Question

13. Write a Pythagorean triplet, whose one member is 16.​

Answer 1

Answer:

Pythagorean triples are a2+b2 = c2 where a, b and c are the three positive integers. These triples are represented as (a,b,c). Here, a is the perpendicular, b is the base and c is the hypotenuse of the right-angled triangle. The most known and smallest triplets are (3,4,5).

Answer 2

Given :

One of the number of the Pythagorean triplet = 16 units

To find :

The Pythagorean triplet.

Solution :

Given on of the sides of the Sides of the Pythagorean triplet is 16 .

So by subtracting x from the given side will give us the other side.i.e, (16 + x)

And by adding x to it will give the third side.i.e,

(16 + x)

We know the Pythagoras theorem and substituting the values in it, we get :

$\boxed{\bf{H^{2} = P^{2} + B^{2}}}$

Where :

• H = Hypotenuse
• B = Base
• P = Height

Now using the Pythagoras theorem and substituting the values in it, we get :

$:\implies \bf{H^{2} = P^{2} + B^{2}}$

$:\implies \bf{(16 + x)^{2} = 16P^{2} + (16 - x)^{2}}$

$:\implies \bf{(16 + x)^{2} = 16^{2} + (16 - x)^{2}}$

Using the identity , we get :

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²

$:\implies \bf{16^{2} + 2 \times 16 \times x + x^{2} = 16^{2} +16^{2} - 2 \times 16 \times x + x^{2}}$

$:\implies \bf{16^{2} + 32x + x^{2} = 16^{2} + 16^{2} - 32x + x^{2}}$

$:\implies \bf{256 + 32x + x^{2} = 256 + 256 - 32x + x^{2}}$

$:\implies \bf{0 = -(256 + 32x + x^{2}) - (256 + 256 - 32x + x^{2})}$

$:\implies \bf{0 = - 256 - 32x - x^{2} - 256 + 256 - 32x + x^{2}}$

$:\implies \bf{0 = 256 - 32x - 32x}$

$:\implies \bf{0 = 256 - 64x}$

$:\implies \bf{64x = 256}$

$:\implies \bf{x = \dfrac{256}{64}}$

$:\implies \bf{x = 4}$

$\boxed{\therefore \bf{x = 4}}$

Hence the value of x is 4 units.

Hence the triplets are 16 , (16 - x) 12 and (16 + x)20 .