NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 14 Math of Space Surface Area and Volume Exercise 14.3

Exercise Set 14.3

Question 1. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer:

Slant height l = 21 m; radius r = 24 ÷ 2 = 12 m.
TSA = πr(l + r) = \( \frac{22}{7} \) × 12 × (21 + 12) = \( \frac{22}{7} \) × 12 × 33 = \( \frac{8712}{7} \) ≈ 1244.57 m²
The total surface area is about 1244.57 m² (with π ≈ 3.14 it is 1243.44 m²).

Question 2. Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Answer:

r = 7 cm, l = 10 cm.
CSA = πrl = \( \frac{22}{7} \) × 7 × 10 = 22 × 10 = 220 cm²
The curved surface area is 220 cm².

Question 3. The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone.
Answer:

h = 16 cm, r = 12 cm. First find the slant height:
l² = h² + r² = 16² + 12² = 256 + 144 = 400, so l = √400 = 20 cm.
Using π ≈ 3.14:
CSA = πrl = 3.14 × 12 × 20 = 240π ≈ 753.6 cm²
TSA = πr(l + r) = 3.14 × 12 × (20 + 12) = 384π ≈ 1205.76 cm²
CSA = 240π ≈ 753.6 cm²; TSA = 384π ≈ 1205.76 cm².

Teacher's Note:
1. πrl needs the slant height l, not the vertical height h; find l first with the Baudhāyana–Pythagoras theorem.
2. 12, 16, 20 is the 3, 4, 5 triple scaled by 4, a useful shortcut.

Question 4. A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.
Answer:

h = 15 cm, V = 1570 cm³; take π ≈ 3.14.
\( \frac{1}{3} \) πr²h = V
\( \frac{1}{3} \) × 3.14 × r² × 15 = 1570
15.7 r² = 1570
r² = \( \frac{1570}{15.7} \) = 100, so r = √100 = 10
The radius of the base is 10 cm.

Question 5. The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the radius of the base, (ii) the total surface area of the cone.
Answer:

(i) πrl = 308
\( \frac{22}{7} \) × r × 14 = 308, i.e. 44r = 308
r = \( \frac{308}{44} \) = 7 cm
(ii) TSA = CSA + πr² = 308 + \( \frac{22}{7} \) × 7² = 308 + 154 = 462 cm²

Question 6. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:

r = 7 cm, h = 24 cm.
l² = h² + r² = 24² + 7² = 576 + 49 = 625, so l = 25 cm.
A cap is open at the bottom, so only its curved surface is made of sheet.
Sheet for one cap = πrl = \( \frac{22}{7} \) × 7 × 25 = 550 cm²
Sheet for 10 caps = 10 × 550 = 5500 cm²
5500 cm² of sheet is needed.

Question 7. What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm.
Answer:

Step 1: Slant height. h = 8 m, r = 6 m, so l² = 8² + 6² = 64 + 36 = 100 and l = 10 m.
Step 2: Area of tarpaulin. It covers only the curved surface (not the floor): CSA = πrl = 3.14 × 6 × 10 = 188.4 m².
Step 3: Length. The tarpaulin is 3 m wide, so length = \( \frac{\text{area}}{\text{width}} \) = \( \frac{188.4}{3} \) = 62.8 m.
Step 4: Add the extra 20 cm = 0.2 m: total length = 62.8 + 0.2 = 63 m.
63 m of tarpaulin is required. (With π = \( \frac{22}{7} \) the length comes to about 63.06 m.)

Teacher's Note:
1. Area ÷ width gives the length of a rectangular strip.
2. Change 20 cm to 0.2 m before adding; adding 20 to 62.8 is a common error.

Question 8. A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
Answer:

6² + 8² = 36 + 64 = 100 = 10², so the triangle is right-angled with perpendicular sides 6 cm and 8 cm.
Rotating it about the 8 cm side gives a right circular cone with h = 8 cm, r = 6 cm and l = 10 cm.
Volume: V = \( \frac{1}{3} \) πr²h = \( \frac{1}{3} \) × 3.14 × 6² × 8 = 3.14 × 96 = 96π ≈ 301.44 cm³
Curved surface area: CSA = πrl = 3.14 × 6 × 10 = 60π ≈ 188.4 cm²
Volume = 96π ≈ 301.44 cm³; curved surface area = 60π ≈ 188.4 cm².

Question 9. Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?
Answer:
Let the cup (a cone standing on its vertex O) have radius r at the top and depth h.

The water also forms a cone, with the same vertex O and height \( \frac{h}{2} \).
In the cross-section, triangles OPQ and OAB are similar (PQ ∥ AB and the angle at O is common), so
\( \frac{PQ}{AB} = \frac{OP}{OA} = \frac{h/2}{h} = \frac{1}{2} \), i.e. the water surface has radius \( \frac{r}{2} \).
Volume of the cup = \( \frac{1}{3} \) πr²h
Volume of water = \( \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 \left(\frac{h}{2}\right) = \frac{1}{3}\pi \times \frac{r^2}{4} \times \frac{h}{2} = \frac{1}{8} \times \frac{1}{3}\pi r^2 h \)
\( \frac{\text{Volume of water}}{\text{Volume of cup}} = \frac{1}{8} \)
The water fills only \( \frac{1}{8} \) of the cup, even though it reaches half the depth — most of the cup's space is near its wide top.

Teacher's Note:
1. Most students guess one half; halving both the radius and the height divides the volume by 2 × 2 × 2 = 8.
2. Similar triangles are what tell us the water surface has radius r/2.