Maximizing the Perimeter of a Right Triangle with a Fixed Hypotenuse

Understanding how to find the perimeter of a right triangle that maximizes its area given a fixed hypotenuse is a classic problem in geometry. This article will delve into the mathematical details to solve this problem in a step-by-step manner using both geometric and calculus-based approaches.

Problem Statement

Given a right triangle with a hypotenuse of 9 meters, the objective is to determine the perimeter when the area of the triangle is maximized.

Geometric Insight: The Isosceles Triangle

A key geometric insight here is that a right triangle with a fixed hypotenuse and maximum area is an isosceles right triangle. In an isosceles right triangle, the two legs are equal, and the relationship between the sides can be derived from the Pythagorean theorem.

Solving the Problem

Let#39;s denote the legs of the triangle as a and b, and the hypotenuse as c.

According to the Pythagorean theorem, we have:

$$a^2 b^2 c^2$$

Given that c 9 meters, we can substitute:

$$a^2 b^2 81$$

For an isosceles right triangle, we have a b. Thus, substituting b a into the equation, we get:

$$2a^2 81$$

Solving for a, we get:

$$a^2 frac{81}{2}$$ $$a b frac{9}{sqrt{2}} frac{9sqrt{2}}{2}$$

Calculating the Perimeter

The perimeter P of the triangle is the sum of all its sides:

$$P a b c$$

Substituting the values, we get:

$$P frac{9sqrt{2}}{2} frac{9sqrt{2}}{2} 9$$ $$P 9sqrt{2} 9$$

Approximating sqrt{2} approx 1.414, we can calculate:

$$P approx 9 times 1.414 9$$ $$P approx 12.726 9$$ $$P approx 21.726 text{ meters}$$

Conclusion

The perimeter of the right triangle when its area is maximized with a hypotenuse of 9 meters is approximately 21.73 meters.

Alternative Approach: Using Calculus

An alternative method involves using calculus to find the maximum area. Here, we express the area A in terms of one variable and then find its derivative to locate the maximum.

The area of the triangle is:

$$A frac{1}{2}ab$$

Substituting b sqrt{81 - a^2}, we get:

$$A frac{1}{2}asqrt{81 - a^2}$$

To find the maximum area, we differentiate A with respect to a and set the derivative to zero:

$$frac{dA}{da} frac{1}{2}left(sqrt{81 - a^2} a(-frac{a}{sqrt{81 - a^2}})right) 0$$

This simplifies to:

$$frac{81 - 2a^2}{sqrt{81 - a^2}} 0$$

Solving this, we get:

$$81 - 2a^2 0$$ $$a^2 frac{81}{2}$$ $$a frac{9sqrt{2}}{2}$$

This confirms the same result as in the geometric approach.

Additional Insights

Problem 2: Among all right triangles inscribed in a semi-circle, the isosceles right triangle has the greatest area.

The reasoning is as follows: The isosceles right triangle with a hypotenuse equal to the diameter of the semi-circle has the maximum height, thus maximizing the area. Given a semi-circle with a diameter of 9 meters, the maximum area can be calculated as follows:

For a right triangle inscribed in a semi-circle with a diameter of 9 meters, the hypotenuse is the diameter, so:

$$a^2 b^2 9^2$$

This is an isosceles right triangle, with a b, so:

$$2a^2 81$$ $$a^2 frac{81}{2}$$ $$a frac{9}{sqrt{2}}$$

The area of the isosceles right triangle is:

$$A frac{1}{2}ab frac{1}{2}left(frac{9}{sqrt{2}}right)^2 frac{81}{4} 20.25 text{ square meters}$$

The perimeter is calculated as:

$$P 9 2 times frac{9}{sqrt{2}} 9 2 times frac{9sqrt{2}}{2} 9 9sqrt{2}$$

Approximating sqrt{2} approx 1.414, we get:

$$P approx 9 9 times 1.414 approx 9 12.726 approx 21.726 text{ meters}$$

This matches our previous calculations, confirming the solution.

Conclusion: The perimeter of an isosceles right triangle inscribed in a semi-circle with a diameter of 9 meters is approximately 21.73 meters.