Class 10 Maths · Original EZT practice
Areas Related to Circles (Class 10 Chapter 11): summary and practice
Areas Related to Circles turns one central angle into an area or length: the arc is (theta/360) of the circumference, the sector is the same fraction of the area, and the segment is the sector minus its triangle. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 22 October 2026:
1. Arc length and sector area
An arc of angle theta in a circle of radius r has length (theta/360) x 2 pi r; the sector area is (theta/360) x pi r^2. Same fraction of the whole, one for length and one for area.
2. Area of a segment
Segment = sector - the triangle formed by the two radii and the chord. The triangle area uses (1/2) r^2 sin theta; for a semicircle the segment is just half the circle.
3. Combinations of plane figures
Shaded regions in exam figures are built by adding and subtracting the areas of circles, sectors, triangles and squares - name every piece before touching a number.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Find the area of a sector
Choose the method: Sector area = (theta/360) x pi r^2.
EZT practice question
A sector of a circle of radius 6 cm has a central angle of 60 degrees. Find its area.
Show the worked solution
- theta/360 = 60/360 = 1/6.
- pi r^2 = 36 pi.
- Area = 36 pi / 6 = 6 pi sq cm, about 18.85 sq cm.
Answer: 6 pi sq cm.
Check: 60 degrees is a sixth of the full turn, and 6 pi is exactly a sixth of 36 pi - the fraction check always works.
Watch for this: Do not square the angle - only r is squared; theta enters as a plain fraction.
Now try it yourself
Sector area for r = 10 cm, angle 90 degrees?
Check your answer
25 pi sq cm.
A quarter of 100 pi.
Find the area of a segment
Choose the method: Segment = sector area - triangle area.
EZT practice question
Find the area of the minor segment cut off by a 60-degree chord in a circle of radius 6 cm.
Show the worked solution
- Sector area = 6 pi sq cm (from the previous question).
- The triangle has two sides 6 with included angle 60 degrees: area = (1/2) x 6 x 6 x sin 60 = 18 x (sqrt3/2) = 9 sqrt3 sq cm.
- Segment = 6 pi - 9 sqrt3, about 18.85 - 15.59.
Answer: 6 pi - 9 sqrt3 sq cm, about 3.26 sq cm.
Check: The segment must be smaller than the sector - 3.26 < 18.85, and positive, which holds.
Watch for this: Do not subtract the whole equilateral-looking triangle with the wrong formula - use (1/2) r^2 sin theta with the radius, not the chord.
Now try it yourself
Segment area for r = 4 cm, angle 90 degrees?
Check your answer
4 pi - 8 sq cm.
Sector 4 pi minus triangle (1/2) x 16 x 1 = 8.
Find the length of an arc
Choose the method: Arc length = (theta/360) x 2 pi r.
EZT practice question
Find the length of the arc subtending 60 degrees at the centre of a circle of radius 6 cm.
Show the worked solution
- theta/360 = 1/6.
- 2 pi r = 12 pi.
- Arc = 12 pi / 6 = 2 pi cm, about 6.28 cm.
Answer: 2 pi cm.
Check: The chord for 60 degrees is exactly r = 6 (equilateral triangle), and the arc must be slightly longer than the chord - 6.28 > 6 holds.
Watch for this: Do not use pi r^2 for length - the arc formula has a single r, like the circumference.
Now try it yourself
Arc length for r = 14 cm, angle 45 degrees?
Check your answer
(1/8) x 28 pi = 3.5 pi cm.
45 degrees is an eighth of the turn.
Special case: the semicircle
Choose the method: For theta = 180 degrees the segment is exactly half the circle - no triangle subtraction needed.
EZT practice question
Find the area of a semicircle of radius 7 cm.
Show the worked solution
- Sector area = (180/360) x pi x 49 = 49 pi / 2.
- The diameter passes through the centre, so the triangle has zero area.
- Nothing to subtract.
Answer: 49 pi / 2 = 24.5 pi sq cm, about 77 sq cm.
Check: Half of pi r^2 = 49 pi - halving again gives the semicircle, 24.5 pi.
Watch for this: Do not subtract a triangle here - the chord is the diameter, which contains the centre, so there is no triangular piece.
Now try it yourself
Area of a semicircle of diameter 14 cm?
Check your answer
24.5 pi sq cm.
Diameter 14 means radius 7 - the same question.
Areas of combined figures
Choose the method: Split the shaded region into named pieces, then add and subtract.
EZT practice question
A square of side 14 cm has four quarter circles of radius 7 cm drawn inward, one at each corner. Find the area left inside the square.
Show the worked solution
- Square area = 14^2 = 196 sq cm.
- Four quarter circles of radius 7 make one full circle: pi x 49 = 49 pi sq cm.
- Remaining area = 196 - 49 pi.
Answer: 196 - 49 pi sq cm, about 196 - 153.9 = 42.1 sq cm.
Check: The circle area 153.9 is less than the square 196, so a positive remainder - the shape stays inside the square.
Watch for this: Do not treat four quarter circles as four circles - four quarters make exactly one.
Now try it yourself
Same square, but two semicircles of radius 7 removed instead - area left?
Check your answer
196 - 49 pi sq cm.
Two semicircles also make one full circle.
Connect arc length and sector area
Choose the method: Sector area = (1/2) x r x arc length - a fast check between the two.
EZT practice question
A sector of radius 6 cm has arc length 2 pi cm. Find its area.
Show the worked solution
- Use area = (1/2) x r x arc = (1/2) x 6 x 2 pi.
- That gives 6 pi sq cm.
- Cross-check via the angle: arc 2 pi out of 12 pi means theta = 60 degrees, so area = 6 pi.
Answer: 6 pi sq cm.
Check: Both routes agree - this relation is the quickest verification tool in the chapter.
Watch for this: The relation is (1/2) r x arc, not r x arc - halve it, like a triangle's area.
Now try it yourself
Sector with r = 10 cm and arc 5 pi cm - area?
Check your answer
25 pi sq cm.
(1/2) x 10 x 5 pi.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Arc length formula?
Show answer and reason
(theta/360) x 2 pi r.
The angle's fraction of the circumference.
2. Sector area formula?
Show answer and reason
(theta/360) x pi r^2.
The angle's fraction of the circle's area.
3. Segment area?
Show answer and reason
Sector - triangle = (theta/360) pi r^2 - (1/2) r^2 sin theta.
Cut the triangle off the sector.
4. Segment of a semicircle?
Show answer and reason
Just half the circle, pi r^2 / 2.
The diameter makes the triangle vanish.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Sector area for r = 10 cm, angle 90 degrees.
- Minor segment for r = 4 cm, angle 90 degrees.
- Arc length for r = 14 cm, angle 45 degrees.
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Questions about this practice page
- What is the difference between a sector and a segment?
- A sector is the pizza slice - bounded by two radii and the arc. A segment is what is left of the slice after the triangular part (the two radii and the chord) is removed: segment = sector - triangle.
- How do I find the area of a minor segment?
- Find the sector area (theta/360) pi r^2, subtract the triangle area (1/2) r^2 sin theta. For a semicircle the triangle vanishes, so the segment is simply pi r^2 / 2.
- What is the formula for arc length?
- Arc = (theta/360) x 2 pi r - the angle's fraction of the full circumference. Handy check: sector area = (1/2) x r x arc length.
- Which chapters come before Areas Related to Circles?
- Circles (Chapter 10) supplies the tangent and chord facts, and Introduction to Trigonometry (Chapter 8) supplies sin theta for the triangle part of a segment - both feed directly into this chapter.
- Are these the NCERT exercise questions for Areas Related to Circles (Class 10)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.