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 (c4)2=(a3)2+(b3)2=(174)2=169  169=a223a+32+b223b+32 169=a2+b26a6b+18 16928918=6a6b  b=a(23)  c2=289=a2+(a(23))2 2a246a+240=0  p=2;q=46;r=240 D=q24pr=46242240=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(a15)(a8)=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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