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Question

# If the area of an isosceles triangle is 60 cm²and the length of each of its equal sides is 13 cm, find its base

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Solution

## Given: Area of the isosceles triangle (ABC) = 60 cm² Length of the each equal side (AB=AC) = 13cm. Draw AD perpendicular from A on BC. Let BC = 2x cm . Then BD = DC= x cm. In ∆ABD AB² = AD² + BD² [by Pythagoras theorem] 13² = AD² + x² 169 = AD² + x² AD² = 169 - x² AD = √ 169 - x² Area of ∆ ABC = ½ base × height 60 = ½ × BC × AD 60 = ½ ( 2x × √ 169 - x² 60 = x ×√ 169 - x² On squaring both sides 60² = (x √ 169 - x²)² 3600 = x²(169 - x²) 3600 = 169x² - x⁴ x⁴ - 169x² +3600= 0 x⁴ -144 x² -25x² +3600= 0 x²(x² -144) -25(x² -144)= 0 (x² -25) (x² -144)= 0 (x² -25) = 0 or (x² -144)= 0 x² = 25 or x² = 144 x = √25 or x = √144 x = 5 or x = 12 When x= 12, then base = 2x = 2×12= 24 cm or When base = 5 , then base = 2x = 2× 5 = 10 cm. Hence, the base is 24 cm or 10 cm

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