This MCQ module is based on: Tangram — Figure it Out (Page 139)
TOPIC 23 OF 41
Tangram — Figure it Out (Page 139)
🎓 Class 6
Mathematics
CBSE
Theory
Ch 6 — Perimeter and Area
⏱ ~18 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Tangram — Figure it Out (Page 139)
Targeting Class 6 level in Mensuration, with Basic difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
Tangram — Figure it Out (Page 139)
Cut out the tangram? pieces given at the end of your textbook. Seven pieces, labelled A to G, fit inside one big square.
The seven-piece tangram puzzle.
Q1. Explore and figure out how many pieces have the same area.
Pieces A and B (the two big triangles) have equal area. Pieces C and E have equal area. Pieces D, F, G each equal one another.
Q2. How many times bigger is Shape D compared to Shape C? What is the relationship between Shapes C, D and E?
Shape D = Shape C in area (both are small triangles). Shape E can be made by joining two of D's area units → E is twice D (or C).
Q3. Which shape has more area: Shape D or Shape F?
They are equal. Trace one over the other and they cover the same region.
Q4. Which shape has more area: Shape F or Shape G?
Equal — both are small triangles congruent to each other.
Q5. How large is the area of Shape A compared to Shape G? Is it twice as big? Four times as big?
Shape A is the big triangle. It can be exactly covered by 4 copies of the small triangle G. So Shape A = 4 × Shape G.
Q6. Can you now figure out the area of the big square formed with all seven pieces?
Taking piece G (smallest triangle) as 1 unit: A = B = 4, C = E = 2, D = F = 1 → total = 4 + 4 + 2 + 2 + 1 + 1 + 1 = 16 units. The big square's area is 16 small-triangle units (equivalent to 8 square-D units since two G's make one D-equivalent square).
Q7. Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle in terms of the area of Shape C?
Area of the rectangle = Area of the big square = 16 small-triangle units = 8 × Shape C (since Shape C = 2 small-triangle units).
Q8. Are the perimeters of the square and the rectangle formed from these 7 pieces the same or different? Explain.
Different. Same area does not imply same perimeter — the rectangle is usually longer and thinner than the square, so its perimeter is larger. A square enclosing a given area has the smallest perimeter among all rectangles of that area.
Figure it Out — Page 144 (Areas by Splitting)
Q1. Find the areas of the figures below by dividing them into rectangles and triangles (results from NCERT):
(a) 24 sq units. (b) 30 sq units. (c) 48 sq units. (d) 16 sq units. (e) 12 sq units.
Method: split each figure into non-overlapping rectangles (and triangles, which equal half of an enclosing rectangle), compute individual areas, and sum.
Method: split each figure into non-overlapping rectangles (and triangles, which equal half of an enclosing rectangle), compute individual areas, and sum.
Figure it Out — Page 145 (9 Unit Squares)
Q1. Smallest perimeter possible with 9 unit squares?
12 units — a 3 × 3 square.
Q2. Largest perimeter possible?
20 units — a 1 × 9 strip.
Q3. Make a figure with perimeter 18 units.
Yes — for example, a 3-unit row on top of a 3-unit row with a jog and one square tacked on; many valid shapes exist.
Q4. Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter?
Yes, except for the smallest perimeter (12 units — uniquely the 3×3 square). For all other perimeters many shapes are possible.
Figure it Out — Page 148 (Area Maze)
Find the missing length/area in each figure.
Area-maze (a) — find the shaded area.
Q (a), (b), (c), (d). Find each missing value.
(a) Top row: 13 + 26 = 39 sq cm over width w. Same width is split 13:26, so the lower rectangles share the same ratio. Lower total = 15 + ? with the same dimensions ratio → ? = 30 sq cm.
(b) ? = 9 sq cm.
(c) ? = 16 sq cm.
(d) ? = 5 cm.
(b) ? = 9 sq cm.
(c) ? = 16 sq cm.
(d) ? = 5 cm.
Figure it Out — Page 149 (Chapter-End)
Q1. Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m.
Total area = 50 + 14 = 64 m². Possible dimensions: 8 m × 8 m (square) or 4 m × 16 m, etc.
Q2. The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
Width = 1000 ÷ 50 = 20 m.
Q3. The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted.
Floor = 5 × 4 = 20 m². Carpet = 3 × 3 = 9 m². Uncovered = 11 m².
Q4. Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?
Garden = 15 × 12 = 180 m². Four beds = 4 × (2 × 1) = 8 m². Lawn = 180 − 8 = 172 m².
Q5. Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the given conditions.
Possible: Shape A = 1 × 18 strip (P = 38); Shape B = 4 × 5 rectangle (P = 18). Also A as 2 × 9 (P = 22) with B as 4 × 5 (P = 18). Corresponding perimeters from NCERT answer: P(A) = 22 or 38 m, P(B) = 18 m.
Q6. On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border?
Given a page of length L cm and width W cm, border dimensions are (L − 2) × (W − 3). Perimeter = 2[(L − 2) + (W − 3)] = 2(L + W) − 10 cm. For a typical 22 × 18 cm page the border is 20 × 15, perimeter = 70 cm.
Q7. Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle, that occupies exactly half the area.
Outer area = 12 × 8 = 96 sq units. Inner rectangle must have area 48 sq units. Possible: 8 × 6, or 12 × 4, or 4 × 12 (inside the outer, leaving a border).
Q8. A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true?
(a) The area of each rectangle is larger than the area of the square.
(b) The perimeter of the square is greater than the perimeters of both rectangles added together.
(c) The perimeters of both the rectangles added together is always 1½ times the perimeter of the square.
(d) The area of the square is always three times as large as the areas of both rectangles added together.
(a) The area of each rectangle is larger than the area of the square.
(b) The perimeter of the square is greater than the perimeters of both rectangles added together.
(c) The perimeters of both the rectangles added together is always 1½ times the perimeter of the square.
(d) The area of the square is always three times as large as the areas of both rectangles added together.
Only (c) is true. If the square has side \(s\), its perimeter is \(4s\). Each rectangle is \(s \times s/2\), with perimeter \(2(s + s/2) = 3s\). Sum of rectangle perimeters = \(6s = \tfrac{6s}{4s}\times 4s = 1.5 \times 4s\). So the two rectangles together have perimeter 1½ times that of the square.
Summary — Perimeter & Area
- The perimeter of a polygon is the sum of the lengths of all its sides.
- The perimeter of a rectangle is twice the sum of its length and width: \(P = 2(L + B)\).
- The perimeter of a square is four times the length of any one of its sides: \(P = 4s\).
- The area of a closed figure is the measure of the region enclosed by the figure.
- Area is measured in square units (cm², m², km²).
- The area of a rectangle is its length times its width: \(A = L \times B\).
- The area of a square is the length of any one of its sides multiplied by itself: \(A = s^2\).
- Two closed figures can have the same area with different perimeters, or the same perimeter with different areas.
- Areas of regions can be estimated (or even determined exactly) by breaking up unit squares, or into more generally-shaped rectangles and triangles whose areas can be calculated.
Activity: Designer's Challenge
L3 Apply
Materials: Graph paper, coloured pencils, ruler
Predict: You have 36 unit squares. Can you create a shape with area 36 but perimeter 26? What about perimeter 30?
- Start with 36 unit squares in a 6 × 6 square arrangement — note its perimeter.
- Rearrange into 4 × 9, 3 × 12, 2 × 18, 1 × 36 strips. Record the perimeter of each.
- Now try L-shapes and T-shapes — experiment.
- Tabulate: shape type → perimeter.
- Conclude: for a fixed area, which shape minimises the perimeter?
6 × 6 gives P = 24; 4 × 9 → P = 26; 3 × 12 → P = 30; 2 × 18 → P = 40; 1 × 36 → P = 74. For a fixed area the square has the smallest perimeter.
Competency-Based Questions
Scenario: Municipal engineers are designing a 40 m × 25 m rectangular park. A footpath of uniform width 1 m runs all around the inside of the boundary. Inside the remaining space they plan a central square lawn of side 20 m, and the rest will be flower beds.
Q1. Apply — what is the area of the footpath?
L3 ApplyOuter rectangle = 40 × 25 = 1000 m². Inner (inside path) = 38 × 23 = 874 m². Path = 1000 − 874 = 126 m².
Q2. Analyse — what area is available for flower beds (inside the path, outside the central lawn)?
L4 AnalyseInside-path region = 874 m². Central lawn = 20 × 20 = 400 m². Flower beds = 874 − 400 = 474 m².
Q3. Evaluate — the engineers estimate ₹200/m² for the lawn, ₹150/m² for flower beds and ₹80/m² for paving. Total cost?
L5 EvaluateLawn = 400 × 200 = ₹80,000. Beds = 474 × 150 = ₹71,100. Path = 126 × 80 = ₹10,080. Total = ₹1,61,180.
Q4. Create — redesign so that the footpath area is exactly 200 m² while keeping the outer dimensions 40 m × 25 m. What uniform width must the path have?
L6 CreateLet path width = w. Inner rectangle = (40 − 2w)(25 − 2w). Path area = 1000 − (40 − 2w)(25 − 2w) = 200. So (40 − 2w)(25 − 2w) = 800. Expand: 1000 − 130w + 4w² = 800, or 4w² − 130w + 200 = 0 → 2w² − 65w + 100 = 0. Solving gives w ≈ 1.61 m (taking the smaller root).
Assertion–Reason Questions
A: Two shapes with the same area must have the same perimeter.
R: Area and perimeter are two different measures of a figure.
R: Area and perimeter are two different measures of a figure.
Answer: (d). A is false — e.g. 1×36 (P=74) and 6×6 (P=24) both have area 36. R is true.
A: When a square of side \(s\) is cut into two equal rectangles, the sum of their perimeters is \(1.5 \times\) the square's perimeter.
R: Each rectangle has perimeter \(3s\); together they have \(6s = 1.5 \times 4s\).
R: Each rectangle has perimeter \(3s\); together they have \(6s = 1.5 \times 4s\).
Answer: (a).
A: For a fixed perimeter among rectangles, the square gives the maximum area.
R: The product of two numbers whose sum is fixed is largest when the two numbers are equal.
R: The product of two numbers whose sum is fixed is largest when the two numbers are equal.
Answer: (a). For a rectangle with perimeter P, L + B = P/2 is fixed, and area L × B is maximised when L = B (i.e. a square).
Frequently Asked Questions — Perimeter and Area
What is Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool in NCERT Class 6 Mathematics?
Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool is a key concept covered in NCERT Class 6 Mathematics, Chapter 6: Perimeter and Area. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool step by step?
To solve problems on Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 6 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 6: Perimeter and Area?
The essential formulas of Chapter 6 (Perimeter and Area) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool important for the Class 6 board exam?
Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool is part of the NCERT Class 6 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool?
Common mistakes in Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool?
End-of-chapter NCERT exercises for Part 4 — Exercises & Summary | Class 6 Maths Ch 6 Perimeter & Area | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.
AI Tutor
Mathematics Class 6 — Ganita Prakash
Ready