An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle. Class 9

An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle. Class 9

Question 1.

An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle. Class 9

Solution:

Let ∆ABC be isosceles triangle with equal sides AB and AC

∴ AB = AC = 15 cm

Perimeter of ∆ABC = 40 cm ... [Given]

∴ 40cm = AB + AC + BC ...[Given]

∴ 40 = 15 + 15 + BC

∴ BC = 10 cm

Draw AD ⊥ BC meeting BC at D

∆ABC is isosceles triangle

∴ AD is perpendicular bisector of ∆ABC

∴ BD = DC= $\frac{1}{2}$ BC = $\frac{10}{5}$ =5 cm

In right ∆ABD,

AD ⊥ BC

By Baudhayana-Pythagoras theorem,

∵ AB² = AD² + BD²

15² = AD² + 5²

225 = AD² + 25

AD² = 200

∴ AD = $\sqrt{200}$ = 10$\sqrt{2}$ cm

∵ Area of triangle = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × AD

= $\frac{1}{2}$ × 10 × $10 \sqrt{2}$ = $50 \sqrt{2}$

∴ Area of AABC = $50 \sqrt{2}$ cm²