Answer:
8 inches and 6 inches
Step-by-step explanation:
The hypotenuse length of 10 is 2 times 5, so there is a possibility that the triangle is a 3-4-5 triangle scaled by a factor of 2. The leg length difference of 2 confirms this. The right triangle of interest has sides of 6 inches, 8 inches, and 10 inches, double the sides of a 3-4-5 triangle.
Since "one leg" is 2 inches longer, it must be the 8-inch leg. The "second leg" must be the 6-inch leg.
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The 3-4-5 right triangle shows up in many algebra and geometry problems, so is useful to remember.
If you want to solve this algebraically, you can assign x as the length of "one leg" and (x -2) as the length of "the second leg." Then the Pythagorean theorem tells you ...
10² = x² +(x -2)²
100 = 2x² -4x +4
x² -2x -48 = 0 . . . . . subtract 100, divide by 2
(x -8)(x +6) = 0 . . . . .factor
x = 8 . . . . . . . . . . . . . the positive solution; the only one of interest
One leg is 8 inches; the second leg is 6 inches.
6 inches and 8 inches.
Given:
Question:
What are the lengths of the legs of the triangle?
The Process:
We will solve a problem regarding right triangles. In the process, there is a relationship between the Pythagorean Theorem with Quadratic Equations.
Let x as the length of one leg of a right triangle. Thus, the length of the second leg is (x - 2).
All length units were expressed in inches.
Let us find out the lengths of the legs of the triangle.
The Phytagorean Theorem: [tex]\boxed{ \ (leg)^2 + (leg)^2 = (hypotenuse)^2 \ }[/tex]
[tex]\boxed{ \ (x-2)^2 + (x)^2 = (10)^2 \ }[/tex]
Remember this, [tex]\boxed{\boxed{ \ (a - b)^2 = a^2 - 2ab + b^2\ } \ \boxed{ \ (a +b)^2 = a^2 + 2ab + b^2\ }}[/tex]
[tex]\boxed{ \ x^2 - 4x + 4 + x^2 = 100 \ }[/tex]
[tex]\boxed{ \ 2x^2 - 4x + 4 - 100= 0 \ }[/tex]
[tex]\boxed{ \ 2x^2 - 4x - 96= 0 \ }[/tex]
Divide by two on both sides.
[tex]\boxed{ \ x^2 - 2x - 48= 0 \ }[/tex]
[tex]\boxed{ \ (x - 8)(x + 6) = 0 \ }[/tex]
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Notes
Of course, we agree with the use of the right intuition, which is to remember Pythagorean Triples, i.e., [tex]\boxed{ \ 3-4-5 \ } \rightarrow \boxed{ \ 6-8-10 \ }[/tex] This method is very convenient when working on multiple-choice tests in a limited amount of time.
