Question

segw ad is median of triangle abc Ab= 12 ad=10 ac=16 then find bc​

Answer 1

given
ab = 12
ad= 10
ac = 16
∆ adc is a right angled triangle
${ac }^{2} = {ad}^{2} + {dc}^{2}$
${16}^{2} = {10}^{2} + {x}^{2}$
${x}^{2} = {16}^{2} - {10}^{2}$
${x}^{2} = 256 - 100$
${x}^{2} = 156$
$x = 2 \sqrt{39}$
$bc = 2(2 \sqrt{39)}$
$bc = 4 \sqrt{39}$