If in a right angled triangle ABC, angle B = 90 degree, AC = 10 cm and radius of incircle is 1 cm, then the perimeter of the triangle is ?
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Draw perpendicular from Incenter O on to AB, BC and AC, meeting them at D, E and F respectively.
BD = BE = incircle radius = 1 cm, as it is a square.
Let CE = CF = x cm
AF = 10 - x => AD = 10 -x
Pythagoras theorem => (10-x +1)^2 + (1+x)^2 = 10^2
x^2 - 10 x + 11 = 0
x = 5 + sqroot(14) or 5 - sqroot(14)
Then AB = 6 - sqroot(14), BC = 6 + sqroot(14), or the otherway round.
Perimeter = 22 cm
BD = BE = incircle radius = 1 cm, as it is a square.
Let CE = CF = x cm
AF = 10 - x => AD = 10 -x
Pythagoras theorem => (10-x +1)^2 + (1+x)^2 = 10^2
x^2 - 10 x + 11 = 0
x = 5 + sqroot(14) or 5 - sqroot(14)
Then AB = 6 - sqroot(14), BC = 6 + sqroot(14), or the otherway round.
Perimeter = 22 cm
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