Class 6 Maths Ganita Prakash Chapter 8 Solutions (NCERT 2026–27) – Playing with Constructions

These Class 6 Maths Ganita Prakash Chapter 8 solutions cover Playing with Constructions from the new NCF textbook (Reprint 2026–27). Every Figure it Out, Math Talk and Try This task is solved step by step, with the construction steps written clearly in words and all measurements verified, so you can master the ruler-and-compass work and revise it quickly.

Class: 6 Subject: Mathematics Book: Ganita Prakash Chapter: 8 Sections: 8.1 Artwork – 8.6 Equidistant Points Session: 2026–27
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Chapter 8 Overview

Chapter 8 of Ganita Prakash, Playing with Constructions, teaches you to draw accurate figures using just a ruler and a compass. It begins with freehand artwork (a person, a wavy wave, eyes), then introduces the circle — the set of all points at a fixed distance (the radius) from a centre. You learn the defining properties of squares and rectangles, how to construct them using perpendiculars and a compass, how to explore points moving along their sides, and how a diagonal divides the corner angles. The chapter closes with the ‘House’ construction, which uses the idea of points equidistant from two given points. The Class 6 Maths Ganita Prakash Chapter 8 solutions below work through every Figure it Out, Math Talk, Think and Try This task step by step.

Key Concepts & Definitions

Curve: any shape that can be drawn on paper with a pencil — this includes straight lines, circles and other figures.

Circle: the set of all points that are the same distance from a fixed point. If you mark every point that is 4 cm from a point P, they together form a circle.

Centre & radius: the fixed point P is the centre of the circle; the equal distance from the centre to any point on the circle is the radius.

Compass: the tool used to draw circles and arcs. Open it against a ruler so the gap between its tip and the pencil equals the radius, keep the tip fixed and rotate the pencil.

Rectangle: a 4-sided figure with corners, sides and angles, where (R1) opposite sides are equal in length and (R2) all angles are 90°.

Square: a 4-sided figure where (S1) all sides are equal and (S2) all angles are 90°. Rotating a square (or rectangle) does not change its sides or angles, so it stays a square (or rectangle).

Diagonal: a line joining two opposite corners of a rectangle, e.g. PR and QS in rectangle PQRS.

Important Construction Ideas (Chapter 8)

Drawing a circle: fix the compass tip at the centre, open the radius against a ruler, and rotate the pencil all the way round.

Valid name of a figure: the corners must be listed in the order you travel around the figure (e.g. PQRS, QRSP, RSPQ, SPQR), starting from any corner — not jumping across, like PQSR.

Constructing a square/rectangle: draw one side, raise a perpendicular (90°) at each end, mark the second side’s length with a ruler or compass, then join.

Side and diagonal given: draw the side, raise a perpendicular at one end, then with the other end as centre draw an arc of radius = diagonal; where it cuts the perpendicular is the next corner.

Equidistant point: a point at distance r from two points B and C is found where two arcs of radius r (centred at B and at C) intersect.

Rectangle that splits into k squares: make its length = k × its breadth (e.g. length = 3 × breadth for 3 identical squares).

Figure it Out — Wavy Wave & the Circle (Section 8.1, Page 188–191)

Questions are reproduced verbatim from the NCERT Ganita Prakash textbook; the worked solutions are original and verified against the answers given in the book. Figure-only tasks are answered in words.

Think (p. 188): Imagine marking all the points of 4 cm distance from the point P. How would they look?

SOLUTION Mark several points that are each 4 cm from P, in every direction. As you add more and more points, they fall along one smooth curve. On joining them, the points form a circle with P as its centre and 4 cm as its radius. You can draw this in one go with a compass opened to 4 cm.

Wavy Wave — Figure it Out (p. 191) 1. What radius should be taken in the compass to get this half circle? What should be the length of AX?

SOLUTION The central line is AB = 8 cm, and the first wave is a half circle (semicircle) drawn on the first part of AB, from A up to the mid-point X. A half circle (semicircle) sits on a diameter, so its diameter is AX. The radius of the half circle = AX ÷ 2. For two equal waves across AB = 8 cm, take AX = 4 cm, so the diameter of each half circle is 4 cm. ∴ radius = 2 cm and AX = 4 cm (the compass is opened to 2 cm and the tip is placed at the mid-point of AX for each wave).

2. Take a central line of a different length and try to draw the wave on it.

SOLUTION (activity) Step 1. Draw a fresh central line, say PQ = 6 cm, and mark its mid-point M (3 cm from each end). Step 2. For the first wave, take the mid-point of PM as centre and radius 1.5 cm; draw a half circle above the line from P to M. Step 3. For the second wave, take the mid-point of MQ as centre and the same radius 1.5 cm; draw a half circle below the line from M to Q. The two equal half circles give a neat wavy curve on the 6 cm line. (Any length works — keep the radius equal to one-quarter of the line for two waves.)

3. Try to recreate the figure where the waves are smaller than a half circle (as appearing in the neck of the figure, ‘A Person’). The challenge here is to get both the waves to be identical. This may be tricky!

SOLUTION (activity) A wave smaller than a half circle is a shallow arc, not a full semicircle. Step 1. Mark the two end points of each wave on the central line, equal distances apart. Step 2. Fix one large radius in the compass (bigger than half the distance between the two end points). Place the tip below the line and draw a short arc joining the two end points — this gives a shallow wave. Step 3. Use the same radius and the same arc-height for the next wave so both waves are identical; only the centre is shifted along. Keeping the radius constant is the trick to making the waves match.

Squares, Rectangles & their Construction (Sections 8.2–8.3, Page 193–201)

Q (p. 193): Which of the following is not a name for this square? 1. PQSR   2. SPQR   3. RSPQ   4. QRSP

SOLUTION A valid name lists the corners in the order you travel around the figure. Going round the square the corners are P → Q → R → S (or the same loop started elsewhere): PQRS, QRSP, RSPQ, SPQR are all valid. In PQSR, after P and Q the next corner jumps to S (skipping R) — this is not the order of travel. ∴ PQSR (option 1) is not a name for the square.

Figure it Out (p. 194) 1. Draw the rectangle and four squares configuration (shown in Fig. 8.3) on a dot paper. What did you do to recreate this figure so that the four squares are placed symmetrically around the rectangle? Discuss with your classmates.

SOLUTION (activity) First draw the central rectangle so its corners sit on the dots. Then attach one square to each side. To keep them symmetric, leave the same one-dot gap diagonally at each side and make the squares the same size, so each square is placed the same way relative to its side of the rectangle. Equal sizes and equal gaps on opposite sides give the symmetric look.

2. Identify if there are any squares in this collection. Use measurements if needed. (figures A, B, C, D on a dot grid)

SOLUTION Check each figure for the two square properties: all four sides equal and all four angles 90°. On measuring (or by counting dot positions), only figure A has all sides equal and all right angles. ∴ A is a square; the others fail at least one property.

3. Draw at least 3 rotated squares and rectangles on a dot grid. Draw them such that their corners are on the dots. Verify if the squares and rectangles that you have drawn satisfy their respective properties.

SOLUTION (activity) Step 1. On a dot grid, draw a tilted (rotated) square — for example, move 2 dots right and 1 dot up for each side, turning 90° each time so the four corners land on dots. Step 2. Draw two more tilted figures the same way, making one a rectangle (different ‘right’ and ‘up’ steps for the two side-directions). Step 3. Verify: measure the four sides and four angles. Each square still has 4 equal sides and 4 right angles; each rectangle still has equal opposite sides and 4 right angles. Rotating does not change lengths or angles, so the properties hold.

Construct (p. 197) — checking rectangle properties

1. Draw a rectangle with sides of length 4 cm and 6 cm. After drawing, check if it satisfies both the rectangle properties.

SOLUTION Step 1. Draw AB = 4 cm with a ruler. Step 2. At A and at B, draw perpendiculars (90°) to AB. Step 3. On each perpendicular mark 6 cm: mark D on the line through A with AD = 6 cm, and C on the line through B with BC = 6 cm. Step 4. Join D to C to complete rectangle ABCD. Check. ∠A = ∠B = ∠C = ∠D = 90° (property R2), and opposite sides are equal: AB = CD = 4 cm, AD = BC = 6 cm (property R1). Both properties hold.

2. Draw a rectangle of sides 2 cm and 10 cm. After drawing, check if it satisfies both the rectangle properties.

SOLUTION Step 1. Draw PQ = 10 cm with a ruler. Step 2. At P and at Q, draw perpendiculars to PQ. Step 3. Mark S on the perpendicular at P with PS = 2 cm, and R on the perpendicular at Q with QR = 2 cm. Step 4. Join S to R to complete rectangle PQRS. Check. ∠P = ∠Q = ∠R = ∠S = 90°, and PQ = SR = 10 cm, PS = QR = 2 cm. Both rectangle properties are satisfied.

3. Is it possible to construct a 4-sided figure in which— • all the angles are equal to 90º but • opposite sides are not equal?

SOLUTION No. If all four angles of a 4-sided figure are 90°, the opposite sides become parallel and the figure closes up only when opposite sides are equal. If you keep all angles 90° but try to make a pair of opposite sides unequal, the last side cannot meet to close the figure. So such a figure cannot be constructed — a 4-sided figure with all right angles is always a rectangle (with equal opposite sides).

Exploration, Diagonals & the House (Sections 8.4–8.6, Page 197–215)

Q (p. 198, Section 8.4): Is there a shorthand way of writing it down? Record the length XY for the given positions of X (from A) and Y (from B) on rectangle ABCD with AB = 7 cm, BC = 4 cm.

SOLUTION Yes — record the three changing quantities (distance of X from A, distance of Y from B, length of XY) in a table:
Distance of X from ADistance of Y from BLength of XY
5 mm3 cm7.4 cm
1 cm1 cm7 cm
2 cm4 cm7.3 cm

Q (p. 199): Have you checked what happens to the length XY when X and Y are placed at the same distance away from A and B, respectively?

SOLUTION When X and Y are the same distance from A and B, the figure ABYX has both pairs of opposite sides equal and stays a rectangle, so XY always equals AB:
Distance of X from ADistance of Y from BLength of XY
5 mm5 mm7 cm
1 cm1 cm7 cm
1 cm 5 mm1 cm 5 mm7 cm

Q (p. 199): In each of these cases, observe (i) how the length XY compares to that of AB, and (ii) the shape of the 4-sided figure ABYX.

SOLUTION (i) XY = AB — the distance XY equals the length of AB (7 cm) in every such case. (ii) ABYX is a rectangle — equal distances make AX and BY equal and both perpendicular to AB, so ABYX has all the rectangle properties.

Q (p. 199): How does the farthest distance between X and Y compare with the length of AC? BD?

SOLUTION X and Y are farthest apart when they sit at the opposite corners of the rectangle, so XY then lies along a diagonal. ∴ the farthest distance between X and Y is equal to AC (or BD), the diagonals of the rectangle.

Construct (p. 199–201) — Breaking Rectangles & more

Construct (p. 199): Construct a rectangle that can be divided into 3 identical squares as shown in the figure.

SOLUTION Three identical squares placed in a row need the rectangle’s length to be three times its breadth. Step 1. Choose a breadth, say 3 cm, so the length must be 3 × 3 = 9 cm. Step 2. Construct rectangle of 9 cm × 3 cm (draw a side, raise perpendiculars at both ends, mark the other side, join). Step 3. Mark points at 3 cm and 6 cm along the long side; from these, draw lines perpendicular to the length. This cuts the rectangle into 3 identical 3 cm × 3 cm squares. (Hint: take length = 3 × breadth.)

Give the lengths of the sides of a rectangle that cannot be divided into — • two identical squares; • three identical squares.

SOLUTION A rectangle splits into two identical squares only if length = 2 × breadth, and into three only if length = 3 × breadth. Choose sides that break those rules. Cannot be divided into two identical squares: take length = 4 cm, breadth = 2.5 cm (since 4 ≠ 2 × 2.5 = 5). (Try for more!) Cannot be divided into three identical squares: take length = 7 cm, breadth = 2 cm (since 7 ≠ 3 × 2 = 6). (Other possibilities also work.)

Construct (p. 201) 4. Square with a Hole. (The circular hole is the same as the centre of the square. Hint: think where the centre of the circle should be.)

SOLUTION The hole must be centred at the centre of the square. Step 1. Construct the square and draw both diagonals (joining opposite corners). They cross at the centre. Step 2. Place the compass tip at this crossing point and draw a small circle of any chosen radius — that is the hole. So the centre of the circle is at the meeting point of the two diagonals (the line segments connecting opposite vertices).

Section 8.5 — Diagonals; Construct (p. 204–211)

Explore (p. 204): How should the rectangle be constructed so that the diagonal divides the opposite angles into equal parts?

SOLUTION A diagonal splits each 90° corner into two angles. They are equal (45° each) only when the two sides meeting at the corner are equal. ∴ make the two adjacent sides equal, i.e. construct a square. In a square the diagonal divides each right angle into two equal 45° parts.

Construct (p. 211) 1. Construct a rectangle in which one of the diagonals divides the opposite angles into 50° and 40°.

SOLUTION Step 1. Draw the base AB of any convenient length. Step 2. At A, draw a ray making 50° with AB — the diagonal AC will lie along it. Step 3. At A, draw AD perpendicular (90°) to AB. The remaining part of ∠A is 90° − 50° = 40° (check). At B, draw BC perpendicular to AB. Step 4. Mark C where the 50° ray meets the perpendicular at B; then draw a perpendicular to BC at C to fix D. Join DC. ABCD is the required rectangle; diagonal AC divides ∠A into 50° and 40° (and the opposite ∠C the same way).

2. Construct a rectangle in which one of the diagonals divides the opposite angles into 45° and 45°. What do you observe about the sides?

SOLUTION Step 1. Draw base AB. Step 2. At A, draw a ray at 45° to AB (the diagonal). At A and B draw perpendiculars to AB. Step 3. Mark C where the 45° ray meets the perpendicular at B, fix D, and join DC. Observation: the diagonal makes equal 45° parts only when the two adjacent sides are equal, so all four sides are equal — the rectangle is a square.

3. Construct a rectangle one of whose sides is 4 cm and the diagonal is of length 8 cm.

SOLUTION Step 1. Draw the side DC = 4 cm. Step 2. At C, draw a perpendicular line l to DC (the other corner B lies on this line). Step 3. With D as centre and radius 8 cm (the diagonal), draw an arc cutting line l at B; then DB = 8 cm. Step 4. Draw perpendiculars to DC at D and to CB at B; where they meet is A. ABCD is the rectangle with side 4 cm and diagonal 8 cm. (Check R1 and R2.)

4. Construct a rectangle one of whose sides is 3 cm and the diagonal is of length 7 cm.

SOLUTION Step 1. Draw the side DC = 3 cm. Step 2. At C, draw a perpendicular line l to DC. Step 3. With D as centre and radius 7 cm, draw an arc cutting l at B (so the diagonal DB = 7 cm). Step 4. Draw perpendiculars at D (to DC) and at B (to CB); their meeting point is A. ABCD is the required rectangle. Verify all angles are 90° and opposite sides equal.

Section 8.6 — The House (Construct, p. 215)

1. Construct a bigger house in which all the sides are of length 7 cm.

SOLUTION Step 1. Draw the base square/rectangle walls with each border line 7 cm, following the house layout (the lower square of side 7 cm, with the small window/door marks as in the figure). Step 2. Let B and C be the two top corners of the walls (BC = 7 cm). To find the roof apex A, take radius 7 cm in the compass and draw an arc from B and another from C. Step 3. The point where the two arcs meet is A, since A is 7 cm from both B and C. Join A to B and A to C. All sides, including the two roof slopes AB and AC, are 7 cm — a bigger house, ready.

2. Try to recreate ‘A Person’, ‘Wavy Wave’, and ‘Eyes’ from the section ‘Artwork’, using ideas involved in the ‘House’ construction.

SOLUTION (activity) The key idea from the House is locating a point at a fixed distance from two given points using two arcs. Wavy Wave: draw equal half-circles/arcs on a base line, keeping the radius constant so the waves match. A Person & Eyes: draw the straight parts first, then fix the compass to a suitable radius and use estimated centres (placed symmetrically) to draw the curved parts. For the eyes, the upper and lower curves use the same radius and symmetric centres (A and B) so the two curves form one symmetrical eye. Repeat for the second eye with the same settings so both eyes are identical.

3. Is there a 4-sided figure in which all the sides are equal in length but is not a square? If such a figure exists, can you construct it?

SOLUTION Yes. A figure with all four sides equal but angles not 90° is a rhombus (a ‘pushed-over’ square). Step 1. Draw AB = 5 cm. Step 2. With A as centre, radius 5 cm, draw an arc; with B as centre, radius 5 cm, draw another arc — mark D and C on them so the figure leans (angles not 90°). Step 3. Use the ‘equidistant’ idea from the House: choose D 5 cm from A, then find C 5 cm from both B and D using two arcs. Join the points. All sides are 5 cm but the angles are not right angles, so it is not a square.

Math Talk, Think & Try This — Answered

These are the in-text reflective and short tasks in the chapter; the determinate ones are answered, the open ones are guided.

Think (p. 194) — Reasoning without measuring Is it possible to reason out if the sides are equal or not, and if the angles are right or not, without using any measuring instruments? Can we do this by only looking at the position of corners in the dot grid? Answer. Yes. On a dot grid the spacing between dots is fixed and equal, so equal numbers of steps mean equal sides, and corners that turn along the grid (right and up) make right angles. The positions of the corners on the grid let us judge equal sides and right angles without any ruler or protractor.
Math Talk (p. 198) — Nearest and farthest X and Y At which positions will the points X and Y be at their closest? When will they be the farthest? How does the minimum distance compare to AB? Answer. X (on AD) and Y (on BC) are closest when they sit directly opposite each other (X and Y at the same distance from A and B); then XY is shortest and equals AB (7 cm). They are farthest when they sit at opposite corners, so XY lies along a diagonal and equals AC (or BD).
Math Talk (p. 204) — Laws about diagonals, angles & sides What general laws did you observe with respect to the angles and sides? How can one be sure the laws will always be true? Answer. In a rectangle the two diagonals are equal in length and the angles a diagonal makes with the two sides are equal to the matching angles at the opposite corner. The two parts of a corner are equal (45° each) only in a square. We can be sure because rotating, and the equal-side reasoning, do not change lengths or angles — the properties follow from the definitions, not from one drawing.
Think (p. 213) — Do we need full circles? Was it necessary to draw two full circles to get the point A? Answer. No. We only need the parts of the circles that cross near A. So point A (or B) can be found just by drawing two short arcs of the right radius from the two centres — the place where the arcs cross is the required point.
Try This (p. 191) — Identical smaller waves Recreate the figure where the waves are smaller than a half circle, making both waves identical. Answer. Use a fixed (large) compass radius and the same arc height for both waves; mark equal end-point spacings on the base line and shift only the centre. Keeping the radius constant makes both shallow arcs identical (see Figure it Out Q3 above).
Explore (p. 200) — Two identical squares What about constructing a rectangle that can be divided into two identical squares? If AF = 4 cm, what must AC be? Answer. For two identical squares side by side, the length must be twice the breadth. With breadth AF = 4 cm, the length AC = 2 × 4 = 8 cm. Construct an 8 cm × 4 cm rectangle and draw one cross-line at its mid-length to get two 4 cm × 4 cm squares. (A compass can transfer the 4 cm length to mark B and C without re-measuring.)

Common Mistakes to Avoid

Watch out for these

  • Naming a figure by jumping across corners (like PQSR) — a valid name follows the order of travel around the figure.
  • Forgetting that a perfect 90° angle is needed at each corner; a slightly slanted ‘perpendicular’ spoils the square/rectangle.
  • Confusing the half-circle’s radius with its diameter — for AX = 4 cm the radius is 2 cm, not 4 cm.
  • Changing the compass radius between two waves or two eyes — keep it constant so the parts are identical.
  • Thinking a 4-sided figure can have all 90° angles with unequal opposite sides — it cannot; that always makes a rectangle.
  • Drawing whole circles when only short arcs are needed to find an intersection point — arcs are quicker and cleaner.
  • Assuming a diagonal always splits a right angle into two equal halves — that happens only in a square.

Practice MCQs & Assertion–Reason

1. All the points that are exactly 4 cm from a point P together form a:

(a) straight line    (b) square    (c) circle    (d) triangle

2. The distance from the centre of a circle to any point on the circle is called its:

(a) diameter    (b) radius    (c) chord    (d) arc

3. Which of the following is NOT a valid name for the square PQRS?

(a) QRSP    (b) RSPQ    (c) PQSR    (d) SPQR

4. In a rectangle, all the angles are:

(a) 45°    (b) 60°    (c) 90°    (d) 120°

5. For a wave drawn as a half circle on a part AX of length 4 cm, the compass radius should be:

(a) 1 cm    (b) 2 cm    (c) 4 cm    (d) 8 cm

6. A rectangle can be divided into 3 identical squares when its length is:

(a) equal to its breadth    (b) 2 × breadth    (c) 3 × breadth    (d) 4 × breadth

7. A diagonal of a rectangle divides a corner angle into two equal parts only when the figure is a:

(a) square    (b) long rectangle    (c) triangle    (d) circle

8. To construct a rectangle with side 4 cm and diagonal 8 cm, after drawing the side and a perpendicular you draw an arc of radius:

(a) 4 cm    (b) 6 cm    (c) 8 cm    (d) 12 cm

9. A point that is 5 cm from both B and C is found at the intersection of:

(a) one circle    (b) two arcs of radius 5 cm centred at B and C    (c) two perpendiculars    (d) a single straight line

10. A 4-sided figure with all sides equal but angles not 90° is a:

(a) square    (b) rectangle    (c) rhombus    (d) circle

Answer key: 1-(c), 2-(b), 3-(c), 4-(c), 5-(b), 6-(c), 7-(a), 8-(c), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: All the points of a circle are at the same distance from its centre.

Reason: That equal distance is called the radius of the circle.

A-R 2. Assertion: PQSR is not a valid name for the square PQRS.

Reason: A valid name lists the corners in the order of travel around the figure.

A-R 3. Assertion: A 4-sided figure can have all angles 90° while its opposite sides are unequal.

Reason: In a rectangle the opposite sides are equal in length.

A-R 4. Assertion: A rotated square is still a square.

Reason: Rotating a figure does not change its side lengths or its angles.

A-R 5. Assertion: When a diagonal of a rectangle divides the opposite angles into 45° and 45°, all its sides are equal.

Reason: A diagonal splits a right angle into two equal parts only when the two adjacent sides are equal, making the rectangle a square.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • All points at a fixed distance from a centre form a circle; that distance is the radius, drawn with a compass.
  • A rectangle has equal opposite sides (R1) and all angles 90° (R2); a square has all sides equal (S1) and all angles 90° (S2).
  • A figure’s name must follow the order of travel around its corners; rotating a square or rectangle keeps its sides and angles.
  • Construct squares/rectangles by drawing a side, raising perpendiculars, marking the other side and joining; transfer lengths with a compass.
  • Given a side and a diagonal, draw the side, a perpendicular, then an arc of radius = diagonal to fix the next corner.
  • A point equidistant from two points is found where two equal arcs cross (used in the ‘House’).
  • A diagonal splits a corner into two equal angles only in a square; the farthest two points along the sides are a diagonal apart.

How to score full marks in this chapter

Always start a construction with a neat rough diagram and mark the known lengths and angles on it. Draw genuine 90° perpendiculars, keep the compass radius unchanged when parts must be identical, and use short arcs instead of full circles to locate a point. After finishing, write the verification line — “all angles 90°, opposite sides equal” — to show the figure satisfies properties R1 and R2 (or S1 and S2).

Frequently Asked Questions

What is Class 6 Maths Ganita Prakash Chapter 8 about?

Chapter 8, Playing with Constructions, teaches drawing accurate figures with a ruler and compass: circles and their radius, the properties of squares and rectangles, constructing them, exploring diagonals, and finding points equidistant from two given points (the ‘House’ construction).

What tools do I need for Chapter 8 constructions?

You mainly need a ruler (to draw and measure straight lines) and a compass (to draw circles and arcs and to transfer lengths). A pencil and an eraser complete the set; some figures also use a set-square for perpendiculars.

Why does a circle form when you mark all points 4 cm from a point P?

Every point that is exactly 4 cm from P lies on one curve at the same distance from P. The set of all such equidistant points is, by definition, a circle with centre P and radius 4 cm.

Are these Class 6 Maths Ganita Prakash Chapter 8 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash textbook for the 2026–27 session, with all measurements verified against the book’s answer key.

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