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If length of equal sides in an isosceles triangle is 10 cm and its area is 48 cm². Find the length of the base.

User Pana
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1 Answer

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LENGTH OF THE BASE

Explanation:

Let base of the isosceles triangle = b cm

And height of the isosceles triangle = h cm

Area of an isosceles triangle = 1/2 × b × h = 48 cm²

Or, 1/2 × b × h = 48

Or, h = 96/b cm

By Pythagoras Theorem,

(b/2)²+(96/b)²=(10)²

Or, b²/4+9216/b²=100

Or, (b⁴+36864)/4b²=100

Or, b⁴+36864=400b²

Or, b⁴-400b²+36864=0

Put b²=p

P²-400p+36864=0, This is a quadratic equation, solving

(i) p=[-(-400)+√{(-400)²-4×36864}]/2×1

p = [400+√12544]/2×1= (400+112)/2 = 512/2

p = 256

But, b² = p = 256

b = √256

b = 16 cm

h = 96/b = 96/16

h = 6 cm

(ii) p=[-(-400) - √{(-400)²-4×36864}]/2×1

p = [400 - √12544]/2×1= (400 - 112)/2 = 288/2

p = 144

But, b² = p = 144

b = √144

b = 12

h = 96/b = 96/12

h = 8

Therefore, when base (b) = 16 cm, h = 6 cm and when base (b) = 12 cm, h = 8 cm.

User Deadlyvices
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