Question
A circular racing track has inner circumference 528 m and outer circumference 616 m. Find the width of the track.
✓
Answer
Outer circumference $=616 \mathrm{~m}$
Radius $(R)=\frac{C}{2 \pi}=\frac{616 \times 7}{2 \times 22} \mathrm{~m}=98 \mathrm{~m}$
Inner circumference $=528 \mathrm{~m}$
$\therefore$ Inner radius $(r)=\frac{528 \times 7}{2 \times 22} \mathrm{~m}=84 \mathrm{~m}$
$\therefore$ Width of track $=R-r=98-84=14 m$
Radius $(R)=\frac{C}{2 \pi}=\frac{616 \times 7}{2 \times 22} \mathrm{~m}=98 \mathrm{~m}$
Inner circumference $=528 \mathrm{~m}$
$\therefore$ Inner radius $(r)=\frac{528 \times 7}{2 \times 22} \mathrm{~m}=84 \mathrm{~m}$
$\therefore$ Width of track $=R-r=98-84=14 m$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the area of a circle whose radius is : 8.4 m
In $-5 x^3 y^2 z^4$; write the coefficient of $y z^2$. Also, write the degree of the given algebraic expression.
→Simplify: $-5ax^2 + 7ax^2 – 12ax^2$
→Simplify and reduce to a simple fraction:
$\frac{4}{5} \text { of } \frac{7}{15} \div \frac{8}{9}$
→$\frac{4}{5} \text { of } \frac{7}{15} \div \frac{8}{9}$
A line $A B$ is of length $6 \mathrm{~cm}$. Another line $C D$ is of length $15 \mathrm{~cm}$. What fraction is: $\frac{1}{5}$ of $C D$ to that of $A B$ ?
→In the figure given below, $A B \| C D$. Find the unknown angles, giving reasons.
→The area of a ring is $528 cm^2$ and the radius of the outer circle is 17 cm . Find : the width of the ring.
→Write the complement of the angle : $(x+12)^{\circ}$
→The weight of a machine is 40 kg. By mistake it was weighed as 40.8 kg. Find the error percent.
Construct an isosceles $\triangle ABC$ such that $AB = AC = 5 cm$ and $\angle A = 60^{\circ}$.
→