Q.8 \(\triangle \)ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of ?ABC. What will be the height from C to AB i.e., CE?

Area of \(\triangle \)ABC = \(\frac{1}{2} \) × base × height

\(\therefore \) \(\frac{1}{2} \)×AB×CE= \(\frac{1}{2} \)×BC×AD

Or, \(\frac{1}{2} \)× 7.5× CE = \(\frac{1}{2} \)× 9×6

Or, CE= \(\frac{54}{7.5} \)= 7.2 cm