Let one angle of a triangle be 60°, the area of triangle is 10`sqrt3`and perimeter is 20 cm. Ifa>b> c where a, b and e denote lengths of sides opposite to vertices A, B and C respectively, then which of the following is (are) correct?

Options

A. (a)Inradius of triangle is `sqrt3`
B. (b)Length of longest side of triangle is 7
C. (c)Circum-radius of triangles is `(7)/(sqrt3)`
D. (d)Radius of largest escribed circle is `/12`

Answer

To solve the given problem step by step, we will use the information provided about the triangle: one angle is 60°, the area is \(10\sqrt{3}\), and the perimeter is 20 cm. We will also denote the sides opposite to angles A, B, and C as a, b, and c respectively, where \(a > b > c\).

### Step 1: Identify the given information
- Angle A = 60°
- Area of triangle = \(10\sqrt{3}\)
- Perimeter = 20 cm
- \(a > b > c\)

### Step 2: Calculate the semi-perimeter (s)
The semi-perimeter \(s\) is given by:
\[ s = \frac{P}{2} = \frac{20}{2} = 10 \text{ cm} \]

### Step 3: Use the formula for the area of a triangle
The area \(A\) of a triangle can also be expressed using the formula:
\[ A = \frac{1}{2}ab\sin(C) \]
For our triangle, we have:
\[ A = \frac{1}{2}bc\sin(60°) = \frac{1}{2}bc\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}bc \]
Setting this equal to the given area:
\[ \frac{\sqrt{3}}{4}bc = 10\sqrt{3} \]
Dividing both sides by \(\sqrt{3}\):
\[ \frac{1}{4}bc = 10 \]
Multiplying by 4:
\[ bc = 40 \]

### Step 4: Set up the equations for sides a, b, and c
From the perimeter, we know:
\[ a + b + c = 20 \]
We also have:
\[ b + c = 20 - a \]
Substituting \(bc = 40\) into the quadratic equation:
Let \(b\) and \(c\) be the roots of the equation \(x^2 - (b+c)x + bc = 0\):
\[ x^2 - (20 - a)x + 40 = 0 \]

### Step 5: Solve the quadratic equation
The quadratic equation can be solved using the quadratic formula:
\[ x = \frac{-(20 - a) \pm \sqrt{(20 - a)^2 - 4 \cdot 40}}{2} \]
This simplifies to:
\[ x = \frac{(20 - a) \pm \sqrt{(20 - a)^2 - 160}}{2} \]

### Step 6: Find the values of a, b, and c
We know \(a + b + c = 20\) and \(bc = 40\). Let's substitute \(b + c = 20 - a\) into the quadratic equation:
\[ x^2 - (20 - a)x + 40 = 0 \]
We can find the discriminant:
\[ D = (20 - a)^2 - 160 \]
For \(b\) and \(c\) to be real numbers, \(D\) must be non-negative:
\[ (20 - a)^2 \geq 160 \]
Solving this inequality will give us the possible values for \(a\).

### Step 7: Calculate the sides
Assuming \(a\) is the largest side, we can test values for \(a\) to find \(b\) and \(c\).

1. If \(a = 8\):
\[ b + c = 20 - 8 = 12 \]
\[ bc = 40 \]
The quadratic becomes:
\[ x^2 - 12x + 40 = 0 \]
The discriminant:
\[ D = 12^2 - 4 \cdot 40 = 144 - 160 = -16 \quad \text{(not valid)} \]

2. If \(a = 7\):
\[ b + c = 20 - 7 = 13 \]
\[ bc = 40 \]
The quadratic becomes:
\[ x^2 - 13x + 40 = 0 \]
The discriminant:
\[ D = 13^2 - 4 \cdot 40 = 169 - 160 = 9 \quad \text{(valid)} \]
Thus, \(b\) and \(c\) can be calculated as:
\[ x = \frac{13 \pm 3}{2} \Rightarrow x = 8 \text{ or } 5 \]
So, \(b = 8\) and \(c = 5\).

### Step 8: Conclusion
The sides of the triangle are:
- \(a = 7\)
- \(b = 8\)
- \(c = 5\)

### Final Verification
- \(a + b + c = 7 + 8 + 5 = 20\) (correct)
- \(bc = 40\) (correct)
- Area calculation confirms \(10\sqrt{3}\).