Finding the Ratio of Sides of Similar Triangles Using Areas

When two triangles are similar, a special relationship exists between their areas and the lengths of their corresponding sides. This property is fundamental in geometry and is often tested in exams. The property states that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Let the two similar triangles be denoted as Triangle 1 and Triangle 2. Let their areas be \(A_1\) and \(A_2\), respectively. Let their corresponding sides be \(s_1\) and \(s_2\), respectively.

According to the property of similar triangles: \[ \frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2 \]

We are given the areas of the two similar triangles: \(A_1 = (7 - 4\sqrt{3})\) cm\(^2\) \(A_2 = (7 + 4\sqrt{3})\) cm\(^2\)

We need to find the ratio of their corresponding sides, which is \(\frac{s_1}{s_2}\). Using the property, we can write: \[ \left(\frac{s_1}{s_2}\right)^2 = \frac{A_1}{A_2} = \frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}} \]

To find the ratio of the sides \(\frac{s_1}{s_2}\), we need to take the square root of the ratio of the areas: \[ \frac{s_1}{s_2} = \sqrt{\frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}}} \]

To simplify the expression under the square root, we can rationalize the denominator inside the square root by multiplying the numerator and the denominator by the conjugate of the denominator, which is \((7 - 4\sqrt{3})\).

Let's simplify the fraction \(\frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}}\): \[ \frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{(7 - 4\sqrt{3})(7 - 4\sqrt{3})}{(7 + 4\sqrt{3})(7 - 4\sqrt{3})} \]

The denominator is in the form \((a+b)(a-b) = a^2 - b^2\), where \(a=7\) and \(b=4\sqrt{3}\). \[ (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - (16 \times 3) = 49 - 48 = 1 \]

The numerator is in the form \((a-b)^2 = a^2 - 2ab + b^2\), where \(a=7\) and \(b=4\sqrt{3}\). \[ (7 - 4\sqrt{3})^2 = 7^2 - 2(7)(4\sqrt{3}) + (4\sqrt{3})^2 = 49 - 56\sqrt{3} + 48 = 97 - 56\sqrt{3} \]

Wait, there seems to be a calculation error in the numerator expansion. Let's re-check the numerator calculation: \[ (7 - 4\sqrt{3})^2 = 7^2 - 2(7)(4\sqrt{3}) + (4\sqrt{3})^2 \] \[ = 49 - 56\sqrt{3} + (16 \times 3) \] \[ = 49 - 56\sqrt{3} + 48 \] \[ = 97 - 56\sqrt{3} \]

This result \(97 - 56\sqrt{3}\) for the numerator over the denominator \(1\) gives \(\frac{s_1}{s_2} = \sqrt{97 - 56\sqrt{3}}\). While this is mathematically correct, it doesn't match the simple form of the options. Let's re-examine the structure of the problem and options. The options are simple linear expressions involving \(\sqrt{3}\). This suggests that the expression inside the square root must be a perfect square of a form like \((a - b\sqrt{3})^2\).

Let's look at the fraction again: \(\frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}}\). The numerator is \((7 - 4\sqrt{3})\) and the denominator is \((7 + 4\sqrt{3})\). We multiplied the top and bottom by \((7 - 4\sqrt{3})\). The denominator became \((7+4\sqrt{3})(7-4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1\). The numerator became \((7 - 4\sqrt{3})(7 - 4\sqrt{3}) = (7 - 4\sqrt{3})^2\).

So, \(\frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{(7 - 4\sqrt{3})^2}{1} = (7 - 4\sqrt{3})^2\).

Now, taking the square root to find the ratio of sides: \[ \frac{s_1}{s_2} = \sqrt{(7 - 4\sqrt{3})^2} \]

The square root of a square is the absolute value. \[ \frac{s_1}{s_2} = |7 - 4\sqrt{3}| \]

To determine if \(7 - 4\sqrt{3}\) is positive or negative, we compare \(7\) and \(4\sqrt{3}\). Compare \(7^2\) and \((4\sqrt{3})^2\). \(7^2 = 49\) \((4\sqrt{3})^2 = 16 \times 3 = 48\) Since \(49 > 48\), we have \(7 > 4\sqrt{3}\). Therefore, \(7 - 4\sqrt{3}\) is a positive value.

So, the absolute value \(|7 - 4\sqrt{3}|\) is simply \(7 - 4\sqrt{3}\).

The ratio of the corresponding sides is \(7 - 4\sqrt{3}\).

Let's verify this with the options provided.

Our calculated ratio \(7 - 4\sqrt{3}\) matches Option 1.

Step-by-Step Solution

1. Identify the given information: Areas of two similar triangles \(A_1 = 7 - 4\sqrt{3}\) and \(A_2 = 7 + 4\sqrt{3}\).
2. Recall the property of similar triangles: The ratio of areas is the square of the ratio of corresponding sides, \(\frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2\).
3. Set up the equation: \(\left(\frac{s_1}{s_2}\right)^2 = \frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}}\).
4. Take the square root of both sides: \(\frac{s_1}{s_2} = \sqrt{\frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}}}\).
5. Rationalize the denominator inside the square root: Multiply numerator and denominator by the conjugate \((7 - 4\sqrt{3})\).
6. Simplify the expression: Numerator: \((7 - 4\sqrt{3})^2\) Denominator: \((7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1\) So, \(\frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{(7 - 4\sqrt{3})^2}{1} = (7 - 4\sqrt{3})^2\).
7. Calculate the square root: \(\frac{s_1}{s_2} = \sqrt{(7 - 4\sqrt{3})^2} = |7 - 4\sqrt{3}|\).
8. Determine the sign of \((7 - 4\sqrt{3})\): Since \(7^2 = 49\) and \((4\sqrt{3})^2 = 48\), \(7 > 4\sqrt{3}\), so \(7 - 4\sqrt{3}\) is positive.
9. The ratio of sides is \(7 - 4\sqrt{3}\).

Revision Table: Similar Triangle Properties

Additional Information: Simplifying Square Roots

In this problem, we encountered an expression under the square root that turned out to be a perfect square. Sometimes, you might need to simplify square roots of the form \(\sqrt{a \pm b\sqrt{c}}\). This can often be done if the expression inside can be written as the square of a binomial like \((\sqrt{x} \pm \sqrt{y})^2\).

Let's consider the structure of \((a - b\sqrt{3})^2\). \((p - q\sqrt{3})^2 = p^2 - 2pq\sqrt{3} + q^2(3) = (p^2 + 3q^2) - (2pq)\sqrt{3}\).

If we had \(\sqrt{7 - 4\sqrt{3}}\), we would look for \(p, q\) such that \(p^2 + 3q^2 = 7\) and \(2pq = 4\). From \(2pq = 4\), we get \(pq = 2\). Possible integer pairs \((p,q)\) are \((1,2)\) or \((2,1)\).

• If \(p=1, q=2\): \(p^2 + 3q^2 = 1^2 + 3(2^2) = 1 + 3(4) = 1 + 12 = 13\). This is not 7.
• If \(p=2, q=1\): \(p^2 + 3q^2 = 2^2 + 3(1^2) = 4 + 3(1) = 4 + 3 = 7\). This matches!

So, \(7 - 4\sqrt{3}\) is the square of \((2 - 1\sqrt{3})\) or \((2 - \sqrt{3})\). \((2 - \sqrt{3})^2 = 2^2 - 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}\).

Also consider the other area \(7 + 4\sqrt{3}\). This looks like \((2 + \sqrt{3})^2\). \((2 + \sqrt{3})^2 = 2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}\).

So the ratio of areas is \(\frac{(2 - \sqrt{3})^2}{(2 + \sqrt{3})^2} = \left(\frac{2 - \sqrt{3}}{2 + \sqrt{3}}\right)^2\).

Taking the square root: \(\frac{s_1}{s_2} = \left|\frac{2 - \sqrt{3}}{2 + \sqrt{3}}\right|\). Since \(2^2=4\) and \((\sqrt{3})^2=3\), \(2 > \sqrt{3}\), so \(2 - \sqrt{3}\) is positive. \(2 + \sqrt{3}\) is also positive. So, \(\frac{s_1}{s_2} = \frac{2 - \sqrt{3}}{2 + \sqrt{3}}\).

Now, rationalize this expression: \[ \frac{2 - \sqrt{3}}{2 + \sqrt{3}} = \frac{(2 - \sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{(2 - \sqrt{3})^2}{2^2 - (\sqrt{3})^2} = \frac{4 - 4\sqrt{3} + 3}{4 - 3} = \frac{7 - 4\sqrt{3}}{1} = 7 - 4\sqrt{3} \]

This confirms our earlier calculation and understanding of the problem. Both methods lead to the same ratio of sides.