Acircle with radius r is inscribed into a right triangle. find the perimeter of the triangle if: the length of the hypotenuse is 24 cm, and r=4 cm;

56 cm

Step-by-step explanation:

The tangents from the 90° angle will form a square with the radii that has a side length of 4. If we call the length of the short side of the right triangle "x", then the tangent lengths are ...

on the short side of the triangle: 4, x-4

on the hypotenuse side of the triangle: x-4, 24-(x-4) = 28-x

on the long side of the triangle: 4, 28-x

The perimeter is twice the sum of the unique tangent lengths:

P = 2(4 + (x-4) + (28-x)) = 2·28

P = 56 . . . . . the perimeter is 56 cm.

Using the Pythagorean theorem on side lengths x and 32-x and hypotenuse 24, we find x = 16-4√2 ≈ 10.34, the length of the short side (in cm).

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.