Right angled triangle

Hypotenuse of a right triangle is 17 cm long. When we decrease length of legs by 3 cm then decrease its hypotenuse by 4 cm. Determine the size of legs.

Correct result:

a =  15 cm
b =  8 cm

Solution:

c2=a2+b2=172=289 (c−4)2=(a−3)2+(b−3)2=(17−4)2=169  169=a2−2⋅3a+32+b2−2⋅3b+32 169=a2+b2−6⋅a−6⋅b+18 169−289−18=−6⋅a−6⋅b  b=−a−(−23)  c2=289=a2+(−a−(−23))2 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=a1=15 cm b=a2=8 cmc^2 = a^2+b^2 = 17^2= 289 \ \\ (c-4)^2 = (a-3)^2+(b-3)^2 = (17-4)^2 = 169 \ \\ \ \\ 169 = a^2 - 2\cdot 3 a + 3^2 + b^2 - 2\cdot 3 b + 3^2 \ \\ 169 = a^2 +b^2 - 6 \cdot a - 6 \cdot b + 18 \ \\ 169 - 289 - 18 = - 6 \cdot a - 6 \cdot b \ \\ \ \\ b = -a - (-23) \ \\ \ \\ c^2 = 289 = a^2 + (-a - (-23))^2 \ \\ 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 = a_1 = 15 \ \text{cm} \ \\ b = a_2 = 8 \ cm

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