The perimeter and the area of a right-angled triangle are 36 cm and 54 square cm respectively. What is the length of the hypotenuse?
15 cm
Finding the Hypotenuse of a Right Triangle Given Perimeter and Area
We are given a right-angled triangle with a perimeter of 36 cm and an area of 54 square cm. Our goal is to determine the length of the hypotenuse of this right triangle.
Let the lengths of the two perpendicular sides (legs) of the right-angled triangle be \(a\) and \(b\), and let the length of the hypotenuse be \(c\).
Using the Given Information
We can write down the given information in terms of equations:
- Perimeter: The sum of the lengths of all sides is 36 cm. \[a + b + c = 36 \quad (1)\]
- Area: The area of a right-angled triangle is half the product of its perpendicular sides. \[\frac{1}{2}ab = 54\] Multiplying both sides by 2, we get: \[ab = 108 \quad (2)\]
- Pythagorean Theorem: For any right-angled triangle, the square of the hypotenuse (\(c\)) is equal to the sum of the squares of the other two sides (\(a\) and \(b\)). \[a^2 + b^2 = c^2 \quad (3)\]
Solving for the Hypotenuse
We need to find the value of \(c\). Let's use the equations we have.
From equation (1), we can express \(a + b\) in terms of \(c\):
\[a + b = 36 - c \quad (4)\]Now, consider the algebraic identity \((a+b)^2 = a^2 + b^2 + 2ab\).
Substitute the values from equations (2), (3), and (4) into this identity:
- Substitute \((a+b)\) from (4): \((36 - c)^2\)
- Substitute \((a^2 + b^2)\) from (3): \(c^2\)
- Substitute \(ab\) from (2): \(108\)
So the identity becomes:
\[(36 - c)^2 = c^2 + 2(108)\]Now, let's expand and simplify the equation:
\[36^2 - 2(36)c + c^2 = c^2 + 216\] \[1296 - 72c + c^2 = c^2 + 216\]Subtract \(c^2\) from both sides:
\[1296 - 72c = 216\]Now, we solve for \(c\). Subtract 216 from both sides:
\[1296 - 216 = 72c\] \[1080 = 72c\]Divide both sides by 72:
\[c = \frac{1080}{72}\]To simplify the division, we can divide both numbers by common factors, for example, 12:
\[c = \frac{1080 \div 12}{72 \div 12} = \frac{90}{6}\]Now, divide 90 by 6:
\[c = 15\]So, the length of the hypotenuse of the right-angled triangle is 15 cm.
Verification (Optional)
We found \(c=15\). From \(a+b = 36-c\), we get \(a+b = 36-15 = 21\). We also know \(ab=108\). We need two numbers that add up to 21 and multiply to 108. These numbers are 9 and 12. Let's assume \(a=9\) cm and \(b=12\) cm.
- Perimeter: \(a+b+c = 9+12+15 = 36\) cm. (Matches the given perimeter)
- Area: \(\frac{1}{2}ab = \frac{1}{2}(9)(12) = \frac{1}{2}(108) = 54\) square cm. (Matches the given area)
- Pythagorean Theorem: \(a^2 + b^2 = 9^2 + 12^2 = 81 + 144 = 225\). And \(c^2 = 15^2 = 225\). Since \(a^2 + b^2 = c^2\), the sides form a right triangle. (Matches the triangle type)
The verification confirms that our calculated hypotenuse length is correct.
| Property | Formula (Right Triangle) | Given Value | Derived Value |
|---|---|---|---|
| Perimeter | \(a+b+c\) | 36 cm | 9 + 12 + 15 = 36 cm |
| Area | \(\frac{1}{2}ab\) | 54 sq cm | \(\frac{1}{2}(9)(12) = 54\) sq cm |
| Hypotenuse (\(c\)) | \(\sqrt{a^2 + b^2}\) | ? | 15 cm |
| Legs (\(a, b\)) | - | - | 9 cm, 12 cm |
Therefore, the length of the hypotenuse is 15 cm.
Revision Table: Key Formulas for Right Triangles
| Formula | Description | Notes |
|---|---|---|
| Pythagorean Theorem: \(a^2 + b^2 = c^2\) | Relates the lengths of the legs (\(a, b\)) to the hypotenuse (\(c\)). | Only applies to right-angled triangles. |
| Area: \(\frac{1}{2} \times \text{base} \times \text{height}\) | For a right triangle, the legs can be the base and height. | Area = \(\frac{1}{2}ab\). |
| Perimeter: Sum of sides | Total length of the boundary. | Perimeter = \(a + b + c\). |
Additional Information on Solving Triangle Problems
Problems involving the perimeter and area of a right triangle often require using a combination of algebraic methods and geometric formulas like the Pythagorean theorem. A common strategy is to express sums and products of the leg lengths in terms of the perimeter and hypotenuse, and then use algebraic identities to form an equation involving only the hypotenuse.
- Recognizing Pythagorean Triples: In this problem, the side lengths (9, 12, 15) form a Pythagorean triple, specifically a multiple of the basic (3, 4, 5) triple (3x3=9, 3x4=12, 3x5=15). Recognizing common triples can sometimes speed up verification.
- Alternative Methods: One could also solve the system of equations \(a+b=21\) and \(ab=108\) directly for \(a\) and \(b\) first (e.g., using substitution or by forming a quadratic equation \(x^2 - (a+b)x + ab = 0\)) and then use \(a\) and \(b\) to find \(c\) via the Pythagorean theorem. Both methods lead to the same result.
- Units: Always pay attention to the units. Perimeter is in cm, area in sq cm, and side lengths in cm.
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