Racing Track with Semicircular Ends: Area | Class 10 Maths

Quick answer: A racing track with semicircular ends is made of two straight strips and two semicircular ends. For the NCERT track (straight parts $106\text{ m}$ long, $60\text{ m}$ apart, track $10\text{ m}$ wide), the inner radius is $30\text{ m}$ and the outer radius is $40\text{ m}$. The distance along the inner edge is $212 + 2\pi(30) = \frac{2804}{7} \approx 400.57\text{ m}$, and the area of the track is $2120 + \pi(40^2 - 30^2) = 4320\text{ m}^2$ (using $\pi = \frac{22}{7}$).

Running tracks are usually made of two straight parts joined at each end by a semicircular curve. To find the distance around such a track, or its area, split it into simple shapes: rectangles for the straight parts and semicircles (or semicircular rings) for the ends.

In this guide, we solve the well-known NCERT racing track problem step by step.

Syllabus note: In the current rationalised NCERT Class 10 textbook, Chapter 11 (Areas Related to Circles) focuses on the areas of sectors and segments. The section on areas of combinations of plane figures, which included this racing track problem, has been removed. The problem is still useful practice; check your current syllabus before your exam.

What You Will Learn

• How to split a racing track into rectangles and semicircles
• How to find the inner and outer radii
• Part (i): the distance around the track along its inner edge
• Part (ii): the area of the track
• A shortcut: two semicircular ends together make one full circle (or ring)

1. Understanding the Track

     ___________________________
    /   ____________________    \
   |   /                    \    |
   |  |    inner field       |   |
   |  |    (60 m across)     |   |
   |   \____________________/    |
    \___________________________/
   straight parts: 106 m each
   track width: 10 m

2. The NCERT Problem

Question

A racing track has semicircular left and right ends. The distance between the two inner parallel line segments is $60\text{ m}$, and they are each $106\text{ m}$ long. If the track is $10\text{ m}$ wide, find:

(i) the distance around the track along its inner edge

(ii) the area of the track (Use $\pi = \frac{22}{7}$)

Part (i): Distance Along the Inner Edge

The inner edge has two straight parts and two semicircles of radius $r = 30\text{ m}$.

Step 1: Straight parts.

$106 + 106 = 212\text{ m}$

Step 2: Curved parts. Two semicircles of the same radius make one full circle:

$2\pi r = 2 \times \frac{22}{7} \times 30 = \frac{1320}{7}\text{ m}$

Step 3: Total distance.

$212 + \frac{1320}{7} = \frac{1484 + 1320}{7} = \frac{2804}{7}\text{ m} \approx 400.57\text{ m}$

Answer (i): The distance along the inner edge is $\frac{2804}{7}\text{ m}$, about $400.57\text{ m}$.

Part (ii): Area of the Track

The track is made of two rectangles (the straight parts) and two semicircular rings (the ends).

Step 1: Area of the two rectangles. Each is $106\text{ m}$ long and $10\text{ m}$ wide:

$2 \times 106 \times 10 = 2120\text{ m}^2$

Step 2: Area of the two semicircular ends. Together they make one full ring with outer radius $40\text{ m}$ and inner radius $30\text{ m}$:

$\pi\left(R^2 - r^2\right) = \frac{22}{7} \times \left(40^2 - 30^2\right) = \frac{22}{7} \times 700 = 2200\text{ m}^2$

Step 3: Total area.

$2120 + 2200 = 4320\text{ m}^2$

Answer (ii): The area of the track is $4320\text{ m}^2$.

3. Summary Table

4. Exam Tips and Common Mistakes

Exam Tip: Use the shortcut two semicircles = one circle for both the distance and the area. It saves time and reduces errors.

Exam Tip: It is fine to leave the answer to part (i) as $\frac{2804}{7}\text{ m}$, unless the question asks for a decimal.

Common Mistake: Adding the track width to the diameter instead of the radius. The outer radius is $30 + 10 = 40\text{ m}$, not $\frac{60 + 10}{2}$.

Common Mistake: Forgetting one of the two straight parts or one of the two ends.

Frequently Asked Questions

What is the distance around the NCERT racing track along its inner edge?

2804/7 m, which is about 400.57 m.

What is the area of the NCERT racing track?

4320 m²: 2120 m² for the straight parts plus 2200 m² for the two curved ends.

Why do the two semicircular ends make one full ring?

Each end is half of a circular ring with the same inner and outer radii, so together they make a complete ring.

How do you find the outer radius of the track?

Add the track width to the inner radius: 30 m + 10 m = 40 m.

Is this problem in the current Class 10 syllabus?

The section on areas of combinations of plane figures has been removed from the rationalised NCERT textbook, but the problem is still good practice.