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 :

$\sqrt{70}$ cm

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