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Measuring Space: Perimeter and Area — Class 9 MCQs with Answers

Class 9 CBSE Mathematics · Chapter 6

90 practice questions · 30 Easy · 30 Medium · 30 Hard · Updated

Practise the most important Class 9 CBSE Mathematics questions from Chapter 6, "Measuring Space: Perimeter and Area". You get 9 timed quizzes made from 90 NCERT-based MCQs, with answers and explanations. The questions are split into 30 Easy, 30 Medium and 30 Hard. Warm up on the basics, then move on to the exam-level questions that top scorers in CBSE Board exams and the JEE & NEET foundation years get right.

To score well in "Measuring Space: Perimeter and Area", focus on fast problem-solving, formula recall and step-by-step working. Each MCQ here is timed and uses exam-style marking (+4 correct, −1 wrong, 0 skipped). This helps you stay accurate when time is short, just as real papers need. Every question has a short explanation, so a wrong answer becomes a quick lesson. It is the fastest way to fix gaps before a test.

Use this chapter for focused revision. Start with the Easy set to check your basics on Measuring Space: Perimeter and Area, then move to Medium and Hard to practise applying them. Your accuracy, streaks and XP save automatically. This chapter also adds to your overall Class 9 Mathematics mastery score. Try the 6 sample questions below: tap an option to see the answer and a worked explanation. Sign in free to start practising.

Measuring Space: Perimeter and Area — important questions & MCQs with answers (Class 9 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

  1. Q1Easy

    Find the area of a rectangle 12 cm by 5 cm.

    A shaded rectangle with its length marked 12 cm along the bottom and its breadth marked 5 cm up the left side.

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    Answer: 60 cm²

    Explanation: Area of a rectangle = length × breadth = 12 × 5 = 60 cm². Adding the sides gives 17 cm², and doubling that (the perimeter) gives 34 cm² — both add the sides instead of multiplying them.

  2. Q2Easy

    Find the perimeter of a rectangle 13 m by 7 m.

    A rectangle with its length marked 13 m along the bottom and its breadth marked 7 m up the left side.

    Show answer

    Answer: 40 m

    Explanation: Perimeter = 2(length+breadth) = 2(13+7) = 40 m. Forgetting to double the sum gives just 20 m, and multiplying the sides instead gives the area, 91 m² — neither is the perimeter.

  3. Q3Medium

    Find the area of a sector of a circle of radius 7 cm with central angle 90°. (Take π = 22/7.)

    A circle of radius 7 cm drawn with a dashed outline. Two radii meet at the centre making an angle of 90°, and the sector between them is shaded. One radius is marked 7 cm.

    Show answer

    Answer: 38.5 cm²

    Explanation: Sector area = (θ/360)×πr² = (90/360) × 22/7 × 7² = 38.5 cm². Forgetting the θ/360 fraction gives the full circle's area, 154 cm², instead of just the 90° slice.

  4. Q4Medium

    Using Heron's formula, find the area of a triangle with sides 13 cm, 14 cm and 15 cm.

    A shaded triangle whose three sides are marked 13 cm, 14 cm and 15 cm.

    Show answer

    Answer: 84 cm²

    Explanation: Heron's formula: s = (13+14+15)/2 = 21 cm. Area = √(s(s−a)(s−b)(s−c)) = √(21×8×7×6) = 84 cm². Using the full perimeter, 42 cm, as s instead of halving it is the usual slip.

  5. Q5Hard

    Find the area of an equilateral triangle of side 6 cm (leave the answer in surd form).

    A shaded equilateral triangle: its three sides carry the same tick mark, and the base is marked 6 cm.

    Show answer

    Answer: 9√3 cm²

    Explanation: Area of an equilateral triangle = (√3/4)×a² = (√3/4)×6² = 9√3 cm². Dropping the √3 gives just 9 (a plain number), and forgetting to divide by 4 gives 36√3 instead.

  6. Q6Hard

    A circular ring has outer radius 14 cm and inner radius 7 cm. Find the area of the ring. (π = 22/7.)

    A shaded circular ring between two circles that share a centre. The outer radius is drawn to the right and marked 14 cm; the inner radius is drawn down-left and marked 7 cm.

    Show answer

    Answer: 462 cm²

    Explanation: Ring area = π(R²−r²) = 22/7×(14²−7²) = 22/7×147 = 462 cm². Squaring the difference of radii instead, π(14−7)² = 154 cm², undercounts badly — subtract the squares, don't square the subtraction.

Key concepts: Measuring Space: Perimeter and Area (Class 9 Mathematics)

Perimeter measures the boundary of a figure and area the space inside it. This chapter gathers the area formulas for the standard polygons, adds Heron's formula for a triangle from its three sides and Brahmagupta's formula for a cyclic quadrilateral, and handles circles, sectors and composite shapes.

Perimeter
The total length of the boundary of a closed figure, measured in units of length. For a circle the boundary length is called the circumference.
Area
The amount of surface a figure encloses, measured in square units. Doubling every length multiplies the area by four, not by two.
Two squares side by side. The small square has side a and its area is marked A. The big square has side 2a, twice as long; lines split it into four squares the size of the small one, each marked A, so its area is 4A: doubling every length multiplies the area by four.
Area of a triangle (base and height)
Half the base times the perpendicular height. The height must be perpendicular to the chosen base, not the slanted side.
A triangle on a horizontal base marked b. A dashed line from the top corner straight down to the base, with a right-angle mark where it meets the base, is marked h, the height. The long sloping side is marked as a slant side, which is not the height. Below: Area = ½ × b × h.
Heron's formula
Gives a triangle's area from its three sides alone, using the semi-perimeter s. Essential when no height is known or easily found.
Semi-perimeter
Half the perimeter, written s. It is the quantity both Heron's and Brahmagupta's formulas are built from.
Area of an equilateral triangle
A special case worth memorising: (√3/4) × side², which is far quicker than applying Heron's formula each time.
Brahmagupta's formula
Gives the area of a cyclic quadrilateral from its four sides, using the semi-perimeter. It reduces to Heron's formula when one side shrinks to zero.
A quadrilateral ABCD drawn inside a circle with all four corners A, B, C and D on the circle, which makes it a cyclic quadrilateral. Its sides are marked a, b, c and d going round. Below: s = (a + b + c + d)/2.
Area of a parallelogram
Base times perpendicular height. A parallelogram and a triangle on the same base and between the same parallels are related by a factor of two.
A parallelogram ABCD on base AB, marked b. A shaded triangle ABE stands on the same base AB with its top corner E on the side DC, so it lies between the same two parallel lines. A dashed line from E straight down to AB, with a right-angle mark, is marked h: the height of both shapes. Below: parallelogram = b × h, triangle ABE = ½ × b × h.
Area of a rhombus
Half the product of its diagonals, since the diagonals of a rhombus bisect each other at right angles.
A rhombus with its four equal sides marked by matching ticks. Its two diagonals cross at O at right angles; each diagonal is cut into two equal halves at O (matching tick marks). Dimension lines mark the horizontal diagonal d₁ and the vertical diagonal d₂. Below: Area = ½ × d₁ × d₂.
Area of a trapezium
Half the sum of the two parallel sides, multiplied by the distance between them.
A trapezium whose top and bottom sides are parallel (matching arrow marks). The top parallel side is marked a and the bottom parallel side b. A dashed line from the top side straight down to the bottom, with a right-angle mark, is marked h, the distance between them. Below: Area = ½ (a + b) × h.
Circumference and area of a circle
Circumference is 2πr and area is πr². Confusing the two is easy — one is a length, the other a square measure.
Sector of a circle
A slice bounded by two radii and an arc. Its arc length and area are the fraction θ/360 of the whole circle's.
A circle with centre O drawn dashed. Two radii OA and OB, each marked r, make an angle θ at the centre, and the slice between them is shaded: that slice is a sector. The curved edge between A and B is labelled arc. Below: arc = (θ/360) × 2πr and area = (θ/360) × πr².
Annulus (ring)
The region between two concentric circles; its area is the outer circle's area minus the inner one's, π(R² − r²).
Two circles with the same centre O. The ring between them is shaded. The radius of the outer circle is drawn to the right and marked R; the radius of the inner circle is drawn down-left and marked r. Below: Area of ring = π(R² − r²).
Composite figures
Shapes built from standard pieces. Split them into known figures and add, or subtract a removed piece from a surrounding shape.
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Measuring Space: Perimeter and Area — FAQs

What are the key concepts in Class 9 Mathematics Measuring Space: Perimeter and Area?+

Perimeter measures the boundary of a figure and area the space inside it. This chapter gathers the area formulas for the standard polygons, adds Heron's formula for a triangle from its three sides and Brahmagupta's formula for a cyclic quadrilateral, and handles circles, sectors and composite shapes. Key ideas include Perimeter, Area, Area of a triangle (base and height), Heron's formula, Semi-perimeter, Area of an equilateral triangle.

What does Class 9 Mathematics Chapter 6 (Measuring Space: Perimeter and Area) cover on XamBaaz?+

It has 90 NCERT-based MCQs on "Measuring Space: Perimeter and Area": 30 Easy, 30 Medium and 30 Hard. Together they make 9 timed quizzes, and you never get the same set twice. Every question has an instant explanation. They help you prepare for CBSE Board exams and the JEE & NEET foundation years.

Are these "Measuring Space: Perimeter and Area" questions free to practise?+

Yes. Sign in with Google to practise "Measuring Space: Perimeter and Area" free. Full unlimited access is ₹999/year. One year from the day you pay. You stay in Class 9 till 31 March; on 1 April your account moves up to Class 10 and the rest of your year carries over. No chapter is charged separately.

How should I revise "Measuring Space: Perimeter and Area" for the exam?+

Start with the Easy quiz to check your basics, then try Medium and Hard to practise applying them. There are 9 timed quizzes on this chapter, so you can come back for a fresh set instead of one you have seen. Read each explanation, retry the questions you miss, and track your accuracy until it stays high.

Are these "Measuring Space: Perimeter and Area" MCQs available with answers?+

Yes. 6 sample questions are on this page with answers: tap an option, or "Show answer", to see the correct option and a step-by-step explanation. Sign in free with Google to start practising, with instant scoring.

Is there negative marking in the "Measuring Space: Perimeter and Area" quizzes?+

Yes. The timed quizzes use exam-style marking: +4 for a right answer, −1 for a wrong one and 0 for a skip. MHT-CET and CBSE board papers have no negative marking. Our mocks for those are scored their way.

What are the important questions from Measuring Space: Perimeter and Area (Class 9 Mathematics)?+

The questions that matter most test Perimeter, Area, Area of a triangle (base and height), Heron's formula, Semi-perimeter, Area of an equilateral triangle. This page shows 6 solved important MCQs with answers and explanations. Sign in to practise all 90 questions on the chapter as timed quizzes.

Is there an online quiz for Measuring Space: Perimeter and Area?+

Yes — Class 9 Mathematics Measuring Space: Perimeter and Area has timed online quizzes at Easy, Medium and Hard levels, with instant scoring and a worked explanation on every question. The first quiz on the chapter is free.

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