Concept:In a right-angled triangle, the square of the longest side (hypotenuse) equals the sum of squares of the other two sides. Consecutive even integers differ by 2.Explanation:Let the side lengths be x, x+2, and x+4 cm, with x+4 as the hypotenuse.Apply Pythagoras theorem: (x+4)2=x2+(x+2)2Expand: x2+8x+16=x2+x2+4x+4Simplify: 8x+16=2x2+4x+4Bring all terms to one side: 0=2x2+4x+4−8x−160=2x2−4x−12Divide by 2: x2−2x−6=0? Wait, careful: 2x2−4x−12=0 → x2−2x−6=0 gives non-integer solutions. Let's re-check: The original existing solution had p2−4p−12=0, which gives p=6 or p=−2. That suggests sides: 6,8,10. But if we use x=6, then x+2=8, x+4=10. So the equation should be: x2+(x+2)2=(x+4)2? Actually we set hypotenuse as x+4. So it's (x+4)2=x2+(x+2)2. Expanding: x2+8x+16=x2+x2+4x+4 → 8x+16=2x2+4x+4 → 0=2x2−4x−12 → divide by 2: x2−2x−6=0? That gives x=1±7​, not 6. So the existing solution equation is wrong? Let's solve correctly: Actually the sides are consecutive even integers, so let them be 2n,2n+2,2n+4. Then (2n+4)2=(2n)2+(2n+2)2 → 4n2+16n+16=4n2+4n2+8n+4 → 16n+16=8n2+8n+4 → 0=8n2−8n−12 → divide by 4: 2n2−2n−3=0 — that doesn't give integer n either. Hmm, the known triple 6,8,10 works: 6,8,10 are consecutive even integers? 6,8,10 are not consecutive even integers? They are even and consecutive? 6,8,10 are spaced by 2, so yes they are consecutive even integers (6, then 8, then 10). So the sides are 6,8,10. That means the numbers are 6,8,10 => 2n=6, then n=3, 2n+2=8. So the equation should yield 2n+4=10. Let's derive properly: let sides be n=3 with a,a+2,a+4 even. Then a → (a+4)2=a2+(a+2)2 → a2+8a+16=a2+a2+4a+4 → 8a+16=2a2+4a+4 → 0=2a2−4a−12. Solving gives a2−2a−6=0, not integer. But the existing solution got a=1±7​ which gives p2−4p−12=0. That came from a different assumption: maybe they let sides be p=6 but then they wrote p,p+2,p+4? That's what I did. Let's check their expansion: they wrote '(p+4)2=p2+(p+2)2' — that's =p2+16+8p=p2+p2+4p+4 → p2+8p+16=2p2+4p+4. Yes, they got 0=p2−4p−12 which factors to p2−4p−12=0. That gives (p−6)(p+2)=0. But we got p=6. There's a discrepancy. Let's recalc: from p2−2p−6=0, LHS: (p+4)2=p2+(p+2)2, RHS: p2+8p+16. So p2+(p2+4p+4)=2p2+4p+4 → bring all: p2+8p+16=2p2+4p+4. Yes! Because 0=2p2+4p+4−p2−8p−16=p2−4p−12, 2p2−p2=p2, 4p−8p=−4p. So 4−16=−12 is correct. I mistakenly wrote p2−4p−12=0 earlier due to arithmetic error. So the equation is correct from the existing solution. So the sides are x2−2x−6=0. Therefore product = 6,8,10.Thus the correct answer is 6×8×10=480.
480
3,4,5
5,12,13
7,24,25
11,60,61
Alternate MethodConcept:In 9,40,41, triangle ABC is angle B so we use the Pythagoras theorem
