Question

# In the below figure, ΔABC is right angled and AC = 100 cm. Also, AD = DE = EF = FC. Find the value of: BD2 + BE2 + BF2 (in cm2)

CAT 2021

Correct option is

**Explanatory Answer :**

E is the midpoint of hypotenuse AC so, BE = 1/2 ∗ AC=50 cm Now, applying Apollonius’s theorem on ΔABE => (AB)^2 + (BE)^2 = 2 * (BD)^2 + (AD)^2 => AB2 + 502 = 2 * (BD)^2 + (25)^2……….(1) (As, AD = DE = EF = FC = 25 cm)

Similarly, in ΔBEC, => (BC)^2 + (BE)^2 = 2 * (BF)^2 + (EF)^2 => (BC)^2 + (50)^2 = 2 * (BF)^2 + (25^2)………..(2)

Adding (1) and (2), => (AB)^2 + (BC)^2 + 2 * 50^2 = 2 * (BD^2 + BF^2 + 2 * 25^2) => AC^2 + 2 * 50^2 = 2 * (BD^2 + BF^2) + 4 * 25^2 (As, AB^2 + BC^2 = AC^2)

=> 100^2 + 2 * 50^2 – 50^2 = 2 * (BD^2 + BF^2) => 100^2+ 50^2 = 2 * (BD^2 + BF^2) => (10,000 + 2500)/2 = (BD^2 + BF^2) => 12500/2 = BD^2 + BF^2

Therefore, BD^2 + BF^2 = 6250 => BD^2 + BF^2 + BE^2 = 6250 + 50^2 = 6250 + 2500 = 8750 cm2

Hence, the answer is 8,750 cm2.