GATE AE 2025 • Q8 MCQ • 2 Marks

General Aptitude

Quantitative Aptitude

Subtopic: Permutations & Combinations

Question Statement:

A regular dodecagon (12-sided regular polygon) is inscribed in a circle of radius r cm as shown in the figure. The side of the dodecagon is d cm. All the triangles (numbered 1 to 12) in the figure are used to form squares of side r cm and each numbered triangle is used only once to form a square. The number of squares that can be formed and the number of triangles required to form each square, respectively, are: Note: The figure shown is representative.

Options:

  • (A) 3; 4
  • (B) 4; 3
  • (C) 3; 3
  • (D) 3; 2

Official Answer Key:

A

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2025 Q.8)** **1. Intuitive Physical Setup & Given Parameters**: - A regular dodecagon (12-sided regular polygon) is inscribed in a circle of radius $r$. - It is partitioned from the center into 12 identical isosceles triangles (numbered 1 to 12). - Each triangle has two sides equal to radius $r$ and apex angle at the center: $\theta = \frac{360^\circ}{12} = 30^\circ$ - All 12 triangles are to be used (each exactly once) to form squares of side $r$. - We need to find: (1) the number of squares formed, and (2) the number of triangles required per square. **2. Step-by-Step Geometric Derivation**: 1. **Area of Each Triangle**: The area of an isosceles triangle with two sides of length $r$ and included angle $30^\circ$ is: $A_{\text{triangle}} = \frac{1}{2} r^2 \sin(30^\circ) = \frac{1}{2} r^2 \left(\frac{1}{2}\right) = \frac{1}{4} r^2$ 2. **Area of a Square of Side $r$**: $A_{\text{square}} = r^2$ 3. **Triangles Required to Form One Square**: $\text{Number of triangles per square} = \frac{A_{\text{square}}}{A_{\text{triangle}}} = \frac{r^2}{\frac{1}{4}r^2} = 4$ 4. **Number of Squares Formed**: With 12 total triangles available: $\text{Number of squares} = \frac{12}{4} = 3\text{ squares}$ 5. Result: **3 squares; 4 triangles per square** (Option A). **3. Comprehensive Options & Distractor Analysis**: - **(A) 3; 4**: ✅ **CORRECT**. Area of each triangle is $r^2/4$; thus 4 triangles form one square of area $r^2$, yielding $12/4 = 3$ squares. - **(B) 4; 3**: ❌ **INCORRECT**. Reverses the two quantities. - **(C) 3; 3**: ❌ **INCORRECT**. Total triangles would be $3 \times 3 = 9 \neq 12$. - **(D) 3; 2**: ❌ **INCORRECT**. Total triangles would be $3 \times 2 = 6 \neq 12$. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **A** - **Exam Takeaway**: Equating areas provides immediate answers: $A_{\triangle} = \frac{1}{2}r^2\sin 30^\circ = \frac{r^2}{4}$. Thus 4 triangles assemble into area $r^2$, creating 3 squares from 12 triangles.

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