Problem 46 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The diameter of the circum-circle of a triangle with sides \(5 \mathrm{~cm}, 6 \mathrm{~cm} ., 7 \mathrm{~cm}\) is (a) \(\frac{3 \sqrt{6}}{2} \mathrm{~cm}\). (b) \(2 \sqrt{6} \mathrm{~cm}\) (c) \(\frac{35}{48} \mathrm{~cm}\) (d) \(\frac{35}{2 \sqrt{6}}\)

Step-by-Step Solution

Answer

The diameter of the circum-circle of a triangle with sides 5 cm, 6 cm, and 7 cm is \(\frac{35}{2\sqrt{6}}\) cm.

Step 1: Compute the Area of the Triangle

The area \(\Delta\) of a triangle with sides \(a, b, c\) can be found using Heron’s formula: \(\Delta = \sqrt{s(s - a)(s - b)(s - c)}\), where \(s = \frac{a + b + c}{2}\) is the semi-perimeter of the triangle. For this triangle, \(s = 9\), so \(\Delta = \sqrt{9(9 - 5)(9 - 6)(9 - 7)} = 3\sqrt{6}\).

Step 2: Compute the Diameter of the Circumcircle

We can use the formula \(d=\frac{abc}{4\Delta}\) to find the diameter of the circumcircle. Here, \(a=5, b=6, c=7\) and \(\Delta = 3\sqrt{6}\). So, \(d=\frac{5 \cdot 6 \cdot 7}{4 \cdot 3\sqrt{6}} = \frac{35}{2\sqrt{6}}\).

Step 3: Simplify the Diameter

We have found the diameter of the circumcircle in its simplest form. It doesn't simplify any further, so our final answer is \(d = \frac{35}{2\sqrt{6}}\).

Key Concepts

Circumcircle diameter

The circumcircle of a triangle is a special circle that passes through all three vertices of the triangle. This means every triangle has one unique circumcircle. The diameter of the circumcircle is a significant aspect. It can be calculated using a simple formula that relates the lengths of the triangle's sides and its area. Here's how it works:
The formula for the circumcircle diameter, denoted as \(d\), is:

• \(d = \frac{abc}{4\Delta}\)

where \(a, b,\) and \(c\) are the lengths of the sides of the triangle and \(\Delta\) is the area of the triangle. This formula is derived by considering the unique radius that can be applied to all sides of the triangle when touching each vertex precisely.

Using the formula mentioned above, if you know the lengths of the sides and the area, finding the circumcircle diameter becomes straightforward. By substituting these values, you can solve for \(d\), giving you insight into the triangle's enclosing circle.

Area of a triangle

Calculating the area of a triangle can be a bit tricky when you do not have the height but just the lengths of the three sides. This challenge is elegantly solved using Heron's Formula. Named after the ancient Greek engineer and mathematician Heron of Alexandria, this method allows us to find the area without the need for an altitude or a height measure.

Here's how Heron's Formula works:

• First, compute the semi-perimeter \(s\) which is half the sum of the triangle's sides: \(s = \frac{a + b + c}{2}\).
• The area, \(\Delta\), is then given by the formula: \(\Delta = \sqrt{s(s-a)(s-b)(s-c)}\).

This extraordinary formula makes it possible to calculate the area with just a few inputs. For our triangle with sides 5 cm, 6 cm, and 7 cm, this computes to \(\Delta = 3\sqrt{6}\).

By mastering Heron's Formula, you can find the area of any triangle given its side lengths, which is not only practical but also a testament to the timelessness of geometric solutions.

Semi-perimeter calculation

The semi-perimeter is a useful concept in geometry, especially when working with triangles and Heron's Formula. It represents half the perimeter of the triangle, essentially averaging the lengths of the sides to simplify area calculations. Here's how you can determine it:

The formula for the semi-perimeter \(s\) of a triangle with sides \(a, b,\) and \(c\) is:

• \(s = \frac{a + b + c}{2}\)

This value is pivotal when using Heron's Formula to calculate the triangle's area. The semi-perimeter serves as an intermediary step to ease the intricacies of dealing with various side lengths.

For our specific example, with sides 5 cm, 6 cm, and 7 cm, the semi-perimeter comes out to be \(s = 9\). This calculation is simple yet crucial, highlighting how smaller steps play an essential role in unlocking more complex solutions in geometry.