Plz solve this que???
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hope this hlp.........
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Given in right ΔABC, ∠B = 90°
Also D is midpoint of BC, that is BD = DC = (BC/2)
In right ΔABC, AC2 = AB2 + BC2 {By Pythagoras theorem] – (1)
In right ΔABD, AD2 = AB2 + BD2
AD2 = AB2 + (BC/2)2
= AB2 + (BC2 /4)
4AD2 = 4AB2 + BC2
BC2 = 4AD2 – 4AB2
Equation (1) becomes,
AC2 = AB2 + [4AD2 – 4AB2]
AC2 = 4AD2 – 3AB2
Also D is midpoint of BC, that is BD = DC = (BC/2)
In right ΔABC, AC2 = AB2 + BC2 {By Pythagoras theorem] – (1)
In right ΔABD, AD2 = AB2 + BD2
AD2 = AB2 + (BC/2)2
= AB2 + (BC2 /4)
4AD2 = 4AB2 + BC2
BC2 = 4AD2 – 4AB2
Equation (1) becomes,
AC2 = AB2 + [4AD2 – 4AB2]
AC2 = 4AD2 – 3AB2
Similar questions
Also D is midpoint of BC, that is BD = DC = (BC/2)
In right ΔABC, AC2 = AB2 + BC2 {By Pythagoras theorem] – (1)
In right ΔABD, AD2 = AB2 + BD2
AD2 = AB2 + (BC/2)2
= AB2 + (BC2 /4)
4AD2 = 4AB2 + BC2
BC2 = 4AD2 – 4AB2
Equation (1) becomes,
AC2 = AB2 + [4AD2 – 4AB2]
AC2 = 4AD2 – 3AB2