The hypotenuse of a right triangle is 11.0, and one leg is 4.0 units shorter than the other.

The hypotenuse of a right triangle is 11.0, and one leg is 4.0 units shorter than the other. Find the dimensions of the figure (rounded to the nearest tenth).

Legs: x and x+4
Hypotenuse: 11

x² + (x+4)² = 11²
x² + x² + 8x + 16 = 121
2x² + 8x - 105 = 0

x = [-8 ± √(64+840)] / 4
x = (-8 ± 2√226) / 4
x = -2 ± 1/2 √226

Since x is length of leg of triangle, then x > 0
x = -2 + 1/2 √226
x + 4 = 2 + 1/2 √226

Legs are:
-2 + 1/2 √226 ≈ 5.516648189 ≈ 5.5
2 + 1/2 √226 ≈ 9.516648189 ≈ 9.5

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Check:

√(5.5² + 9.5²) = √120.5 ≈ 10.9772492 ≈ 11

ok

One leg 4 units shorter than other
means

let x = 1 leg
then
x-4 equals the second leg
and solve

x² + (x-4)² = 11²

x² + x² - 8x + 16 = 121

I'll allow you to solve this yourself

a^2 +b^2 =c^2; where a and b are the legs and c is the hypotenuse . Plug in a calculate.

a^2 + (a-4)^2 = 11^2

Solve for a.

b is 4 units shorter than a, as stated.