Question

Regular polygons A and B have number of sides in the ratio \(1:2\) and interior angles in the ratio \(3:4\).Then the number of sides of B equals

Answer 1

Given: Regular polygons A and B have their interior angles in the ratio \(3 : 4\).

Let the number of sides of polygon A be \(n\).
Then, the number of sides of polygon B is \(2n\) (since the ratio of number of sides is \(1 : 2\)).

Interior angle of a regular polygon with \(n\) sides is given by:

\[\text{Interior angle} = \frac{(n - 2) \times 180^\circ}{n}\]

So for polygon A: \(\frac{(n - 2) \times 180}{n}\)
For polygon B: \(\frac{(2n - 2) \times 180}{2n}\)

Given ratio of interior angles: 

\[\frac{(n - 2) \times 180}{n} : \frac{(2n - 2) \times 180}{2n} = 3 : 4\]

Step 1: Eliminate 180 from both sides

\[\frac{n - 2}{n} : \frac{2n - 2}{2n} = 3 : 4\]

Step 2: Cross-multiply the ratios

\[4(n - 2) = 3(2n - 2)\]

Expand both sides: \(4n - 8 = 6n - 6\)
Rearranging: \(2n = 2 \Rightarrow n = 1\) — Wait! That seems wrong — let's recheck.

Correct Step:

Go back to: \(4(n - 2) = 3(2n - 2)\)
\(4n - 8 = 6n - 6\)
Rearranging: \(4n - 6n = -6 + 8 \Rightarrow -2n = 2 \Rightarrow n = -1\) — No again? Check again.

Actually: There is a mistake in the above simplification. Let's solve correctly:

Simplification Again:

Start again from: \(4(n - 2) = 3(2n - 2)\)
\(4n - 8 = 6n - 6\)
\(4n - 6n = -6 + 8 \Rightarrow -2n = 2 \Rightarrow n = -1\)
Still not right — this implies negative sides, which is invalid.

Let’s correct with clean math:

From original: 

\[\frac{(n - 2) \times 180}{n} : \frac{(2n - 2) \times 180}{2n} = 3 : 4\]

Remove common factor 180: 

\[\frac{n - 2}{n} : \frac{2n - 2}{2n} = 3 : 4\]

Cross multiply: 

\[4(n - 2) = 3(2n - 2) \Rightarrow 4n - 8 = 6n - 6 \Rightarrow -2n = 2 \Rightarrow n = -1\]

There is inconsistency. Let’s retry fresh assuming polygon A has n sides and B has m sides.

Correct clean approach:

Let polygon A have \(n\) sides, and polygon B have \(m\) sides. Their interior angles: \(\frac{(n - 2) \cdot 180}{n}\) and \(\frac{(m - 2) \cdot 180}{m}\)

Given ratio: 

\[\frac{(n - 2)}{n} : \frac{(m - 2)}{m} = 3 : 4\]

Cross-multiplying: 

\[4(n - 2) \cdot m = 3(m - 2) \cdot n\]

Now, assume \(m = 2n\) (as per question). Substituting:

\[4(n - 2)(2n) = 3(2n - 2)(n) \Rightarrow 8n(n - 2) = 3n(2n - 2)\]

Expand both sides: 

\[8n^2 - 16n = 6n^2 - 6n \Rightarrow 2n^2 - 10n = 0 \Rightarrow n(n - 5) = 0 \Rightarrow n = 0 \text{ or } n = 5\]

Reject \(n = 0\) (a polygon cannot have 0 sides), so \(n = 5\).

Therefore, Polygon A has 5 sides, and Polygon B has: \(2 \times 5 = \boxed{10}\) sides.

Answer 2

Let the number of sides of polygons A and B be \(n\) and \(2n\), respectively.

The sum of interior angles of a polygon is given by:

\((n - 2) \times 180^\circ\)

So, each interior angle of a regular polygon is:

\(\frac{(n - 2) \times 180^\circ}{n}\)

Now, set up the ratio of interior angles:

\(\frac{\frac{(n - 2) \times 180}{n}}{\frac{(2n - 2) \times 180}{2n}} = \frac{3}{4}\)

Simplify numerator and denominator:

\(\frac{\frac{(n - 2) \times 180}{n}}{\frac{2(n - 1) \times 180}{2n}} = \frac{3}{4}\)

Cancel out \(180\) and simplify the expression:

\(\frac{n - 2}{n - 1} = \frac{3}{4}\)

Solving the equation:

\(4(n - 2) = 3(n - 1) \\ 4n - 8 = 3n - 3 \\ n = 5\)

Therefore:

• Polygon A has \(n = 5\) sides
• Polygon B has \(2n = 10\) sides

Final Answer: \(\boxed{10}\)