If the perimeter of a right-angled triangle is 30 cm and the hypotenuse is 13 cm, then what is the area of the triangle?
30 cm 2
Solving the Right-Angled Triangle Area Problem
This problem asks us to find the area of a right-angled triangle given its perimeter and the length of its hypotenuse. Let's break it down using the properties of right triangles.
Understanding the Terms
- Right-angled triangle: A triangle with one angle exactly 90 degrees.
- Perimeter: The total distance around the triangle, which is the sum of the lengths of all three sides.
- Hypotenuse: The side opposite the right angle in a right-angled triangle. It is always the longest side.
- Area: The amount of space inside the triangle. For a right-angled triangle, the area is half the product of the lengths of the two shorter sides (the legs or base and height).
Setting up the Problem
Let the lengths of the two shorter sides (the legs) of the right-angled triangle be \(a\) and \(b\), and the length of the hypotenuse be \(c\).
We are given:
- Perimeter \(P = 30\) cm
- Hypotenuse \(c = 13\) cm
The perimeter is the sum of the sides: \(P = a + b + c\). Substituting the given values:
\(30 = a + b + 13\)
Subtracting 13 from both sides, we get the sum of the two shorter sides:
\(a + b = 30 - 13\)
\(a + b = 17\) cm
Using the Pythagorean Theorem
In a right-angled triangle, the Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides:
\(a^2 + b^2 = c^2\)
We know \(c = 13\), so:
\(a^2 + b^2 = 13^2\)
\(a^2 + b^2 = 169\)
Finding the Product of the Sides (ab)
We have two relationships involving \(a\) and \(b\):
- \(a + b = 17\)
- \(a^2 + b^2 = 169\)
We need to find the area, which is given by the formula: Area \(A = \frac{1}{2} \times a \times b\). This means we need to find the value of the product \(ab\).
Consider the equation \(a + b = 17\). If we square both sides of this equation, we get:
\((a + b)^2 = 17^2\)
Expanding the left side \((a + b)^2\), we use the algebraic identity \((x+y)^2 = x^2 + 2xy + y^2\):
\(a^2 + 2ab + b^2 = 289\)
Now, notice that we have \(a^2 + b^2\) in this equation. We know from the Pythagorean theorem that \(a^2 + b^2 = 169\). We can substitute this value into the expanded equation:
\(169 + 2ab = 289\)
Now, we can solve for \(2ab\):
\(2ab = 289 - 169\)
\(2ab = 120\)
To find \(ab\), divide both sides by 2:
\(ab = \frac{120}{2}\)
\(ab = 60\)
Calculating the Area of the Triangle
The area of a right-angled triangle is given by \(\frac{1}{2} \times \text{base} \times \text{height}\), which is \(\frac{1}{2} \times a \times b\).
We just found that \(ab = 60\). Substitute this value into the area formula:
Area \(A = \frac{1}{2} \times 60\)
Area \(A = 30\)
Since the sides are in centimeters, the area is in square centimeters (cm\(^2\)).
The area of the triangle is 30 cm\(^2\).
Verification (Optional but helpful)
We found \(a+b=17\) and \(ab=60\). We can try to find \(a\) and \(b\). These values are the roots of the quadratic equation \(x^2 - (a+b)x + ab = 0\), which is \(x^2 - 17x + 60 = 0\). Factoring this equation, we look for two numbers that multiply to 60 and add up to 17. These numbers are 5 and 12.
So, the sides \(a\) and \(b\) are 5 cm and 12 cm (in any order). Let's check if this forms a valid right triangle with a hypotenuse of 13 cm:
Pythagorean theorem: \(5^2 + 12^2 = 25 + 144 = 169\). \(13^2 = 169\). \(169 = 169\). The sides 5 cm, 12 cm, and 13 cm form a valid right triangle.
Perimeter check: \(5 + 12 + 13 = 17 + 13 = 30\) cm. This matches the given perimeter.
Area check: \(\frac{1}{2} \times 5 \times 12 = \frac{1}{2} \times 60 = 30\) cm\(^2\). This matches our calculated area.
The dimensions of the right-angled triangle are 5 cm, 12 cm, and 13 cm. This is a common Pythagorean triple (5-12-13).
| Given | Formula Used | Result |
|---|---|---|
| Perimeter = 30 cm, Hypotenuse = 13 cm | Perimeter = \(a + b + c\) | \(a + b = 17\) cm |
| Sides \(a, b\), Hypotenuse \(c=13\) | Pythagorean Theorem: \(a^2 + b^2 = c^2\) | \(a^2 + b^2 = 169\) |
| \(a + b = 17\), \(a^2 + b^2 = 169\) | \((a+b)^2 = a^2 + 2ab + b^2\) | \(2ab = (a+b)^2 - (a^2+b^2) = 17^2 - 169 = 289 - 169 = 120\) |
| Product of legs \(ab = 60\) | Area of right triangle = \(\frac{1}{2} ab\) | Area = \(\frac{1}{2} \times 60 = 30\) cm\(^2\) |
Revision Table: Key Formulas for Right Triangles
| Concept | Formula | Notes |
|---|---|---|
| Perimeter (P) | \(P = a + b + c\) | Sum of all three sides (legs \(a, b\), hypotenuse \(c\)) |
| Pythagorean Theorem | \(a^2 + b^2 = c^2\) | Relates the legs (\(a, b\)) to the hypotenuse (\(c\)) |
| Area (A) | \(A = \frac{1}{2} \times a \times b\) | Half the product of the two legs (base and height) |
| Algebraic Identity | \((a+b)^2 = a^2 + 2ab + b^2\) | Useful for problems involving sum and sum of squares |
Additional Information: Triangle Properties
Understanding right-angled triangles and their properties is fundamental in geometry. Here are a few related points:
- Pythagorean Triples: Sets of three integers \(a, b, c\) that satisfy \(a^2 + b^2 = c^2\), forming the sides of a right triangle. Common examples include (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25). Recognizing these can sometimes speed up solving problems. In our case, the sides 5, 12, and 13 form a well-known triple.
- Other Area Formulas: While \(\frac{1}{2} \times \text{base} \times \text{height}\) is standard, the area of any triangle can also be found using Heron's formula if all three side lengths are known: \(A = \sqrt{s(s-a)(s-b)(s-c)}\), where \(s\) is the semi-perimeter (\(s = (a+b+c)/2\)). However, for a right triangle, the base and height are simply the two legs, making the \(\frac{1}{2} ab\) formula more direct if \(a\) and \(b\) are known or can be found.
- Geometric Interpretation: The identity \((a+b)^2 = a^2 + 2ab + b^2\) can be visualized geometrically as a square with side length \((a+b)\) divided into two squares (sides \(a\) and \(b\)) and two rectangles (sides \(a\) and \(b\)). This helps understand the relationship between \(a+b\), \(a^2+b^2\), and \(ab\).
This problem effectively combined the concepts of perimeter, the Pythagorean theorem, and basic algebra to find the area of the right-angled triangle.
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