If $15^{2} + a^{2} = b^{2},$ then find $a$ and $b$ respectively.

8, 9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8, 16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8, 17

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9, 16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
8, 17

For an integer $m \geq 2,$ the numbers $m^{2} - 1, 2 m$ and $m^{2} + 1$ form a Pythagorean triplet.

As the given number (15) is odd, it can't be of the form $2 m .$

$\Rightarrow m^{2} - 1 = 15$

$\Rightarrow m^{2} = 16 = 4^{2}$

$\Rightarrow m = 4$

$2 m = 2 \times 4 = 8$

$\Rightarrow m^{2} + 1 = 4^{2} + 1 = 16 + 1 = 17$

So, $a = 8$ and $b = 17.$

Suggest Corrections

Similar questions

15^{2}

a^{2}

b^{2}

a - b = 6

a b = 16

a^{2} + b^{2}

Find a and b from $15^{2}$ + $a^{2}$ = $b^{2}$ .

If a + b = 10 and ab = 16, find the value of $a^{2} - a b + b^{2} a n d a^{2} + a b + b^{2}$ .

If a - b = 6 and ab = 16; find $a^{2} + b^{2}$

Join BYJU'S Learning Program