Question

The hypotenuse of a right triangle is 20m. If the difference between the length of the other sides is 4m. Find the sides.​

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

$\underline \mathsf {ATQ:-}$

$\mathsf {x^2 + y^2 = 20^2}$

$\mathsf {x^2 + y^2 = 400}$

$\mathsf {also,\: x - 4 = 4}$

$\implies \mathsf {x = 404 + y}$

$\mathsf {(4 + y)^2 + y^2 = 400}$

$\implies \mathsf {2y^2 + 8y - 384 = 0}$

$\implies \mathsf {(y + 16) (y - 12) = 0}$

$\implies \mathsf {y = 12y = -16 \:(N.P)}$

$\therefore \mathsf \blue{12\:Cm\: and \:16\:Cm}$

Answer 2

Answer:

• The other side = 16 m

• And it's length = 12 m

Given :

• The hypotenuse of a right triangle is 20m.

• The difference between the length of the other sides is 4m.

To find :

• The sides =?

Step-by-step explanation:

The hypotenuse of a right triangle = 20m. [Given]

Let, The other side (Perpendicular) be x.

Then, Length of a right triangle (Base) be x - 4.

According to the given question :

By Pythagoras theorem,

H² = P² + B²

Substituting the values in the above formula, we get,

➟ (20)^2 = x^2 + ( x - 4)^2

➟ 400 = x^2 + x^2 + 16 - 8x

➟ 400 = 2x^2 + 16 - 8x

➟ 400 - 16 = 2x^2 + 8x

➟ 384 = 2x^2 + 8x

➟ 2x^2 + 8x - 384 = 0

➟ 2(x^2 + 4x - 192) = 0

➟ x^2 + 4x - 192 = 0/2 = 0

➟ x^2 + 4 - 192 = 0

➟ x^2 + 4x - 192 = 0

➟ x^2 + (16 - 12)x - 192 = 0

➟ x^2 + 16 - 12x - 192 = 0

➟ x(x + 16) - 12 ( x + 16) = 0

➟ (x + 16) (x - 12) = 0

Factors are, 16 and - 12

∴ x + 16 = 0

➟ x = 0 - 16

➟ x = -16 [ignore negative]

So, x = 16

And, x - 12 = 0

➟ x = 0 + 12

➟ x = 12

Hence, the other side = 16 m

And it's length = 12 m