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

Let polygon A have $a$ sides.

Since the ratio of sides is $1:2$, polygon B has $2a$ sides.

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For any regular polygon with $n$ sides, each interior angle equals:

$\text{Interior angle} = \tfrac{(n-2) \times 180°}{n}$

This works because the sum of all interior angles $=(n-2) \times 180°$, and since it's regular, all angles are equal, so each angle $=$ Total sum $÷$ Number of angles.

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For polygon A (with $a$ sides):

$\text{Interior angle of A} = \dfrac{(a-2) \times 180°}{a}$

For polygon B (with $2a$ sides):

$\text{Interior angle of B} = \dfrac{(2a-2) \times 180°}{2a} = \dfrac{180°(a-1)}{a}$

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Given that interior angles are in ratio 3:4:

$\dfrac{\text{Interior angle of A}}{\text{Interior angle of B}} = \dfrac{3}{4}$

Substituting our expressions:

$\Rightarrow \dfrac{\frac{(a-2) \times 180°}{a}}{\frac{180°(a-1)}{a}} = \dfrac{3}{4}$

$\Rightarrow \dfrac{a-2}{a-1} = \dfrac{3}{4}$

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Cross multiplying:

$4(a-2) = 3(a-1)$

$4a - 8 = 3a - 3$

$4a - 3a = 8 - 3$

$a = 5$

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Since polygon B has $2a$ sides:

$\text{Number of sides of B} = 2a = 2 \times 5 = 10$