Measuring Space: Perimeter and Area: CBSE Class 9 Mathematics practice questions

C. On the bends the outer lanes follow bigger circles, so their path is longer; starting ahead makes every runner cover the same distance

On the straight parts all lanes have the same length, but on the bends the outer lanes run along circles of larger radius, and the length of a circle grows with its radius ($C = 2\pi r$). The stagger moves the outer starting lines forward by exactly that extra length so that all runners cover the same distance. It is not a reward for speed, the finish line is common to all, and every lane has the same width (1.22 m), so width is not the reason.

A. $22$ cm

Arc length $= 2\pi r \times \dfrac{\theta}{360} = 2 \times \dfrac{22}{7} \times 21 \times \dfrac{60}{360} = 132 \times \dfrac{1}{6} = 22$ cm. Traps: using $\pi r$ instead of $2\pi r$ (gives $11$), using the area formula $\pi r^2 \times \dfrac{\theta}{360} = 231\ \text{cm}^2$ (square units, not a length), or stopping at the full circumference $132$ cm.

C. $\pi < 2\sqrt{3} \approx 3.46$

Outer hexagon: perimeter $= 6a = \dfrac{12}{\sqrt{3}} = 4\sqrt{3} \approx 6.93$. The circle lies inside it, so $2\pi < 4\sqrt{3}$, hence $\pi < 2\sqrt{3} \approx 3.46$. Traps: forgetting that the circle has perimeter $2\pi$ (giving $4\sqrt{3}$), using the inner hexagon's perimeter 6 (that gives the lower bound $\pi > 3$), or quoting one side $a$.