Right-angled triangle

Determine the content of a right triangle whose side lengths form successive members of an arithmetic progression and the radius of the circle described by the triangle is 5 cm.

Result

S =  24 cm2

Solution:

r=5 cm c=2⋅ r=2⋅ 5=10 cm c2=a2+b2 a=c−2d b=c−d c2=(c−2d)2+(c−d)2   −5d2+60d−100=0 5d2−60d+100=0  a=5;b=−60;c=100 D=b2−4ac=602−4⋅5⋅100=1600 D>0  d1,2=−b±D2a=60±160010 d1,2=60±4010 d1,2=6±4 d1=10 d2=2   Factored form of the equation:  5(d−10)(d−2)=0 d<c d=d2=2 b=c−d=10−2=8 cm a=c−2⋅ d=10−2⋅ 2=6 cm S=a⋅ b/2=6⋅ 8/2=24=24 cm2r = 5 \ cm \ \\ c = 2 \cdot \ r = 2 \cdot \ 5 = 10 \ cm \ \\ c^2 = a^2 + b^2 \ \\ a = c-2d \ \\ b = c-d \ \\ c^2 = (c-2d)^2 + (c-d)^2 \ \\ \ \\ \ \\ -5d^2 +60d -100 = 0 \ \\ 5d^2 -60d +100 = 0 \ \\ \ \\ a = 5; b = -60; c = 100 \ \\ D = b^2 - 4ac = 60^2 - 4\cdot 5 \cdot 100 = 1600 \ \\ D>0 \ \\ \ \\ d_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 60 \pm \sqrt{ 1600 } }{ 10 } \ \\ d_{1,2} = \dfrac{ 60 \pm 40 }{ 10 } \ \\ d_{1,2} = 6 \pm 4 \ \\ d_{1} = 10 \ \\ d_{2} = 2 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 5 (d -10) (d -2) = 0 \ \\ d<c \ \\ d = d_{ 2 } = 2 \ \\ b = c-d = 10-2 = 8 \ cm \ \\ a = c-2 \cdot \ d = 10-2 \cdot \ 2 = 6 \ cm \ \\ S = a \cdot \ b/2 = 6 \cdot \ 8/2 = 24 = 24 \ cm^2

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