what is x short leg for right triangle

when given long leg b is 1/2x +11
when given hypotenuse c is 2x + 1

i put (-1/2x + 11) +(2x + 1) = x and came out with 20 but that did not work out when squared A + b = C.

I estimated x/a = 8, then b = 15 and c = 17 and that works 8^2 + 15^2 = 17^2 or 64 + 225 = 289 for A^2 +b^2 = c^2
could you please show me how to do it the right way?
thanks

You are not even using the correct formula.

it is A^2 + B^2 = C^2 , where C is the hypotenuse, so

x^2 + ((1/2)x + 11)^2 = (2x+1)^2
x^2 + (1/4)x^2 + 11x + 121 = 4x^2 + 4x + 1
times 4 to get rid of my fraction

4x^2 + x^2 + 44x + 484 = 16x^2 + 16x + 4
-11x^2 + 28x + 480 = 0
11x^2 - 28x - 480 = 0
(x-8)(11x + 60) = 0
x = 8 or x = -60/11 , but x can't be negative

so x = 8

one side is 8
the other is (1/2)x+11 = 15
and the hypotenuse is 2x+1 = 17

Reiny, thank you so much, I really needed that help.

To find the value of the short leg, x, we will use the Pythagorean theorem, which states that for a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse.

Let's start by rewriting the given equations:

Long leg, b = 1/2x + 11
Hypotenuse, c = 2x + 1

We can substitute these values into the Pythagorean theorem equation:

x^2 + (1/2x + 11)^2 = (2x + 1)^2

Now, let's simplify and solve for x:

Expand the expression on the left side:

x^2 + (1/2)(1/2x + 11)(1/2x + 11) = (2x + 1)^2
x^2 + (1/4)(x + 22)^2 = (2x + 1)^2

Expand the squares:

x^2 + (1/4)(x^2 + 2*22x + 22^2) = (4x^2 + 4x + 1)

Multiply through by 4 to eliminate the fraction:

4x^2 + x^2 + 44x + 484 = 16x^2 + 16x + 4

Simplify and rearrange terms:

0 = 11x^2 + 27x - 480

Now, we have a quadratic equation. We can solve this equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

For our equation, a = 11, b = 27, and c = -480. Plugging these values into the formula:

x = (-27 ± sqrt(27^2 - 4 * 11 * -480)) / (2 * 11)

x = (-27 ± sqrt(729 + 21120)) / 22

x = (-27 ± sqrt(21849)) / 22

Now, calculate the square root:

x = (-27 ± 147) / 22

x1 = (-27 + 147) / 22 = 120 / 22 = 60/11 (approximately 5.45)
x2 = (-27 - 147) / 22 = -174 / 22 = -87/11 (approximately -7.91)

Since we are dealing with the length of a side, x cannot be negative. Therefore, x ≈ 5.45.

So, the approximate value of the short leg, x, is 5.45.

To find the value of x, the short leg of the right triangle, given the long leg and the hypotenuse, you can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the sum of the squares of the two legs is equal to the square of the hypotenuse. So, in this case, we can use the following equation:

a^2 + b^2 = c^2

where 'a' represents the short leg, 'b' represents the long leg, and 'c' represents the hypotenuse.

Using the given information, we know that the long leg, b, is equal to 1/2x + 11, and the hypotenuse, c, is equal to 2x + 1.

Substituting these values into the equation, we get:

a^2 + (1/2x + 11)^2 = (2x + 1)^2

Now, let's solve this equation step by step:

Expand the squares:
a^2 + (1/4)x^2 + 11^2 + 2 * (1/2x)(11) = 4x^2 + 4x + 1

Simplify and combine like terms:
a^2 + (1/4)x^2 + 121 + (11/x) = 4x^2 + 4x + 1

Multiply the entire equation by 4 to clear the fraction:
4a^2 + x^2 + 484 + (44/x) = 16x^2 + 16x + 4

Rearrange the terms:
15x^2 - 4x - x^2 + 16a^2 - 144 - (44/x) = 0

Combine like terms:
14x^2 - 4x - x^2 + 16a^2 - 144 - (44/x) = 0

Combine the x and x^2 terms:
13x^2 - 4x + 16a^2 - 144 - (44/x) = 0

Now, you can solve this quadratic equation by factoring, completing the square, or using the quadratic formula.