# 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 \ \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

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