RTriangle 17

The hypotenuse of a right triangle is 17 cm. If you decrease both two legs by 3 cm you will reduce the hypotenuse by 4 cm. Determine the length of this legs.

Correct result:

a =  15 cm
b =  8 cm

Solution:

172=a2+b2 (17−4)2=(a−3)2+(b−3)2  289=a2+b2 169=a2−6a+9+b2−6b+9  289=a2+b2 151=a2+b2−6a−6b  151=289−6a−6b 138=6a+6b 23=a+b  289=a2+b2 289=a2+(23−a)2 289=a2+232−46a+a2 2a2−46a+240=0 2a2−46a+240=0  p=2;q=−46;r=240 D=q2−4pr=462−4⋅2⋅240=196 D>0  a1,2=−q±D2p=46±1964 a1,2=46±144 a1,2=11.5±3.5 a1=15 a2=8   Factored form of the equation:  2(a−15)(a−8)=0   a=15 cm b=8 cm17^2 = a^2 + b^2 \ \\ (17-4)^2 = (a-3)^2 + (b-3)^2 \ \\ \ \\ 289=a^2+b^2 \ \\ 169 = a^2-6a+9+b^2-6b+9 \ \\ \ \\ 289=a^2+b^2 \ \\ 151 = a^2+b^2-6a-6b \ \\ \ \\ 151 = 289 - 6a-6b \ \\ 138 = 6a + 6b \ \\ 23 = a+b \ \\ \ \\ 289=a^2+b^2 \ \\ 289=a^2+(23-a)^2 \ \\ 289=a^2+23^2-46a+a^2 \ \\ 2a^2 - 46a +240 = 0 \ \\ 2a^2 -46a +240 =0 \ \\ \ \\ p=2; q=-46; r=240 \ \\ D = q^2 - 4pr = 46^2 - 4\cdot 2 \cdot 240 = 196 \ \\ D>0 \ \\ \ \\ a_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 46 \pm \sqrt{ 196 } }{ 4 } \ \\ a_{1,2} = \dfrac{ 46 \pm 14 }{ 4 } \ \\ a_{1,2} = 11.5 \pm 3.5 \ \\ a_{1} = 15 \ \\ a_{2} = 8 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2 (a -15) (a -8) = 0 \ \\ \ \\ \ \\ a = 15 \ cm \ \\ b= 8 \ cm

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