# GeometryIf the hypothenuse of right triangle of 26 cm long and one cathetus 14 cm longer than the other , find the lengths of the legs of triangle.

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It is given that the hypotenuse of a right triangle is 26 cm long and one leg is 14 cm longer than the other.

Let the length of the shorter leg be x , the other leg is x + 14

AS it is a right triangle use hypotenuse theorem to write

x^2 + (x + 14)^2 = 26^2

=> x^2 + x^2 + 14^2 + 28x = 26^2

=> 2x^2 + 28x + 14^2 - 26^2 = 0

=> x^2 + 14x -240 = 0

=> x^2 + 24x - 10x - 240 = 0

=> x(x + 24) - 10(x + 24) = 0

=> (x - 10)(x + 24) = 0

We only take x = 10 as length cannot be negative

**The lengths of the legs of the triangle are 10 and 24.**

We'll note:

x = shorter leg

y = longer leg

z = hypothenuse

We'll apply Pythagorean Theorem in the given right triangle:

z^2 = x^2 + y^2

But y is 14 cm longer than x.

y = x + 14

We'll re-write Pythagorean Theorem:

z^2 = x^2 + ( x + 14)^2

But z = 26. We'll substitute z in the Pythagorean identity and we'll expand the square:

26^2 = x^2 + x^2 + 28x+ 196

We'll combine like terms:

676 = 2x^2 + 28x + 196

We'll divide by 2:

x^2 + 14a + 98 - 338 = 0

We'll combine like terms:

a^2 + 14x- 240 = 0

We'll apply quadratic formula:

x1 = [-14+sqrt(196 + 960)]/2

x1 = (-14+34)/2

x1 = 10

x2 = (-14-34)/2

x2 = -24

Since we are talking about a distance, which is always positive, we'll reject x2 = -24.

One cathetus of triangle is x = 10 cm and the other leg, y = 14 + 10 cm.

**y = 24 cm**