The measures (in cm) of sides of a right angled triangle are given by consecutive integers. Its area (in `cm^(2)` ) is

Options

A. 9
B. 8
C. 5
D. 6

Answer

To solve the problem of finding the area of a right-angled triangle with sides that are consecutive integers, we can follow these steps:

### Step 1: Identify the sides of the triangle
Since we are looking for a right-angled triangle with sides that are consecutive integers, we can denote the sides as \(a\), \(a+1\), and \(a+2\). However, we need to ensure that they satisfy the Pythagorean theorem, which states that for a right-angled triangle with sides \(a\), \(b\), and \(c\) (where \(c\) is the hypotenuse), the following holds true:
\[ a^2 + b^2 = c^2 \]

### Step 2: Check possible sets of consecutive integers
The only set of consecutive integers that can form a right-angled triangle is \(3\), \(4\), and \(5\). We can verify this:
- Let \(a = 3\), \(b = 4\), and \(c = 5\).
- Check the Pythagorean theorem:
\[ 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \]
This confirms that \(3\), \(4\), and \(5\) can indeed form a right-angled triangle.

### Step 3: Calculate the area of the triangle
The area \(A\) of a right-angled triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case, we can take \(3\) as the base and \(4\) as the height:
\[ A = \frac{1}{2} \times 3 \times 4 \]
Calculating this gives:
\[ A = \frac{1}{2} \times 12 = 6 \text{ cm}^2 \]

### Final Answer
Thus, the area of the right-angled triangle with sides that are consecutive integers is \(6 \text{ cm}^2\).

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