Problem: ABABAB is the hypotenuse of a right triangle ABCABCABC. Median AD=7\mathrm{AD}=7AD=7 and median BE=4\mathrm{BE}=4BE=4. The length of ABA BAB is:
Answer Choices:
A. 101010
B. 535 \sqrt{3}53​
C. 525 \sqrt{2}52​
D. 2132 \sqrt{13}213​
E. 2152 \sqrt{15}215​ Solution:
16=a2+b2416=a^{2}+\dfrac{b^{2}}{4} 16=a2+4b2​
49=a24+b249=\dfrac{a^{2}}{4}+b^{2} 49=4a2​+b2
65=5/4(a2+b2)65=5 / 4\left(a^{2}+b^{2}\right) 65=5/4(a2+b2)
a2+b2=c2=52;c=213a^{2}+b^{2}=c^{2}=52 ; \quad c=2 \sqrt{13} a2+b2=c2=52;c=213​