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.

Correct 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  102=(10−2d)2+(10−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 cm2r=5 \ \text{cm} \ \\ c=2 \cdot \ r=2 \cdot \ 5=10 \ \text{cm} \ \\ c^2=a^2 + b^2 \ \\ a=c-2d \ \\ b=c-d \ \\ c^2=(c-2d)^2 + (c-d)^2 \ \\ \ \\ 10^2=(10-2d)^2 + (10-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 \ \text{cm} \ \\ a=c-2 \cdot \ d=10-2 \cdot \ 2=6 \ \text{cm} \ \\ S=a \cdot \ b/2=6 \cdot \ 8/2=24 \ \text{cm}^2

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