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Let a, b, c be the lengths of sides BC, CA, AB respectively of a triangle ABC. If p is the perimeter and q is the area of the triangle, then what is \(p(p-2a) \tan \left(\frac{A}{2}\right)\) equal to ?
Concept:The perimeter of the triangle = sum of all sidesArea of triangle = (1/2)base × heightCalculation:Since it is not mentioned, ? ABC is which type of triangle. So, to make calculation easyLet us assume, ? ABC is a right-angle triangle where ?A is 90°.Let a = 5, b = 4, and c = 3 (Pythagorean triple)Using the above concept, Perimeterp = 3 + 4 + 5? p = 12 Area of triangleq = (1/2) 3 × 4? q = 6Hence, the required value\(\Rightarrow p(p-2a) \tan \frac{A}{2}\) = 12(12 - 2 × 5)tan (90°/2) [? ?A = 90°]? p(p ? 2a)tan(A/2)= 24 [? tan 45° = 1From option 44q = 4 × 6 = 24? 4q is the correct answer.Alternate MethodConcept:The perimeter of the triangle = sum of all sides = a + b + cArea of scalene triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\), Half-angle formulae: \(\displaystyle tan(\frac{A}{2})=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \)Where 2s = a + b + c and a, b, c are the sides of the triangle.Calculation:If p be the perimeter of the triangle,p = a + b + c ------(1)q = \(\sqrt{s(s-a)(s-b)(s-c)}\) ------(2)where a, b, c are the sides of the triangle.Now, \(\displaystyle p(p-2a) \tan \left(\frac{A}{2}\right)\) = (a + b+ c) (a + b+ c - 2a) \(\displaystyle \tan \left(\frac{A}{2}\right) \)? (a + b+ c) (a + b+ c - 2a) \(\displaystyle \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \)Multiplying and dividing \(\sqrt{s(s - a)}\), we get, ? (a + b+ c) (a + b+ c - 2a) \(\displaystyle \sqrt{\frac{(s-b)(s-c)}{s(s-a)}\times\frac{s(s-a)}{ s(s-a)}} \)? (a + b+ c) (a + b+ c - 2a) \(\displaystyle \sqrt{\frac{s(s-a)(s-b)(s-c)}{s^2(s-a)^2}} \)? (a + b+ c) (a + b+ c - 2a) \(\displaystyle \frac{\sqrt{s(s-a)(s-b)(s-c)}}{s(s-a)}\) From equation (1) & (2)? 2s (b + c - a) \(\displaystyle \frac{q}{s(s-a)}\) [? b + c - a = 2(s - a)] ? 2s 2(s - a) \(\displaystyle \frac{q}{s(s-a)}\)? 4q? \(\displaystyle p(p-2a) \tan \left(\frac{A}{2}\right)\) = 4q
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