# ∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). 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?

**Solution:**

In triangle ABC, AD = 6 cm and BC = 9 cm

Area of triangle ABC = 1/2 × Base × Height

= 1/2 × BC × AD

= 1/2 × 9 cm × 6 cm

= 27 cm^{2}

Now, Area of triangle ABC = 1/2 × AB × CE

27 cm^{2 }= 1/2 × 7.5 cm × CE

CE = (2 × 27 cm^{2}) / 7.5 cm

CE = 54 cm^{2} / 7.5 cm

CE = 7.2 cm

**Video Solution:**

## ∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). 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?

### Class 7 Maths NCERT Solutions - Chapter 11 Exercise 11.2 Question 8

**Summary:**

∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. 27 cm^{2} is the area of ∆ABC and 7.2 cm will be the height from C to AB.