One angle of an isosceles Δ is 120° and radius of its incircle = √3. Then the area of the triangle i

Q: Wait, why do we use the Sine Rule here? Can't we just use basic trigonometry?

You absolutely can! The Sine Rule is just a formal way of applying the same trigonometric principles. If you drop an altitude from the vertex $A$ to the base $BC$, you create two right-angled triangles. You will find that the base $a$ is $2x \cos(30^\circ)$, which is $2x \cdot \frac{\sqrt{3}}{2} = x\sqrt{3}$. The Sine Rule just gets you there faster. Both methods are perfectly valid!

Q: Why is the area formula $\frac{1}{2}x^2 \sin(120^\circ)$ valid?

That is the standard formula for the area of a triangle given two sides and the included angle: $\text{Area} = \frac{1}{2}ab \sin(C)$. Here, our sides are $x$ and $x$, and the included angle is $120^\circ$. It is a very efficient way to calculate the area without needing to find the height explicitly.

Q: Is there a simpler way to find the semi-perimeter?

The semi-perimeter $s$ is defined as $\frac{a+b+c}{2}$. In our case, the sides are $x$, $x$, and $a$. So, $s = \frac{x+x+a}{2} = \frac{2x+a}{2}$. Since we found $a = x\sqrt{3}$, substituting this gives $s = \frac{2x + x\sqrt{3}}{2} = x(1 + \frac{\sqrt{3}}{2})$. It is as simple as that!

Question

One angle of an isosceles $\Delta$ is $120^\circ$ and radius of its incircle $= \sqrt{3}$. Then the area of the triangle in sq. units is

Options:

(A) $7 + 12\sqrt{3}$

(B) $12 - 7\sqrt{3}$

(C) $12 + 7\sqrt{3}$

(D) $4\pi$

Sigma Percentile · Accepted Answer

### The Geometry of the Incircle: A Journey into Symmetry

My dear student, welcome to this beautiful exploration of geometry. Today, we are not just solving a problem; we are uncovering the hidden architecture of an isosceles triangle.

Imagine you are standing before a triangle with one angle of $120^\circ$. It is an obtuse, elegant shape, and tucked perfectly inside it is an incircle with a radius of $\sqrt{3}$. Our goal is to find the area of this triangle.

### Phase 1: The Geometry of the Triangle

First, let us orient ourselves. We have an isosceles triangle, let us call it $\Delta ABC$, with $\angle A = 120^\circ$.

Because it is isosceles, the angles opposite the equal sides must be equal. The sum of angles in any triangle is $180^\circ$. If $\angle A = 120^\circ$, then $\angle B + \angle C = 180^\circ - 120^\circ = 60^\circ$. Since $\angle B = \angle C$, each must be $30^\circ$.

This $120^\circ-30^\circ-30^\circ$ triangle is a classic, beautiful configuration.

Now, let us define our variables. Let the equal sides be $AB = AC = x$, and let the base be $BC = a$.

To link these, we use the Sine Rule:

$$
\frac{a}{\sin 120^\circ} = \frac{x}{\sin 30^\circ}
$$

Substituting the values $\sin 120^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$, we get:

$$
\frac{a}{\frac{\sqrt{3}}{2}} = \frac{x}{\frac{1}{2}}
$$

This simplifies beautifully to $a = x\sqrt{3}$.

### Phase 2: The Bridge to the Inradius

We need the area $\Delta$ and the semi-perimeter $s$ to use the inradius formula $r = \frac{\Delta}{s}$.

The area of our triangle is:

$$
\Delta = \frac{1}{2} \cdot AB \cdot AC \cdot \sin A = \frac{1}{2} x^2 \sin 120^\circ = \frac{\sqrt{3}}{4} x^2
$$

Next, the semi-perimeter $s$ is:

$$
s = \frac{x + x + a}{2} = \frac{2x + x\sqrt{3}}{2} = \frac{x(2 + \sqrt{3})}{2}
$$

### Phase 3: The Inradius Connection

Now, we bring in the given inradius $r = \sqrt{3}$. The formula $r = \frac{\Delta}{s}$ acts as our bridge.

Substituting our expressions, we have:

$$
\sqrt{3} = \frac{\frac{\sqrt{3}}{4} x^2}{\frac{x(2 + \sqrt{3})}{2}}
$$

Do not be intimidated by this fraction! Watch as the terms cancel out. The $\sqrt{3}$ on both sides vanishes, and the $x$ in the numerator cancels with one $x$ in the denominator.

We are left with:

$$
1 = \frac{x}{4} \cdot \frac{2}{2 + \sqrt{3}} \Rightarrow 1 = \frac{x}{2(2 + \sqrt{3})}
$$

Solving for $x$, we find $x = 2(2 + \sqrt{3}) = 4 + 2\sqrt{3}$.

### Phase 4: The Final Calculation

We are almost there. We have the side length $x$, and we need the area $\Delta = \frac{\sqrt{3}}{4} x^2$.

Substituting $x = 2(2 + \sqrt{3})$, we get:

$$
\Delta = \frac{\sqrt{3}}{4} [2(2 + \sqrt{3})]^2 = \frac{\sqrt{3}}{4} \cdot 4 \cdot (2 + \sqrt{3})^2
$$

Expanding the square, we obtain:

$$
\Delta = \sqrt{3} (4 + 3 + 4\sqrt{3}) = \sqrt{3} (7 + 4\sqrt{3})
$$

Distributing the $\sqrt{3}$, we get $7\sqrt{3} + 4(3) = 12 + 7\sqrt{3}$.

And there it is! The area of our triangle is **$12 + 7\sqrt{3}$** square units. It is a testament to the elegance of geometry that such a complex-looking problem resolves into such a clean, beautiful result.

Animated Solution for Mathematics - Trigonometry: One angle of an isosceles $Δ$ is $12 0^{\circ}$ and radius of its incircle $= 3$ . Then the area of the triangle in sq. units is

Select Answer:

Visualized Solution

Visualizing the Isosceles Triangle

Defining the Sides

Applying the Sine Rule

Solving for Base $a$

Area of the Triangle

Simplifying the Area

Finding the Semi-perimeter

The Inradius Formula

Substituting into the Formula

Simplifying the Equation

Solving for $x$

Calculating the Final Area

Expanding the Square

Final Result

The Sigma Insight: Properties of Triangles

The Geometry of the Incircle

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } One angle of an isosceles Δ is 120∘ and radius of its incircle =3​. Then the area of the triangle in sq. units is