A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of

Q: Why is the diagonal of the square equal to the diameter of the circle?

When a square is inscribed in a circle, its vertices lie on the circle's circumference. The diagonal of the square passes through the center of the circle, connecting two opposite vertices. Since the distance between two opposite points on the circle passing through the center is the diameter, the diagonal must equal the diameter.

Q: Can I use the formula $r = \frac{s}{2\sqrt{3}}$ for the equilateral triangle directly?

Absolutely! That is a fantastic shortcut. For an equilateral triangle of side $s$, the inradius is indeed $r = \frac{s}{2\sqrt{3}}$. Using this saves time, but understanding the derivation from $r = \frac{\Delta}{s_{semi}}$ is crucial for deeper conceptual clarity.

Q: Why do we need to calculate $m + n^2$ instead of just $m$ or $n$?

This is a classic JEE-style question designed to test your ability to handle multiple steps and algebraic expressions. It ensures you can correctly compute both the area and perimeter of the inscribed square and then perform the final algebraic manipulation without making a calculation error.

Question

A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are $m$ and $n$, respectively, then $m + n^2$ is equal to

Options:

(A) 408

(B) 414

(C) 396

(D) 312

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before a perfectly symmetric equilateral triangle with side length $s = 12$. We aim to find the properties of a square inscribed within the incircle of this triangle.

The radius $r$ of the incircle is determined by the relationship between the area of the triangle $\Delta$ and its semi-perimeter $s_{semi}$ using the formula $r = \frac{\Delta}{s_{semi}}$.

First, we calculate the area of the equilateral triangle:

$$
\Delta = \frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4} 	imes 144 = 36\sqrt{3}
$$

Next, we find the semi-perimeter $s_{semi}$:

$$
s_{semi} = \frac{12 + 12 + 12}{2} = 18
$$

Dividing the area by the semi-perimeter, we obtain the radius of the incircle:

$$
r = \frac{36\sqrt{3}}{18} = 2\sqrt{3}
$$

### The Square Within the Circle

We now place a square with side length $a$ inside this circle. The diagonal of the square is equal to the diameter of the circle, which is $2r$.

Given the diagonal of a square is $a\sqrt{2}$, we establish the following relationship:

$$
a\sqrt{2} = 2r
$$

Substituting $r = 2\sqrt{3}$ into the equation:

$$
a\sqrt{2} = 2(2\sqrt{3}) = 4\sqrt{3}
$$

Solving for the side length $a$:

$$
a = \frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}
$$

### The Final Payoff

The area $m$ of the square is the square of its side length:

$$
m = a^2 = (2\sqrt{6})^2 = 4 	imes 6 = 24
$$

The perimeter $n$ of the square is four times the side length:

$$
n = 4a = 4(2\sqrt{6}) = 8\sqrt{6}
$$

Finally, we compute the value of $m + n^2$:

$$
m + n^2 = 24 + (8\sqrt{6})^2
$$

Calculating the square of $8\sqrt{6}$:

$$
(8\sqrt{6})^2 = 64 	imes 6 = 384
$$

Adding these values together, we arrive at the final result:

$$
24 + 384 = **408**
$$

Animated Solution for Mathematics - Trigonometry: A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are $m$ and $n$ , respectively, then $m + n^{2}$ is equal to

Select Answer:

Visualized Solution

Equilateral Triangle Setup

Inscribed Circle

Formula for Inradius

Calculate $Δ$ and $s_{se mi}$

Find the Inradius $r$

Square Inscribed in Circle

Diagonal = Diameter

Equation for Side $a$

Solve for $a$

Calculate Area $m$

Calculate Perimeter $n$

Setup $m + n^{2}$

Final Computation

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Square Within the Circle

The Final Payoff

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Select Answer:

Visualized Solution

Equilateral Triangle Setup

Inscribed Circle

Formula for Inradius

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ssemi​

Find the Inradius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Square Inscribed in Circle

Diagonal = Diameter

Equation for Side .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Calculate Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Calculate Perimeter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Setup .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n2