An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides? Class 9
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An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides? Class 9
Question 1.
An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides? Class 9
Solution:
Let ∆ABC be isosceles triangle with base BC and equal sides AB and AC.
Let h be height of triangle
Using,
Area of triangle = $\frac{1}{2}$ × base × height
60 = $\frac{1}{2}$ × BC × AD
60 = $\frac{1}{2}$ × 10 × h
60 = 5h
∴ h = 12 cm
∴ AD = 12cm
∆ABC is isosceles triangle
∴ BD = DC = $\frac{10}{2}$ = 5 cm
In right-angled ∆ABD,
AD ⊥ BC
∴ By Baudhayana-Pythagoras theorem
AB² = AD² + BD²
∴ AB² = 12² + 5²
= 144 + 25 = 169
∴ AB = $\sqrt{169}$ = 13 cm
∴ Length of equal sides = 13 cm each