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The radius of the circle of the area of ∆ abc is 84cm 2 |
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Answer» Answer: please mark me as brainlist Step-by-step explanation: Given: AREA of ∆ABC = 84 cm2 PB = 8 cm PA = 6 cm AB = 14 cm BQ = 8 cm (Common tangent with PB) AR = 6 cm (Common tangent with PA) Since QC & CR are Common tangent Let QC = CR = x AC = 6 + x BC = 8 + x So ΔROA and ΔPOA are congruent ΔPOB and ΔQOB are congruent ΔQOC and ΔROC are congruent Area of ΔROA = ΔPOA = 12 cm2 Area of ΔPOB = ΔQOB = 16 cm2 Area of ΔQOC = ΔROC = 2x The SUM areas of all the small triangles = Area of ∆ABC ⇒ 2 × 12 + 2 × 16 + 2 × 2x = 84 ⇒ 4x = 28 ⇒ x = 7 cm Answer: Length of AC = 13 cm & Length of BC = 15 cm |
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