The altitude drawn to the base of an isosceles triangle is 6 cm andthe perimeter of the triangle is 36 cm. The area (in cm²) of the triangle is
Let AB = AC = a cm. BD = DC = b cm. Altitude of isosceles triangle is also median. In right ∆ADC, 6² = a² - b² 36 = a² - b² ………… (i) Perimeter = 36 a + a + 2b = 36 2a + 2b = 36 a + b = 18 ………. (ii) On dividing (i) & (ii) we get, (a² - b² )/(a+b) = 36/18 = 2 ((a+b) (a-b))/((a+b)) = 2 a – b = 2 a + b = 18 on solving, 2a = 20 a = 10 b = 8 BC = 16 Area of ∆ABC = 1/2 × AD × BC = 1/2 × 6 × 16 = 48 cm²
The term ‘corpus delicti’ means
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