the radius of the circle of the area of ∆ abc is 84cm 2
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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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