NCERT Solutions Ganita Prakash (Part 2) Chapter 4 Nets of a cylinder and a cone — In-text Questions

Book page 82 Updated on2026-09-05

Q1.

What is the net of a cylinder?

Answer

Two circles and one rectangle. Unroll the curved surface after cutting it once along the height, and lay the two circular ends flat.

The curved surface flattens into a rectangle; the two ends are circles of radius r.

Why the curved surface flattens perfectly: a cylinder is bent in only one direction. Cut it along a line parallel to the axis and it opens out flat with nothing stretched or torn — like unrolling a mat.

Q2.

What are the sidelengths of the rectangle obtained?

Answer

One side is the height h of the cylinder; the other is the circumference of the base, 2πr.

Rectangle = h × 2πr
Curved surface area = 2πrh
Total surface area = 2πrh + 2 × πr² = 2πr(h + r)

Why the circumference and not the diameter: the cut edge of the rectangle was originally wrapped once right round the circular rim. Unrolling does not change its length, so the rectangle is exactly as long as the rim is round — that is 2πr.

Check it yourself: wrap a strip of paper once round a tin, mark where it overlaps and cut. Now measure the strip and the diameter of the tin. The strip is always about 3.14 times the diameter.

Q3.

How will the net of a cone look?

Answer

A circle for the base and a sector of a larger circle for the curved surface.

The sector has radius equal to the slant height l of the cone, and its arc is exactly as long as the base circle, 2πr.

Sector radius = l
Arc length = 2πr
Angle of the sector = (2πr ⁄ 2πl) × 360° = (r⁄l) × 360°

For example, a cone with r = 3 cm and l = 9 cm opens out into a sector of angle (3⁄9) × 360° = 120°, that is one-third of a full circle.

Q4.

If the cone is slit open along the line l and then unrolled, what will we get?

Answer

A sector of a circle with centre O, where O is the apex of the cone.

Every point on the rim of the base is joined to O by a slant line, and all these slant lines have the same length l. Slitting along one of them and unrolling does not change any of those lengths, so in the flat figure every rim point is still at distance l from O. Points at a fixed distance from O lie on a circle centred at O — so the curved boundary of the net is an arc of a circle with centre O.

Sector radius = l for every rim point
Arc length = 2πr, the circumference of the base
Curved surface area = ½ × arc × radius = ½ × 2πr × l = πrl

Why unrolling preserves lengths: like the cylinder, a cone bends in only one direction. Rolling and unrolling stretches nothing, so every distance measured along the surface is the same before and after. This is exactly the property the chapter will use again for the ant and the laddu.

Q5.

What surface do you construct by using the above net, in which O is not the centre of the boundary circle? Make a physical model to help you answer this question!

Answer

You get an oblique cone — a cone that leans over, with its apex not above the centre of its base.

In the ordinary net, every point of the boundary is the same distance l from O, so when it is rolled up all the slant lines are equal and the apex sits directly above the centre of the base. That is a right circular cone.

Now take a net whose boundary is a circle whose centre is somewhere other than O. Different boundary points are now at different distances from O, so when the surface is rolled up the slant lines have different lengths — long on one side, short on the other. The apex is pulled towards the short side, and the cone tilts.

	All boundary points at distance l from O	Distances from O unequal
Slant lines	All equal	Different lengths
Solid formed	Right circular cone	Oblique (slanting) cone
Apex sits	Above the centre of the base	Off to one side

Tip: Cut a paper disc, mark a point O well away from its centre, cut along a line from O to the boundary and overlap the two cut edges. The paper will curl into a tilted funnel — a leaning ice-cream cone.

Q6.

Draw a net with appropriate measurements that can be folded into a triangular prism. Verify that it works by making an actual cutout.

Answer

Take a prism whose ends are equilateral triangles of side 5 cm and whose length is 9 cm.

Net = 3 rectangles of 9 cm × 5 cm, joined in a row
+ 2 equilateral triangles of side 5 cm, one on each end of the middle rectangle
Total length of the strip = 3 × 5 = 15 cm

Three rectangles wrap round to make the sides; the two triangles fold in to close the ends.

Surface area = 3 × (9 × 5) + 2 × (√3⁄4) × 5²
= 135 + 2 × 10.83 ≈ 156.7 cm²

Check it yourself: the three rectangles must be equal in width (5 cm each) or the prism will not close, and the triangle side must equal that width exactly. Roll the strip into a triangular tube first, then fold the ends in.

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