Ganita Manjari Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume End-of-Chapter Exercises Solutions

The container is about the size of a small tub. A scoop is a ball a little bigger than a golf ball. A guess: about 40 to 50 scoops.

Assume a scoop is a sphere of radius 2.5 cm (diameter 5 cm):

Scoops are not perfect spheres and ice cream is a little compressed, so real numbers are about 40 to 50. Our guess of 45 is close.

We add 3 faces of size a × a (3a² cubes), then 3 edges of a cubes each (3a), and the corner cube (1).

Check a = 1: 3 + 3 + 1 = 7, and 2³ − 1³ = 7 ✓. Check a = 2: 12 + 6 + 1 = 19 and 27 − 8 = 19 ✓.

• Walls: perimeter 36 m × height 4.5 m = 162 m². Take away about 10% for doors and windows: 146 m².
• Wall thickness 20 cm = 0.2 m, so wall volume ≈ 146 × 0.2 ≈ 29 m³.
• One brick with its mortar: 20 cm × 10 cm × 10 cm = 0.002 m³.

So about 15,000 bricks.

A is 25,000 mm³ more. The big cube has 125 cubes of 10 mm in it (since 5 × 5 × 5 = 125), so B has only 100 of those 125.

1 km² = 1,000,000 m², so 2 × 10⁹ m² is 2000 km². That is a square of side about 45 km, which is small compared with a large city or a small state.

A sphere of radius r fits in a cylinder of radius r and height 2r.

If the first circle has radius R, its length is 2πR. The second has radius R + d and length 2π(R + d). It is 1 m longer:

R cancelled out, so d does not depend on the size of the sphere.

It is surprising that the answers are all the same, and that the Earth needs only 1 m of extra string to lift the string 16 cm all around. The reason is that a circle's length grows in direct proportion to its radius (circumference = 2πr), so adding 1 m to the length adds the same 1 over 2π to the radius for every circle.

The diagonal of the square is the diameter 2r. A square with diagonal d has area d²2, so the area is (2r)²2 = 2r². The height of the pyramid is h.

The volume decreases by 4πr² − 4πr + 4π3 cubic units. Check with r = 1: the sphere disappears, and the change is 4π − 4π + 4π3 = 4π3, which is the full volume ✓.

V = 8 gives a = 2 and S = 24, and 6 × 8^2/3 = 6 × 4 = 24 ✓.

S = 24 gives a = 2 and V = 8, and (246)^3/2 = 4^3/2 = 8 ✓.

The water must rise by 20 − 16 = 4 cm. The volume of that extra layer is:

The glass is 25 cm tall and the water ends at 20 cm, so it does not overflow ✓.

1. Open the pen and take out the refill (the thin tube of ink).
2. Measure the length of the ink column with a ruler: for example 10 cm.
3. Measure the inner diameter of the tube: for example 1 mm, so the radius is 0.5 mm = 0.05 cm.
4. The ink is a cylinder: V = πr²h.

So a new pen holds about 0.08 cm³ (0.08 mL) of ink, which is less than a quarter of a teaspoon.

Assume a chapati is a thin cylinder of radius 8 cm and thickness 1.5 mm = 0.15 cm:

• Treat the outer shell as a hemisphere of outer radius R, and the empty hollow as a hemisphere of inner radius r.
• Measure R across the cut face (for example 6 cm) and r across the hollow (for example 4.5 cm).
• Flesh ≈ 23π(R³ − r³) = 23 × 3.14 × (216 − 91.1) ≈ 261 cm³ for one half.

Books are cuboids, so they can be stacked with almost no gaps. Rounding down for the edges of the room, the number is about 6 lakh books.

A person's volume is about 0.07 m³ (a 70 kg person has a volume of about 70 litres). The total is 8 × 10⁹ × 0.07 = 5.6 × 10⁸ m³.

If each person has 0.5 m × 0.5 m × 1.7 m = 0.425 m³ of space, the total is 3.4 × 10⁹ m³ and the side is ∛(3.4 × 10⁹) ≈ 1500 m.

The layer is the space between the sphere of radius R = 6371 and the sphere of radius R + t.

A quick check: the layer is thin, so t ≈ V4πR² = 1.38 × 10⁹4π × 6371² ≈ 2.71 km ✓.

The volume is half of 1080 and the surface area is twice 636 ✓.

Choose a cuboid with volume V and surface area S. Then look for whole-number edges a, b, c with abc = V2 and 2(ab + ac + bc) = 2S. A long thin shape has a large surface area for its volume, so making one edge 1 often works.

Edges 1, 6, 90 give 6 × 90 = 540 cm³ = 10802 ✓ and 2(6 + 90 + 540) = 1272 = 2 × 636 ✓.

1. Draw a simple plan of your house: list the rooms as cuboids.
2. Measure each room's length, width and height (a tape or your paces: one pace is about 0.75 m).
3. Calculate the volume of each room: V = l × w × h.
4. Add the room volumes, and add something for walls and floors if you want the volume of the whole house.
5. State your assumptions: ignore furniture, treat slanted ceilings as flat at the average height.

The house holds about 170 m³ of air, which is about 1,70,000 litres.