Question

Find a and b

${14}^2+a^2=b^2$

• A. 38, 40
• B. 48, 58
• C. 46, 50
• D. 48, 50

Answer 1

The correct option is D

48, 50

${14}^2+a^2=b^2$

Let the Pythagorean triplet be

$(m^2+1),2m,(m^2-1)$

Now try equating 14 with any of the three variables from above.

Observe that equating 14 with $(m^2+1)and(m^2-1)$ is not effective as 13 and 15 are not perfect square.

$\therefore 2m=14$

$\Rightarrow m=7$

$\Rightarrow m^2+1=7^2+1=50$

$\Rightarrow m^2-1=7^2-1=48$

$\therefore {14}^2+{48}^2={50}^2$

$\Rightarrow a=48andb=50$