Class 9 Maths · Original EZT practice
Measuring Space: Perimeter and Area (Class 9 Chapter 6): summary and practice
How much fencing does a field need, and how much land does it hold? This chapter measures space: perimeters of shapes, the circle's constant C/D ratio and why pi cannot be a fraction, the length of an arc, then areas of rectangles, parallelograms and triangles. 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. Perimeter and the C/D ratio
Perimeter is total boundary length. For circles the ratio circumference/diameter never changes - wrap a thread, measure, divide - and always lands near 3.14. That constant is pi, so C = pi x D = 2 pi r.
2. Pi is irrational; length of an arc
Aryabhata (499 CE) gave 3.1416 and called it asanna - approximate; Lambert proved pi irrational in 1761, so 22/7 is only convenient. An arc is a fraction of the circle: length = (angle/360) x 2 pi r.
3. Perimeter puzzles and area of a rectangle
A semicircle's perimeter includes the closing diameter: for radius 7 cm, 22 + 14 = 36 cm. Area counts unit squares: a 12 cm by 5 cm rectangle holds 60, so area = length x breadth.
4. Parallelogram, triangle, squaring and circle areas
Snip-and-slide turns a parallelogram into a rectangle: area = base x height. Two triangles make a parallelogram: area = (1/2) x base x height. The chapter closes with Baudhayana's squaring of a rectangle (Shulbasutra, 800 BCE) and the area of a circle, pi r^2.
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.
Add up a perimeter
Choose the method: Perimeter is the total length of the boundary.
EZT practice question
Find the perimeter of a triangle with sides 5 cm, 12 cm and 13 cm.
Show the worked solution
- Perimeter means walking the whole boundary once.
- Add the sides: 5 + 12 + 13.
- That gives 30 cm.
Answer: 30 cm.
Check: Runners on a 400 m track cover exactly one perimeter of the field - boundary length, nothing inside.
Watch for this: Perimeter is a length, not an area - the answer is in cm, not sq cm.
Now try it yourself
A square has side 9 cm. Perimeter?
Check your answer
36 cm.
4 x 9 = 36.
Use the C/D ratio
Choose the method: C = pi x D, and pi is about 3.14 for every circle.
EZT practice question
A circle has diameter 14 cm. Using pi = 22/7, find its circumference.
Show the worked solution
- The ratio circumference/diameter is the same for every circle - the number pi.
- C = pi x D = (22/7) x 14.
- That gives 44 cm.
Answer: 44 cm.
Check: Estimate with a thread: wrap a reel 20 times, divide by 20 and by the diameter - the answer lands near 3.14 whatever the reel.
Watch for this: 22/7 is only an approximation - pi itself is irrational (Lambert, 1761); Aryabhata's 3.1416 was already called asanna, approximate.
Now try it yourself
Circumference of a circle of radius 14 cm (pi = 22/7)?
Check your answer
88 cm.
C = 2 x (22/7) x 14 = 88.
Measure an arc
Choose the method: An arc is the same fraction of the circumference as its angle is of 360 degrees.
EZT practice question
Find the length of a 90-degree arc of a circle of radius 7 cm.
Show the worked solution
- 90/360 is one quarter of the full circle.
- Length = (90/360) x 2 x (22/7) x 7.
- That is (1/4) x 44 = 11 cm.
Answer: 11 cm.
Check: The full circumference here is 44 cm, and a quarter of 44 is 11 - the fraction method just formalises that.
Watch for this: Use the arc's CENTRAL angle as the fraction - 90 degrees is a quarter, 60 degrees a sixth.
Now try it yourself
What fraction of the circumference is a 60-degree arc?
Check your answer
One-sixth.
60/360 = 1/6.
Perimeter of a semicircle
Choose the method: Read the boundary - include the closing diameter.
EZT practice question
Find the perimeter of a semicircle of radius 7 cm.
Show the worked solution
- Half the circumference: (1/2) x 2 x (22/7) x 7 = 22 cm.
- The diameter closes the shape: 2 x 7 = 14 cm.
- Perimeter = 22 + 14 = 36 cm.
Answer: 36 cm.
Check: Half the circumference alone (22 cm) leaves the straight edge out - the boundary of a semicircle is the arc PLUS the diameter.
Watch for this: The classic trap: answering half the circumference. The diameter is part of the boundary.
Now try it yourself
Why is a semicircle's perimeter more than half the circle's circumference?
Check your answer
The diameter also bounds it.
Boundary, not just curve.
Area of a parallelogram
Choose the method: Snip a right triangle off one end, slide it across - you get a rectangle.
EZT practice question
A parallelogram has base 9 cm and height 4 cm. Find its area.
Show the worked solution
- Snip-and-slide makes a rectangle of the same base and height.
- Area = base x height = 9 x 4.
- That is 36 sq cm.
Answer: 36 sq cm.
Check: Height means the perpendicular distance between the parallel sides - the slant side would overstate it.
Watch for this: Do not multiply by the slant side - only the perpendicular height counts.
Now try it yourself
A parallelogram has base 15 cm and area 90 sq cm. Height?
Check your answer
6 cm.
90 / 15 = 6.
Area of a triangle
Choose the method: Two identical triangles make a parallelogram.
EZT practice question
Find the area of a triangle with base 10 cm and height 6 cm.
Show the worked solution
- Two copies of the triangle form a parallelogram of the same base and height.
- That parallelogram's area is 10 x 6 = 60 sq cm.
- One triangle is half: (1/2) x 10 x 6 = 30 sq cm.
Answer: 30 sq cm.
Check: Area = (1/2) x base x height - the doubling picture is why the half appears. The chapter extends the same cut-and-paste spirit to squaring a rectangle (Baudhayana) and to the circle's pi r^2.
Watch for this: Do not forget the half - base x height is the parallelogram, not the triangle.
Now try it yourself
Area of a triangle of base 12 cm and height 5 cm?
Check your answer
30 sq cm.
(1/2) x 12 x 5 = 30.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Perimeter?
Show answer and reason
Total length of the boundary.
For polygons, just add the sides.
2. C/D ratio?
Show answer and reason
Circumference over diameter - pi, about 3.14.
The same for every circle; C = pi x D.
3. Arc length?
Show answer and reason
(angle/360) x 2 pi r.
An arc is a fraction of the full circumference.
4. Area of parallelogram vs triangle?
Show answer and reason
base x height vs half of that.
Two triangles make one parallelogram.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Find the perimeter of a pentagon with all sides 6 cm.
- Find the length of a 120-degree arc of a circle of radius 21 cm.
- Find the area of a triangle of base 12 cm and height 5 cm.
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Questions about this practice page
- What is the C/D ratio of a circle?
- The ratio of circumference to diameter - the same for every circle, about 3.14. That constant is pi, so C = pi x D = 2 pi r.
- Is pi equal to 22/7?
- No - 22/7 is only a convenient approximation. Aryabhata gave 3.1416 and called it asanna (approximate), and Lambert proved in 1761 that pi is irrational, so no fraction equals it exactly.
- How do you find the length of an arc?
- Take the arc's central angle as a fraction of 360 degrees and multiply by the circumference: (angle/360) x 2 pi r. A 90-degree arc of radius 7 cm is (1/4) x 44 = 11 cm.
- What is the area of a parallelogram?
- Base x height, where height is the perpendicular distance between the parallel sides. A triangle on the same base and height has half that area.
- Are these the NCERT exercise questions for Class 9 Maths Chapter 6?
- 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.