Solving the Area of a Right-Angled Triangle Given Its Perimeter and Hypotenuse

When faced with a mathematical problem involving a right-angled triangle and given constraints such as the perimeter and hypotenuse, it is essential to harness the properties of such triangles and apply appropriate formulas. This article delves into a step-by-step resolution of the example: If the perimeter of a right-angled triangle is given to be 60 cm and the length of the hypotenuse is 24 cm, what is the area of the triangle? This problem will be solved using the principles of geometry and algebra.

Understanding the Problem

In a right-angled triangle, the sides are typically denoted as a and b (the legs) and c (the hypotenuse). The given constraints are as follows:

The first step is to express the sum of the sides in terms of the given perimeter:

Step 1: Expressing the Sum of the Sides

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

P a b c 60 cm

Given the hypotenuse c 24 cm, we can substitute this value into the equation:

a b 24 60 cm

Simplifying, we find:

a b 60 - 24 36 cm

Step 2: Setting Up the Equations

Next, we utilize the Pythagorean theorem a2 b2 c2. Given that c 24 cm, we have:

a2 b2 242 576 cm2

Now, we can express one variable in terms of the other using the equation derived from the perimeter:

b 36 - a

Substituting this into the Pythagorean theorem, we get:

a2 (36 - a)2 576

Expanding and simplifying:

a2 1296 - 72a a2 576

2a2 - 72a 1296 576

2a2 - 72a 720 0

a2 - 36a 360 0

Using the quadratic formula:

a (-b ± sqrt{b2 - 4ac}) / (2a)

Here, a 1, b -36, and c 360 (note that the quadratic formula uses negative b in this case):

a (36 ± sqrt{(-36)2 - 4(1)(360)}) / (2(1))

a (36 ± sqrt{1296 - 1440}) / 2

a (36 ± sqrt{-144}) / 2

The discriminant is negative, indicating no real solutions, which suggests an inconsistency in the problem setup.

Conclusion

The inconsistency in the problem setup implies that with the given values for the perimeter and hypotenuse, a right-angled triangle cannot be formed. This is because the quadratic equation derived from the given constraints does not yield any real solutions. Therefore, it is crucial to recheck the provided values for the perimeter and hypotenuse to ensure they are consistent with the formation of a valid right-angled triangle.

Additionally, it is worth noting that for a given hypotenuse, there are infinitely many pairs of legs (a and b) that can form a right-angled triangle, each leading to a different area. A simple example is when the area is minimized or maximized, but these specific values would require independent verification and calculation.