Question

In an equilateral triangle ABC, the side BC is trisected at D. Then AD² is equal to

(a) \(\frac{9}{7}AB^2\) 

(b) \(\frac{7}{9}AB^2\)

(c) \(\frac{3}{4}AB^2\)

(d) \(\frac{4}{5}AB^2\)

Answer 1

(b) \(\frac{7}{9}AB^2\)

ABC being an equilateral triangle, AB = BC = AC. 

Also BC being trisected at D 

⇒ BD = \(\frac{1}{3}\)BC

Let AF be drawn perpendicular to BC ⇒ BF = FC. 

Also E is given as the other point of trisection, so BD = DE = EC 

Also, BD = 2DF                     …(ii) 

Now AB² = BF² + AF² and 

AD² = DF² + AF² 

Now AB² = BF² + AF

⇒ \(AB^2 = \bigg(\frac{1}{2}BC\bigg)^2+AF^2\)

⇒ \(AB^2 = \frac{1}{4}BC^2+AF^2\)           (∵ BC = AB)

⇒ \(AB^2 = \frac{1}{4}AB^2+AF^2\) ⇒ AF² = \(\frac{3}{4}\) AB²            ....(iii)

Also, AD² = DF² + AF²

= \(\bigg(\frac{1}{3}\times\frac{1}{3}BC\bigg)^2+AF^2\)                                   (From (i) and (ii))

= \(\bigg(\frac{1}{6}BC\bigg)^2+AF^2\)

= \(\frac{1}{36}BC^2+AF^2\)

= \(\frac{1}{36}AB^2+\frac{3}{4}AB^2\)             (∵ BC = AB and putting the value from (iii))

= \(\frac{28}{36}AB^2\) ⇒ AD² = \(\frac{7}{9}AB^2\)