $2\sqrt{13}$

$AD=7,BE=4LetAC=b,BC=a{b}^2+\frac{a^2}{4}={AD}^2=49------\left(1\right)a^2+\frac{b^2}{4}={BE}^2=164a^2+b^2=64------\left(2\right)\left(2\right)-\left(1\right),⟹4a^2-\frac{a^2}{4}=15⟹\frac{{15a}^2}{4}=15⟹a^2=4\therefore a=2\left(1\right)⟹b^2+1=49⟹b^2=48\therefore b=4\sqrt{3}\therefore AB=\sqrt{a^2+b^2}=\sqrt{48+4}=\sqrt{52}=2\sqrt{13}$