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Class 9th Mathematics Chapter 8 Mensuration: Area and Perimeter – PYQ

Previous year questions (PYQ) with answers for Class 9th Mathematics Chapter 8 Mensuration: Area and Perimeter, based on the NCERT CBSE syllabus.

Topics covered in Mensuration: Area and Perimeter

Frequently asked questions

Where can I find previous year questions (PYQ) for Class 9th Mathematics Chapter 8 Mensuration: Area and Perimeter?

This page lists previous year questions with answers for Mensuration: Area and Perimeter. Topics covered: Heron's formula, Area of triangle, Area of quadrilateral, Area of trapezium, Area of circle.

Are Mensuration: Area and Perimeter PYQs free?

Yes, Notebook gives free previous year questions for Class 9 Mathematics.

How to prepare Mensuration: Area and Perimeter for Class 9 Mathematics exams?

Read the Mensuration: Area and Perimeter notes first, revise the key topics (Heron's formula, Area of triangle, Area of quadrilateral, Area of trapezium, Area of circle), then practise the important questions and previous year questions (PYQ) for this chapter.

Where can I find Class 9th Mathematics Chapter 8 Mensuration: Area and Perimeter pyq for quick revision?

You are on the right page: this pyq page for Mensuration: Area and Perimeter covers Class 9 Mathematics for the CBSE exam 2026-27. Use the tabs on this chapter for notes, important questions and PYQ.

Is Class 9th Mathematics Chapter 8 Mensuration: Area and Perimeter important for the CBSE exam 2026-27?

Mensuration: Area and Perimeter is part of the Class 9 Mathematics NCERT CBSE syllabus. Notebook gives notes, important questions and PYQs for it, free.

Also see: Class 9th Mathematics Chapter 8 Mensuration: Area and Perimeter Notes · Class 9th Mathematics Chapter 8 Mensuration: Area and Perimeter Important Questions

Mensuration: Area and Perimeter

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1 Mark CBSE 2023
One angle of a parallelogram is 70°. Find the remaining three angles.
Ans: Opposite angle = 70°. Adjacent angles are supplementary: 180°−70° = 110°. Remaining angles: 70°, 110°, 70°, 110°.
3 Marks CBSE 2022
Prove that a diagonal of a parallelogram divides it into two congruent triangles.
Ans: In parallelogram ABCD, diagonal AC divides it into △ABC and △ACD. AB∥CD (opposite sides), so ∠BAC = ∠DCA (alternate interior angles). AD∥BC, so ∠DAC = ∠BCA. AC = AC (common). By ASA: △ABC ≅ △CDA. Hence proved.
5 Marks CBSE 2021
State and prove the Mid-Point Theorem. Use it to show that the line joining midpoints of a quadrilateral forms a parallelogram.
Ans: Mid-Point Theorem: Line joining midpoints of two sides of a triangle is parallel to the third side and half its length. (Full proof as stated earlier.) In quadrilateral ABCD, let P, Q, R, S be midpoints of AB, BC, CD, DA. In △ABD: SR∥BD and SR=BD/2. In △BCD: PQ∥BD and PQ=BD/2. So PQ∥SR and PQ=SR. Hence PQRS is a parallelogram (one pair of opposite sides equal and parallel).
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