The lengths of the sides of a right-angled triangle are consecutive even integers (in cm). What is the product of these integers?
480
Finding the Product of Consecutive Even Integer Sides of a Right-Angled Triangle
Let's break down this problem step by step to find the lengths of the sides of the right-angled triangle and then calculate their product. We are given that the lengths are consecutive even integers.
Consecutive even integers are even numbers that follow each other, like 2, 4, 6 or 10, 12, 14. They differ by 2.
Setting up the Side Lengths
Let the smallest even integer representing one side be \(x\).
Since the sides are consecutive even integers, the other two sides will be \(x+2\) and \(x+4\).
In a right-angled triangle, the longest side is always the hypotenuse. The longest side among \(x\), \(x+2\), and \(x+4\) is \(x+4\).
Applying the Pythagorean Theorem
The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (legs). Mathematically, this is expressed as \(a^2 + b^2 = c^2\), where \(a\) and \(b\) are the lengths of the legs and \(c\) is the length of the hypotenuse.
In our case, the legs are \(x\) and \(x+2\), and the hypotenuse is \(x+4\). So, we can write the equation:
\(x^2 + (x+2)^2 = (x+4)^2\)
Solving the Equation for x
Now, let's expand and solve the equation:
\(x^2 + (x^2 + 4x + 4) = (x^2 + 8x + 16)\)
Combine like terms on the left side:
\(2x^2 + 4x + 4 = x^2 + 8x + 16\)
Move all terms to one side to form a quadratic equation:
\(2x^2 - x^2 + 4x - 8x + 4 - 16 = 0\)
\(x^2 - 4x - 12 = 0\)
Now, we need to solve this quadratic equation. We can factor it:
\((x - 6)(x + 2) = 0\)
This gives us two possible values for \(x\):
- \(x - 6 = 0 \Rightarrow x = 6\)
- \(x + 2 = 0 \Rightarrow x = -2\)
Since the length of a side of a triangle must be a positive value, we discard \(x = -2\).
Thus, the value of \(x\) is 6.
Determining the Side Lengths
Using \(x=6\), the lengths of the sides are:
- First side: \(x = 6\) cm
- Second side: \(x+2 = 6+2 = 8\) cm
- Third side: \(x+4 = 6+4 = 10\) cm
Let's check if these lengths are consecutive even integers: 6, 8, 10. Yes, they are.
Let's also check if they form a right-angled triangle using the Pythagorean theorem:
\(6^2 + 8^2 = 36 + 64 = 100\)
\(10^2 = 100\)
Since \(6^2 + 8^2 = 10^2\), the triangle with sides 6 cm, 8 cm, and 10 cm is indeed a right-angled triangle.
Calculating the Product of the Integers
The question asks for the product of these three integers (the side lengths).
Product = \(6 \times 8 \times 10\)
Product = \(48 \times 10\)
Product = \(480\)
Summary of Steps
We assumed the sides were consecutive even integers \(x, x+2, x+4\). Using the Pythagorean theorem, we set up the equation \(x^2 + (x+2)^2 = (x+4)^2\). Solving this quadratic equation, we found the valid side lengths to be 6, 8, and 10 cm. Finally, we calculated the product of these lengths.
| Step | Description | Result |
|---|---|---|
| 1 | Represent sides as consecutive even integers | \(x, x+2, x+4\) |
| 2 | Apply Pythagorean Theorem | \(x^2 + (x+2)^2 = (x+4)^2\) |
| 3 | Solve the equation for \(x\) | \(x=6\) |
| 4 | Find side lengths | 6, 8, 10 cm |
| 5 | Calculate the product | \(6 \times 8 \times 10\) |
| 6 | Final Product | 480 |
Revision Table: Right Triangle Side Product
| Concept | Key Idea | Formula/Example |
|---|---|---|
| Consecutive Even Integers | Even numbers in sequence, differ by 2. | \(n, n+2, n+4, ...\) |
| Right-Angled Triangle | A triangle with one angle measuring 90 degrees. | |
| Hypotenuse | The side opposite the right angle; the longest side. | In \(a^2+b^2=c^2\), \(c\) is the hypotenuse. |
| Pythagorean Theorem | Relationship between the sides of a right-angled triangle. | \(a^2 + b^2 = c^2\) |
| Solving Quadratic Equations | Finding the values of the variable that satisfy the equation. | Factoring, quadratic formula. |
Additional Information: Pythagorean Triples
The set of integers (6, 8, 10) is an example of a Pythagorean triple. A Pythagorean triple is a set of three positive integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\). Common Pythagorean triples include (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25).
Our triple (6, 8, 10) is actually a multiple of the basic (3, 4, 5) triple, specifically \(2 \times (3, 4, 5)\). Multiplying a Pythagorean triple by any positive integer results in another Pythagorean triple.
This problem cleverly uses the properties of both consecutive integers and right-angled triangles, solved using the fundamental Pythagorean theorem.
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