A Question is given followed by two Statements I and II. Consider the Question and the Statements and mark the correct option.Question : A circle is inscribed in the equilateral triangle $\Delta$. What is the radius of the circle ?Statement-I : Area of $\Delta$ is equal to $16\sqrt{3} \text{ cm}^2$.Statement-II : Perimeter of $\Delta$ is 24 cm.Which one of the following is correct in respect of the above Question and the Statements ?

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CDS - 2026

A Question is given followed by two Statements I and II. Consider the Question and the Statements and mark the correct option.
Question : A circle is inscribed in the equilateral triangle $\Delta$. What is the radius of the circle ?
Statement-I : Area of $\Delta$ is equal to $16\sqrt{3} \text{ cm}^2$.
Statement-II : Perimeter of $\Delta$ is 24 cm.
Which one of the following is correct in respect of the above Question and the Statements ?

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CDS 2 2026 Maths Question Paper (13-Sep-2026)

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The Question can be answered by using one of the statements alone, but cannot be answered using the other statement alone

The Question can be answered by using either Statement alone

The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

The Question cannot be answered even by using both the Statements together

The correct answer is

The Question can be answered by using either Statement alone

Analyzing Statement I: Equilateral Triangle Area

The question asks for the radius of a circle inscribed in an equilateral triangle ($\Delta$). Let the side length of the triangle be '$a$' and the inradius be '$r$'.

Statement I provides the area of $\Delta$ as $16\sqrt{3} \text{ cm}^2$. The formula for the area of an equilateral triangle is:

$\text{Area} = \frac{\sqrt{3}}{4} a^2$

Using the given area:

$16\sqrt{3} = \frac{\sqrt{3}}{4} a^2$

Solving for '$a$':

$a^2 = \frac{4 \times 16\sqrt{3}}{\sqrt{3}} = 64$ $a = \sqrt{64} = 8 \text{ cm}$

The inradius '$r$' of an equilateral triangle can be calculated using the side length '$a$' with the formula:

$r = \frac{a}{2\sqrt{3}}$

Substituting $a = 8$ cm:

$r = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \text{ cm}$

Conclusion: Statement I alone is sufficient to determine the side length and subsequently the inradius of the circle.

Analyzing Statement II: Equilateral Triangle Perimeter

Statement II provides the perimeter of $\Delta$ as 24 cm.

The formula for the perimeter of an equilateral triangle is:

$\text{Perimeter} = 3a$

Using the given perimeter:

$24 = 3a$

Solving for '$a$':

$a = \frac{24}{3} = 8 \text{ cm}$

Using the inradius formula for an equilateral triangle ($r = \frac{a}{2\sqrt{3}}$) with $a = 8$ cm:

$r = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \text{ cm}$

Conclusion: Statement II alone is sufficient to determine the side length and subsequently the inradius of the circle.

Final Conclusion

Both Statement I and Statement II independently provide enough information to calculate the side length of the equilateral triangle, which allows for the calculation of the inscribed circle's radius. Therefore, the question can be answered using either statement alone.

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