# In Fig. 9.65, triangle AEC is right-angled at E, B is a point on EC, BD is the altitude of triangle ABC, AC = 25 cm, BC = 7 cm and AE = 15 cm. Find the area of triangle ABC and the length of DB.

**Solution:**

Given, the triangle AEC is right angled at E.

B is a point on EC

BD is the altitude of triangle ABC.

AC = 25 cm

BC = 7 cm

AE = 15 cm

We have to find the area of triangle ABC and the length of DB.

In triangle AEC,

By using __Pythagorean theorem__,

AC² = AE² + EC²

(25)² = (15)² + EC²

625 = 225 + EC²

EC² = 625 - 225

EC² = 400

Taking __square root__,

EC = 20 cm

Now, EB = EC - BC

= 20 - 7

EB = 13 cm

Area of triangle = 1/2 × base × height

Area of triangle AEC = 1/2 × EC × AE

= 1/2 × 20 × 15

= 10(15)

= 150 cm²

__Area of triangle__ AEB = 1/2 × EB × AE

= 1/2 × 13 × 15

= 97.5 cm²

Area of triangle ABC = area of triangle AEC - area of triangle AEB

= 150 - 97.5

= 52.5 cm²

Therefore, the area of triangle ABC is 52.5 cm².

Area of triangle ABC = 1/2 × AC × BD

52.5 = 1/2 × 25 × BD

BD = 52.5(2)/25

BD = 4.2 cm

Therefore, the length of DB is 4.2 cm.

**✦ Try This: **The perimeter of the given figure is

**☛ Also Check: **NCERT Solutions for Class 7 Maths Chapter 11

**NCERT Exemplar Class 7 Maths Chapter 9 Problem 122**

## In Fig. 9.65, triangle AEC is right-angled at E, B is a point on EC, BD is the altitude of triangle ABC, AC = 25 cm, BC = 7 cm and AE = 15 cm. Find the area of triangle ABC and the length of DB.

**Summary:**

In Fig. 9.65, triangle AEC is right-angled at E, B is a point on EC, BD is the altitude of triangle ABC, AC = 25 cm, BC = 7 cm and AE = 15 cm. The area of triangle ABC is 52.5 cm² and the length of DB is 4.2 cm

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