Finding the Area of a Triangle Given Side Lengths

The problem asks us to find the area of a triangle with specific side lengths involving variables \(a\) and \(b\). The given side lengths are:

• Side 1: \(\sqrt{a^2+b^2}\)
• Side 2: \(\sqrt{(2a)^2+b^2} = \sqrt{4a^2+b^2}\)
• Side 3: \(\sqrt{a^2+(2b)^2} = \sqrt{a^2+4b^2}\)

We can approach this problem by trying to represent the triangle using coordinate geometry. Let's consider if these side lengths correspond to distances between simple points.

Observe the structure of the side lengths. They resemble the distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \). Let's try placing one vertex at the origin \((0,0)\).

Consider the vertices of the triangle to be \(A(0,0)\), \(B(a,b)\), and \(C(-a, 2b)\). Let's calculate the distances between these points:

• Distance AB: \( \sqrt{(a-0)^2 + (b-0)^2} = \sqrt{a^2+b^2} \)
• Distance AC: \( \sqrt{(-a-0)^2 + (2b-0)^2} = \sqrt{(-a)^2 + (2b)^2} = \sqrt{a^2+4b^2} \)
• Distance BC: \( \sqrt{(-a-a)^2 + (2b-b)^2} = \sqrt{(-2a)^2 + (b)^2} = \sqrt{4a^2+b^2} \)

These calculated distances match the given side lengths of the triangle. Therefore, the triangle can be represented by the vertices \(A(0,0)\), \(B(a,b)\), and \(C(-a, 2b)\).

Calculating Triangle Area Using Coordinates

The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the determinant formula:

Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)

Substituting the coordinates \(A(0,0)\), \(B(a,b)\), and \(C(-a, 2b)\):

• \( (x_1, y_1) = (0,0) \)
• \( (x_2, y_2) = (a,b) \)
• \( (x_3, y_3) = (-a, 2b) \)

Area \( = \frac{1}{2} |0(b - 2b) + a(2b - 0) + (-a)(0 - b)| \)

Simplify the expression inside the absolute value:

Area \( = \frac{1}{2} |0(-b) + a(2b) + (-a)(-b)| \)

Area \( = \frac{1}{2} |0 + 2ab + ab| \)

Area \( = \frac{1}{2} |3ab| \)

Since area must be positive, and assuming \(a\) and \(b\) are real numbers related to lengths, \(ab\) can be positive or negative, but the area magnitude is \(|3ab|\). If \(a\) and \(b\) are dimensions, they might be positive, leading to area \( = \frac{3}{2}ab \). Even if \(a\) or \(b\) were negative in the coordinate system, the absolute value ensures a positive area.

The area of the triangle is \(\frac{3}{2}ab\) square units.

Checking Options

Let's compare our calculated area with the given options:

Our calculated area, \(\frac{3}{2}ab\), matches Option 4.

Conclusion

By representing the triangle using coordinate geometry with vertices at \( (0,0) \), \( (a,b) \), and \( (-a, 2b) \), we successfully matched the given side lengths. Using the coordinate area formula, we calculated the area of the triangle to be \(\frac{3}{2}ab\) square units.

Revision Table: Key Concepts

Additional Information: Alternative Methods for Triangle Area

While coordinate geometry worked well here due to the specific form of the side lengths, other methods exist for finding triangle area:

• Heron's Formula: If the side lengths \(s_1, s_2, s_3\) are known, calculate the semi-perimeter \(S = \frac{s_1+s_2+s_3}{2}\). The area is then \( \sqrt{S(S-s_1)(S-s_2)(S-s_3)} \). This would be algebraically complex with the given side lengths.
• Base and Height: Area \( = \frac{1}{2} \times \text{base} \times \text{height} \). Finding the height relative to a chosen base can be challenging if the triangle isn't a right triangle and vertices/equations of lines aren't easily determined.
• Trigonometry: Area \( = \frac{1}{2} s_1 s_2 \sin(\theta) \), where \( \theta \) is the angle between sides \(s_1\) and \(s_2\). Finding an angle would require using the Law of Cosines with the given side lengths, which would also be algebraically intensive.

The coordinate geometry approach proved efficient because the structure of the side lengths conveniently suggested a simple set of coordinates for the vertices.