Using the identity (m^2 - 1, 2m, m^2 + 1) gives a Pythagorean triple, for m = 5 the triple is 24, 10, 26. Verify and state the largest member of the triple.

Question

Accepted Answer

26. For m = 5: m^2 - 1 = 24, 2m = 10, m^2 + 1 = 26. Largest is 26. (10^2 + 24^2 = 100 + 576 = 676 = 26^2.)

Using the identity $(m^{2} - 1, 2 m, m^{2} + 1)$ gives a Pythagorean triple, for $m = 5$ the triple is 24, 10, 26.
Verify and state the largest member of the triple.

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More Class 8 Properties of Square Numbers practice

Using the identity (m2−1,2m,m2+1) gives a Pythagorean triple, for m=5 the triple is 24, 10, 26.Verify and state the largest member of the triple.

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 8 Properties of Square Numbers practice

Using the identity $(m^{2} - 1, 2 m, m^{2} + 1)$ gives a Pythagorean triple, for $m = 5$ the triple is 24, 10, 26.
Verify and state the largest member of the triple.