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Solving for the Perimeter of a Right Triangle: Hypotenuse and Height Differences
Solving for the Perimeter of a Right Triangle: Hypotenuse and Height Differences
In mathematics, right triangles are fundamental to understanding various geometric and algebraic principles. One common problem involves finding the perimeter of a right triangle when given the relationships between the base, height, and hypotenuse. In this article, we'll explore how to solve such a problem step-by-step.
Problem Statement
Consider a right triangle where the hypotenuse is 1 cm longer than the base, and the height is 7 cm shorter than the base. We need to determine the perimeter of this triangle.
Step-by-Step Solution
Let's denote the base of the triangle as ( b ) and the height as ( a ). The hypotenuse, denoted as ( h ), can be expressed as:
( h b 1 ) ( a b - 7 )According to the Pythagorean theorem, the relationship between the sides of a right triangle is given by:
Equation from Pythagorean Theorem
( h^2 a^2 b^2 )
Substitute the given expressions for ( h ) and ( a ) in the equation:
Substitution
( (b 1)^2 (b - 7)^2 b^2 )
Expand and simplify:
Expansion and Simplification
( b^2 2b 1 b^2 - 14b 49 b^2 )
( b^2 2b 1 2b^2 - 14b 49 )
( b^2 - 16b 48 0 )
Solving the Quadratic Equation
To solve the quadratic equation ( b^2 - 16b 48 0 ), we can use the quadratic formula:
Quadratic Formula
( b frac{-(-16) pm sqrt{(-16)^2 - 4 cdot 1 cdot 48}}{2 cdot 1} )
( b frac{16 pm sqrt{256 - 192}}{2} )
( b frac{16 pm sqrt{64}}{2} )
( b frac{16 pm 8}{2} )
( b 12 ) or ( b 4 )
Since the height ( a b - 7 ) and ( a ) cannot be negative, we discard ( b 4 ):
Valid Solution
( b 12 )
Substitute ( b 12 ) back into the expressions for ( h ) and ( a ):
( h b 1 12 1 13 ) ( a b - 7 12 - 7 5 )Now, we calculate the perimeter of the triangle:
Perimeter Calculation
( text{Perimeter} b a h 12 5 13 30 ) cm
Conclusion
The perimeter of the right triangle, where the hypotenuse is 1 cm longer than the base and the height is 7 cm shorter than the base, is (boxed{30}) cm.