Question

If $a=3, b=5, c=7$ are the sides of a triangle ABC, then its circumradius is

• A. $7/\sqrt{3}$
• B. $15/2$
• C. $15\sqrt{3}/4$
• D. $\sqrt{3}/2$

Hint:

For finding the circumradius, you have two primary formulas: $R = \frac{abc}{4\Delta}$ and $2R = \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Choose the method based on the information given. If side lengths are simple integers, the Law of Cosines is often quick for finding an angle.

Answer 1

The Correct Option is A

We can find the circumradius $R$ using two main methods.
Method 1: Using the formula $R = \frac{abc}{4\Delta}$, where $\Delta$ is the area.
First, calculate the semi-perimeter $s$:
$s = \frac{a+b+c}{2} = \frac{3+5+7}{2} = \frac{15}{2}$.
Next, calculate the area $\Delta$ using Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
$\Delta = \sqrt{\frac{15}{2}(\frac{15}{2}-3)(\frac{15}{2}-5)(\frac{15}{2}-7)} = \sqrt{\frac{15}{2}(\frac{9}{2})(\frac{5}{2})(\frac{1}{2})}$.
$\Delta = \sqrt{\frac{15 \cdot 9 \cdot 5}{16}} = \sqrt{\frac{3 \cdot 5 \cdot 9 \cdot 5}{16}} = \frac{\sqrt{9 \cdot 25 \cdot 3}}{4} = \frac{3 \cdot 5 \sqrt{3}}{4} = \frac{15\sqrt{3}}{4}$.
Now, use the circumradius formula:
$R = \frac{abc}{4\Delta} = \frac{3 \cdot 5 \cdot 7}{4 \cdot (\frac{15\sqrt{3}}{4})} = \frac{105}{15\sqrt{3}} = \frac{7}{\sqrt{3}}$.
Method 2: Using the Law of Cosines and the Sine Rule.
First, find the cosine of one of the angles, for example, angle C.
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{3^2+5^2-7^2}{2(3)(5)} = \frac{9+25-49}{30} = \frac{34-49}{30} = \frac{-15}{30} = -\frac{1}{2}$.
This means angle $C = 120^\circ$.
Now, use the extended Sine Rule: $\frac{c}{\sin C} = 2R$.
$R = \frac{c}{2\sin C} = \frac{7}{2\sin(120^\circ)} = \frac{7}{2(\frac{\sqrt{3}}{2})} = \frac{7}{\sqrt{3}}$.
Both methods yield the same result.