In ΔABC, ∠B = 90°, AD and CE are the medians drawn form A and C, respectively. If AC = 10 cm and AD = $\sqrt{55}$ cm, then the length of CE is :
Answer & explanation
Correct answer: option 2
We have,
∠B = 90°
AC = 10 cm
AD = √55 cm
In right angled triangle ABC,
We know that,
= AC2 = AB2 + BC2
In right angled triangle ABD,
= AD2 = AB2 + BD2
= AD2 = AB2 + \(\frac{BC^2}{4}\) ...(a)
In right angled triangle CBE,
= CE2 = BE2 + BC2
= CE2 = \(\frac{AB^2}{4}\) + BC2 ....(b)
Adding a and b,
= AD2 + CE2 = AB2 + BC2/4 + AB2/4 + BC2
= AD2 + CE2 = \(\frac{5}{4}\) × (AB2 + BC2)
= AD2 + CE2 = \(\frac{5}{4}\) × AC2
= CE2 = \(\frac{5}{4}\) × 102 - (√55)2
CE = √70 cm