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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