Hi.
1. I have a right triangle here. I know that I am to use the Pythagorean Theorem: a^2 + b^2 = c^2.
In this case, side a is (x + 4) and b is x. The hypotenuse is 4 + (x + 4)
So here we have the equation: (x + 4)^2 + x^2 = 4 + (x + 4)^2.
I need to find the length of a (the short leg), the length of a, and the length of the hypotenuse. I have tried several times to work out this equation with no luck.
2. I have another right triangle here. I know that I am to use the Pythagorean Theorem: a^2 + b^2 = c^2.
In this case, side a is x and b is 1/2x + 11. The hypotenuse is 2x + 1. I need to find the length of a (the short leg), the length of a, and the length of the hypotenuse.
x^2 + (1/2x + 11)^2 = (2x + 1)^2
Thank you!
# 2: * I need to find the length of a (the short leg), the length of b, and the length of the hypotenuse.
just curious what grade is this math
This is Algebra. Haha.. You can take algebra between 8th and 11th grade. Something like that.. Not really a grade thing.. Just a subject. Why do you ask?
1. To solve the equation (x + 4)^2 + x^2 = 4 + (x + 4)^2, we can simplify it step by step:
Step 1: Expand the squares:
(x^2 + 8x + 16) + x^2 = 4 + (x^2 + 8x + 16)
Step 2: Combine like terms on both sides:
2x^2 + 8x + 16 = 4 + x^2 + 8x + 16
Step 3: Simplify:
2x^2 + 8x + 16 = x^2 + 8x + 20
Step 4: Move all terms to one side:
2x^2 + 8x + 16 - (x^2 + 8x + 20) = 0
Step 5: Combine like terms:
x^2 - 4 = 0
Step 6: Solve for x:
x^2 = 4
Taking the square root of both sides:
x = ±2
Now, we can substitute the value of x back into the given expressions to find the lengths.
For side a (the short leg):
a = x = ±2
For side b:
b = x = ±2
For the hypotenuse:
c = 4 + (x + 4) = 4 + (±2 + 4) = 10 (when x = 2) or 6 (when x = -2)
Therefore, the lengths of side a are 2 and -2, the lengths of side b are 2 and -2, and the lengths of the hypotenuse are 10 and 6.
2. To solve the equation x^2 + (1/2x + 11)^2 = (2x + 1)^2, we can follow a similar process:
Step 1: Expand the squares:
x^2 + (1/4x^2 + 11x + 121) = 4x^2 + 4x + 1
Step 2: Combine like terms on both sides:
x^2 + 1/4x^2 + 11x + 121 = 4x^2 + 4x + 1
Step 3: Simplify:
5/4x^2 - 11x + 120 = 4x^2 + 4x + 1
Step 4: Move all terms to one side:
5/4x^2 - 4x^2 - 11x + 4x + 120 - 1 = 0
Step 5: Combine like terms:
1/4x^2 - 7x + 119 = 0
Step 6: This quadratic equation cannot be easily factored, so we can use the quadratic formula to solve it:
x = [-(-7) ± √((-7)^2 - 4(1/4)(119))] / (2(1/4))
x = (7 ± √(49 - 119/4)) / (1/2)
x = (7 ± √(196 - 119)) / (1/2)
x = (7 ± √77) / (1/2)
x = 14 ± 2√77
Now, we can substitute the values of x back into the given expressions to find the lengths.
For side a:
a = x = 14 ± 2√77
For side b:
b = 1/2x + 11 = (1/2)(14 ± 2√77) + 11 = 7 ± √77 + 11 = 18 ± √77
For the hypotenuse:
c = 2x + 1 = 2(14 ± 2√77) + 1 = 28 ± 4√77 + 1 = 29 ± 4√77
Therefore, the lengths of side a are 14 + 2√77, 14 - 2√77, the lengths of side b are 18 + √77, 18 - √77, and the lengths of the hypotenuse are 29 + 4√77, 29 - 4√77.