Question
The area of an isosceles triangle ABC is 8√5 cm². In
this triangle, sides AB and BC are equal in length, and the base AC has a length of 8 cm. Determine the perimeter of triangle ABCSolution
Let AB = BC = 'x' cm Area of isosceles triangle = {(c/4) X √(4a2 - c2)} [Where 'a' is the length of two equal sides and 'c' is the length of third (unequal) side] 8√5 = (8/4) X {√(4 X x2 - 82)} Or, (4√5) = √(4x2 - 64) Squaring both sides. (16 X 5) = 4x2 - 64 Or, 144 = 4x2 Or, x2 = 36 So, x = 6 cm (Since, length cannot be negative therefore, we will take the positive root only) So, required perimeter = 6 + 6 + 8 = 20 cm
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