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.