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The perimeter of an isosceles triangle i...

The perimeter of an isosceles triangle is 32 cm. The ratio of one of the equal sides to its base is 3:2. Find the area of the triangle.

A

`48cm^(2)`

B

`28 sqrt3 cm^(2)`

C

`32 sqrt2 cm^(2)`

D

`44cm^(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the area of the isosceles triangle given the perimeter and the ratio of its sides, we can follow these steps: ### Step 1: Define the sides of the triangle Let the equal sides of the isosceles triangle be represented as \(3R\) each, and the base as \(2R\). ### Step 2: Set up the perimeter equation The perimeter of the triangle is given as 32 cm. Therefore, we can write the equation: \[ 3R + 3R + 2R = 32 \] This simplifies to: \[ 8R = 32 \] ### Step 3: Solve for \(R\) Now, divide both sides by 8 to find \(R\): \[ R = \frac{32}{8} = 4 \] ### Step 4: Calculate the lengths of the sides Now that we have \(R\), we can find the lengths of the sides: - Equal sides: \(3R = 3 \times 4 = 12\) cm - Base: \(2R = 2 \times 4 = 8\) cm ### Step 5: Calculate the semi-perimeter The semi-perimeter \(s\) is half of the perimeter: \[ s = \frac{32}{2} = 16 \text{ cm} \] ### Step 6: Apply Heron's formula to find the area Heron's formula for the area \(A\) of a triangle is given by: \[ A = \sqrt{s(s-a)(s-b)(s-c)} \] where \(a\), \(b\), and \(c\) are the sides of the triangle. Here, \(a = 12\), \(b = 12\), and \(c = 8\). Substituting the values into Heron's formula: \[ A = \sqrt{16(16-12)(16-12)(16-8)} \] Calculating each term: - \(s - a = 16 - 12 = 4\) - \(s - b = 16 - 12 = 4\) - \(s - c = 16 - 8 = 8\) Now substituting these values: \[ A = \sqrt{16 \times 4 \times 4 \times 8} \] ### Step 7: Simplify the expression Calculating the product inside the square root: \[ A = \sqrt{16 \times 4 \times 4 \times 8} = \sqrt{2048} \] ### Step 8: Further simplify \(\sqrt{2048}\) We can factor \(2048\) into \(1024 \times 2\): \[ A = \sqrt{1024 \times 2} = \sqrt{1024} \times \sqrt{2} = 32\sqrt{2} \text{ cm}^2 \] ### Final Answer Thus, the area of the isosceles triangle is: \[ \text{Area} = 32\sqrt{2} \text{ cm}^2 \]
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