Important Questions for Class 9 Maths (Ganita Manjari I) Chapter 6 Measuring Space: Perimeter and Area - 2026-27

The important questions for Class 9 Maths Chapter 6 focus on strengthening students' understanding of perimeter, area and their practical applications. Each question includes a solution to help students apply formulas, improve accuracy and prepare effectively for exams.

For more chapter-wise practice across subjects, students can explore Class 9 Important Questions.

Class 9 Maths Chapter 6 Important Questions MCQs

Q1. A circular park has a diameter of 28 m. What is the circumference of the park? Use $$\pi=\frac{22}{7}$$.

(a) 44 m

(b) 88 m

(c) 154 m

(d) 176 m

Answer: (b) 88 m

Explanation:

Circumference of a circle = $$\pi d$$

Given diameter = 28 m

$$C=\frac{22}{7}\times28$$

$$C=88$$ m

Final Answer: The circumference of the circular park is 88 m.

Q2. A circular track has a radius of 14 m. What is the length of an arc that subtends an angle of 90° at the centre?

(a) 11 m

(b) 22 m

(c) 44 m

(d) 88 m

Answer: (b) 22 m

Explanation:

Arc length = $$\frac{\theta}{360^\circ}\times2\pi r$$

Given:

Radius = 14 m

Central angle = 90°

$$\text{Arc length}=\frac{90}{360}\times2\times\frac{22}{7}\times14$$

$$=\frac{1}{4}\times88$$

$$=22$$ m

Final Answer: The arc length is 22 m.

Q3. Which of the following statements about π is correct?

(a) π is exactly equal to $$\frac{22}{7}$$.

(b) π is a rational number.

(c) π is an irrational number.

(d) The value of π increases when the radius of a circle increases.

Answer: (c) π is an irrational number.

Explanation:

The value of π is the ratio of the circumference of a circle to its diameter.

$$\pi=\frac{C}{D}$$

This ratio remains constant for all circles.

π is irrational because it cannot be expressed exactly as a fraction of two integers. Its decimal expansion is non-terminating and non-repeating.

Final Answer: π is an irrational number.

Q4. A parallelogram has a base of 15 cm and a perpendicular height of 8 cm. What is its area?

(a) 60 cm²

(b) 100 cm²

(c) 120 cm²

(d) 240 cm²

Answer: (c) 120 cm²

Explanation:

Area of a parallelogram = Base × Height

Given:

Base = 15 cm

Height = 8 cm

Area = 15 × 8

Area = 120 cm²

Final Answer: The area of the parallelogram is 120 cm².

Class 9 Maths Chapter 6 Very Short Answer Questions

Q5. What is the relationship between the circumference and diameter of a circle? Write the formula used to calculate the circumference.

Answer:

The ratio of the circumference of a circle to its diameter is always constant. This constant is called π (pi).

$$\frac{C}{D}=\pi$$

Therefore:

$$C=\pi d$$

Since diameter = 2 × radius, we can also write:

$$C=2\pi r$$

Here, C is the circumference, d is the diameter, and r is the radius.

Q6. A circle has a circumference of 44 cm. Find its radius using $$\pi=\frac{22}{7}$$.

Solution:

Circumference = $$2\pi r$$

Given circumference = 44 cm

$$44=2\times\frac{22}{7}\times r$$

$$r=\frac{44\times7}{44}$$

$$r=7$$ cm

Final Answer: The radius of the circle is 7 cm.

Q7. Find the semi-perimeter of a triangle whose three sides measure 9 cm, 10 cm and 11 cm.

Solution:

Semi-perimeter = $$\frac{a+b+c}{2}$$

Given:

a = 9 cm, b = 10 cm, c = 11 cm

$$s=\frac{9+10+11}{2}$$

$$s=\frac{30}{2}$$

$$s=15$$ cm

Final Answer: The semi-perimeter of the triangle is 15 cm.

Q8. A triangle ABC has an area of 48 cm². AD is a median drawn from vertex A to side BC. Find the areas of triangles ABD and ACD.

Solution:

A median divides a triangle into two triangles of equal area.

Therefore:

Area of △ABD = Area of △ACD

Each area = $$\frac{48}{2}$$

= 24 cm²

Final Answer:

Area of △ABD = 24 cm²

Area of △ACD = 24 cm²

Class 9 Maths Chapter 6 Short Answer Questions

Q9. A circular garden has a radius of 21 m. Find its circumference and the total cost of fencing the garden at ₹25 per metre. Use $$\pi=\frac{22}{7}$$.

Solution:

Step 1: Find the circumference.

Circumference = $$2\pi r$$

Given radius = 21 m

$$C=2\times\frac{22}{7}\times21$$

$$C=132$$ m

Step 2: Calculate the fencing cost.

Cost per metre = ₹25

Total cost = 132 × 25

Total cost = ₹3,300

Final Answer:

Circumference of the garden = 132 m

Total cost of fencing = ₹3,300

Q10. A sector of a circle has a radius of 14 cm and a central angle of 75°. Find its arc length and total perimeter. Use $$\pi=\frac{22}{7}$$.

Solution:

Given:

Radius = 14 cm

Central angle = 75°

Step 1: Calculate the arc length.

Arc length = $$\frac{\theta}{360^\circ}\times2\pi r$$

$$=\frac{75}{360}\times2\times\frac{22}{7}\times14$$

$$=\frac{75}{360}\times88$$

$$=\frac{55}{3}$$ cm

Arc length = $$18\frac{1}{3}$$ cm

Step 2: Calculate the perimeter of the sector.

Perimeter = Arc length + 2 × Radius

$$P=\frac{55}{3}+2(14)$$

$$P=\frac{55}{3}+28$$

$$P=\frac{139}{3}$$ cm

$$P=46\frac{1}{3}$$ cm

Final Answer:

Arc length = $$18\frac{1}{3}$$ cm

Perimeter of sector = $$46\frac{1}{3}$$ cm

Q11. A triangular field has sides of lengths 13 m, 14 m and 15 m. Find the area of the field using Heron's formula.

Solution:

Given:

a = 13 m

b = 14 m

c = 15 m

Step 1: Find the semi-perimeter.

$$s=\frac{a+b+c}{2}$$

$$s=\frac{13+14+15}{2}$$

$$s=21$$ m

Step 2: Apply Heron's formula.

$$A=\sqrt{s(s-a)(s-b)(s-c)}$$

Substituting the values:

$$A=\sqrt{21(21-13)(21-14)(21-15)}$$

$$A=\sqrt{21\times8\times7\times6}$$

$$A=\sqrt{7056}$$

$$A=84$$ m²

Final Answer: The area of the triangular field is 84 m².

Q12. A rectangular playground has a perimeter of 100 m. Its length is 30 m. Find its breadth and area. If the playground is divided into two equal parts along its length, find the area of each part.

Solution:

Step 1: Find the breadth.

Perimeter of rectangle = 2(Length + Breadth)

$$100=2(30+b)$$

$$50=30+b$$

$$b=20$$ m

Step 2: Find the area of the playground.

Area = Length × Breadth

= 30 × 20

= 600 m²

Step 3: Find the area of each equal part.

Area of each part = $$\frac{600}{2}$$

= 300 m²

Final Answer:

Breadth = 20 m

Total area = 600 m²

Area of each equal part = 300 m²

Q13. A car tyre has a diameter of 56 cm. Find the distance travelled by the car when the tyre completes 250 revolutions without slipping. Express the answer in metres.

Solution:

Given:

Diameter = 56 cm

Number of revolutions = 250

Step 1: Find the circumference of the tyre.

Circumference = $$\pi d$$

$$C=\frac{22}{7}\times56$$

$$C=176$$ cm

Step 2: Find the distance travelled.

In one revolution, the tyre travels 176 cm.

Distance in 250 revolutions:

= 176 × 250

= 44,000 cm

Step 3: Convert centimetres into metres.

Distance = $$\frac{44000}{100}$$

= 440 m

Final Answer: The car travels 440 m in 250 revolutions.

Class 9 Maths Chapter 6 Long Answer Questions

Q14. Prove that a median of a triangle divides it into two triangles of equal area.

Solution:

Given:

In △ABC, AD is a median drawn from vertex A to side BC.

To Prove:

Area of △ABD = Area of △ACD

Step 1: Use the definition of a median.

Since AD is a median, D is the midpoint of BC.

Therefore:

BD = DC

Step 2: Consider the perpendicular height.

Let h be the perpendicular distance from A to the line BC.

Both triangles ABD and ACD have the same height h.

Step 3: Find the area of △ABD.

Area of △ABD = $$\frac{1}{2}\times BD\times h$$

Step 4: Find the area of △ACD.

Area of △ACD = $$\frac{1}{2}\times DC\times h$$

Step 5: Compare the areas.

Since BD = DC:

$$\frac{1}{2}\times BD\times h=\frac{1}{2}\times DC\times h$$

Therefore:

$$\text{Area of }\triangle ABD=\text{Area of }\triangle ACD$$

Final Answer: A median divides a triangle into two triangles of equal area. Hence proved.

Q15. An equilateral triangular park has a side length of 12 m. Find its area using Heron's formula. Verify the result using the height of the triangle and the Baudhāyana–Pythagoras theorem.

Solution:

Given:

Side of the equilateral triangle = 12 m

Method 1: Using Heron's Formula

Step 1: Calculate the semi-perimeter.

$$s=\frac{12+12+12}{2}$$

$$s=18$$ m

Step 2: Apply Heron's formula.

$$A=\sqrt{s(s-a)(s-b)(s-c)}$$

$$A=\sqrt{18(18-12)(18-12)(18-12)}$$

$$A=\sqrt{18\times6\times6\times6}$$

$$A=\sqrt{3888}$$

$$A=36\sqrt{3}$$ m²

Method 2: Using the Baudhāyana–Pythagoras Theorem

Step 3: Draw an altitude.

Let AD be the altitude from vertex A to side BC.

Since the triangle is equilateral, the altitude bisects BC.

Therefore:

BD = DC = 6 m

AB = 12 m

Step 4: Find the height.

Using the Baudhāyana–Pythagoras theorem in right triangle ABD:

$$AB^2=AD^2+BD^2$$

$$12^2=h^2+6^2$$

$$144=h^2+36$$

$$h^2=108$$

$$h=6\sqrt{3}$$ m

Step 5: Calculate the area.

Area of triangle = $$\frac{1}{2}\times\text{Base}\times\text{Height}$$

$$A=\frac{1}{2}\times12\times6\sqrt{3}$$

$$A=36\sqrt{3}$$ m²

Both methods give the same result.

Final Answer: The area of the equilateral triangular park is $$36\sqrt{3}$$ m², approximately 62.35 m².

Q16. A rectangular field is 60 m long and 40 m wide. A circular pond of radius of 7 m is constructed inside the field. Find (i) the area of the rectangular field, (ii) the area of the pond, (iii) the remaining area, and (iv) the cost of covering the remaining area with grass at ₹20 per square metre.

Solution:

Given:

Length of field = 60 m

Breadth of field = 40 m

Radius of pond = 7 m

Cost of grass = ₹20 per m²

Step 1: Find the area of the rectangular field.

Area = Length × Breadth

= 60 × 40

= 2,400 m²

Step 2: Find the area of the circular pond.

Area of circle = $$\pi r^2$$

$$A=\frac{22}{7}\times7^2$$

$$A=\frac{22}{7}\times49$$

$$A=154$$ m²

Step 3: Find the remaining area.

Remaining area = Area of field − Area of pond

= 2,400 − 154

= 2,246 m²

Step 4: Calculate the cost of covering the remaining area with grass.

Cost = Remaining area × Rate

= 2,246 × 20

= ₹44,920

Final Answer:

(i) Area of rectangular field = 2,400 m²

(ii) Area of circular pond = 154 m²

(iii) Remaining area = 2,246 m²

(iv) Cost of covering the remaining area with grass = ₹44,920

How to Prepare for Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area?

Preparing for Class 9 Maths Chapter 6 becomes easier when students understand the difference between perimeter and area and learn to apply formulas correctly. Regular practice helps improve accuracy and confidence in solving measurement problems.

Follow these simple tips for exam preparation:

Understand the Concepts: Learn how perimeter and area are calculated for different shapes.

Revise Important Formulas: Practise formulas for squares, rectangles, triangles and other shapes covered in the chapter.

Solve Numerical Problems: Work on questions involving dimensions, measurements and real-life situations.

Check Units: Remember to use appropriate units for perimeter and square units for area.

Practise Different Questions: Solve MCQs, short-answer and long-answer questions to improve problem-solving skills.

Benefits of Vedantu's Class 9 Maths Chapter 6 Important Questions PDF

Vedantu's Class 9 Maths Chapter 6 Important Questions PDF helps students revise Measuring Space: Perimeter and Area through focused practice and clear solutions. It brings important problems together, making revision simpler and more organised.

Here's how the PDF supports students:

Chapter-Focused Practice: Revise important concepts without searching through multiple resources.

Step-by-Step Solutions: Understand how to apply formulas and solve problems correctly.

Easy Revision: Revisit important questions before class tests and school exams.

Self-Assessment: Compare answers, identify mistakes and focus on areas needing improvement.

Convenient PDF Access: Download and practise questions anytime, even without an internet connection.

Students can use Vedantu's Important Questions with Answers PDF alongside their NCERT textbook to strengthen their understanding and prepare confidently for the 2026-27 exams.

Important Study Materials for Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area

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