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A circle is inscribed in a triangleABC If AB=12cm,AC=10cm,BC=8cm Find AD,BE and CF |
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Answer» Let AD be x cm, BE = y cm, CF = z cmThen, AD = AF = x cm [Since tangents from\xa0an external point to\xa0the circle\xa0are equal] BD = BE = y cm [Since tangents from\xa0an external point to\xa0the circle\xa0are equal] CE = CF = z cm [Since tangents from\xa0an external point to\xa0the circle\xa0are equal]According to question,AD + BD = 12 cm => x\xa0+ y = 12 ..........(i)BE + EC =\xa08 cm => y\xa0+\xa0z =\xa08 ..........(ii)CF + AC =\xa010 cm => x\xa0+\xa0z =\xa010 ..........(iii)Adding eq.(i), (ii) and (iii), we get2(x + y + z) = 30 => x + y + z = 15 ..........(iv)Solving eq. (iv) and (ii), we get x = 7 cm => AD = 7 cmSolving eq. (iv) and (iii), we get\xa0y =\xa05 cm => BE =\xa05 cmSolving eq. (iv) and (i), we get\xa0z =\xa03 cm => CF =\xa03 cm Ans. Given : A Circle inscribed in A\xa0triangle ABCAB = 12cm, AC = 10cm, BC = 8cmWe Know that Tangent drawn from an external point are equal in length.\xa0Let AD = x\xa0Then AD = AE = x [Tangent from Same point]Similarly\xa0BD = BF = y\xa0CE = CF = z\xa0AB = AD + BD => x + y =\xa012 (1)AC = AE + EC\xa0=> x + z = 10 (2)BC = BF + FC=> y + z = 8 (3)Adding all three equations, we get2(x+y+z) = 30=> x + y + z = 15 (4)from (1) x + y = 12then z = 3 cmSimilarly y = 5 cm and z = 7cm |
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