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To solve the problem of finding the value of [tex]\( x \)[/tex] and the length of the hypotenuse of a right triangle with given sides 10 inches, [tex]\(9x\)[/tex] inches, and [tex]\( (x + 10) \)[/tex] inches, we can use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse.
The legs of the right triangle are:
- [tex]\(10\)[/tex] inches
- [tex]\(9x\)[/tex] inches
The hypotenuse is:
- [tex]\( (x + 10) \)[/tex] inches
According to the Pythagorean theorem:
[tex]\[ \text{(leg 1)}^2 + \text{(leg 2)}^2 = \text{hypotenuse}^2 \][/tex]
Substituting the given lengths:
[tex]\[ 10^2 + (9x)^2 = (x + 10)^2 \][/tex]
Calculating each term:
[tex]\[ 100 + (81x^2) = (x^2 + 20x + 100) \][/tex]
Rearrange the equation to form a standard quadratic equation:
[tex]\[ 81x^2 - x^2 + 100 - 100 - 20x = 0 \][/tex]
[tex]\[ 80x^2 - 20x = 0 \][/tex]
Factor out the common term:
[tex]\[ 20x(4x - 1) = 0 \][/tex]
This gives us two possible solutions:
[tex]\[ 20x = 0 \][/tex]
or
[tex]\[ (4x - 1) = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \][/tex]
or
[tex]\[ 4x = 1 \][/tex]
[tex]\[ x = \frac{1}{4} \][/tex]
Since [tex]\( x \)[/tex] must be a positive number, we discard [tex]\( x = 0 \)[/tex] and keep:
[tex]\[ x = 0.25 \][/tex]
Now that we have [tex]\( x \)[/tex], we can find the length of the hypotenuse:
[tex]\[ \text{Hypotenuse} = x + 10 \][/tex]
[tex]\[ \text{Hypotenuse} = 0.25 + 10 \][/tex]
[tex]\[ \text{Hypotenuse} = 10.25 \][/tex]
Thus, the length of [tex]\( x \)[/tex] is [tex]\( 0.25 \)[/tex] and the hypotenuse measures [tex]\( 10.25 \)[/tex] inches.
The legs of the right triangle are:
- [tex]\(10\)[/tex] inches
- [tex]\(9x\)[/tex] inches
The hypotenuse is:
- [tex]\( (x + 10) \)[/tex] inches
According to the Pythagorean theorem:
[tex]\[ \text{(leg 1)}^2 + \text{(leg 2)}^2 = \text{hypotenuse}^2 \][/tex]
Substituting the given lengths:
[tex]\[ 10^2 + (9x)^2 = (x + 10)^2 \][/tex]
Calculating each term:
[tex]\[ 100 + (81x^2) = (x^2 + 20x + 100) \][/tex]
Rearrange the equation to form a standard quadratic equation:
[tex]\[ 81x^2 - x^2 + 100 - 100 - 20x = 0 \][/tex]
[tex]\[ 80x^2 - 20x = 0 \][/tex]
Factor out the common term:
[tex]\[ 20x(4x - 1) = 0 \][/tex]
This gives us two possible solutions:
[tex]\[ 20x = 0 \][/tex]
or
[tex]\[ (4x - 1) = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \][/tex]
or
[tex]\[ 4x = 1 \][/tex]
[tex]\[ x = \frac{1}{4} \][/tex]
Since [tex]\( x \)[/tex] must be a positive number, we discard [tex]\( x = 0 \)[/tex] and keep:
[tex]\[ x = 0.25 \][/tex]
Now that we have [tex]\( x \)[/tex], we can find the length of the hypotenuse:
[tex]\[ \text{Hypotenuse} = x + 10 \][/tex]
[tex]\[ \text{Hypotenuse} = 0.25 + 10 \][/tex]
[tex]\[ \text{Hypotenuse} = 10.25 \][/tex]
Thus, the length of [tex]\( x \)[/tex] is [tex]\( 0.25 \)[/tex] and the hypotenuse measures [tex]\( 10.25 \)[/tex] inches.
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