Ganit Prakash - Class-X - Right Circular Cone
Let us work out 16 Solid Geometry
📘 Exercise 16 Solutions (Q1 - Q12)
Right Circular ConeGiven:
Base radius \( (r) \) = 15 cm
Slant height \( (l) \) = 24 cm
1. Curved Surface Area (CSA):
Formula for CSA of a cone = \( \pi r l \)
\[\begin{array}{l} \text{CSA} = \frac{22}{7} \times 15 \times 24 \\ = \frac{22 \times 360}{7} \\ = \frac{7920}{7} \\ = 1131\frac{3}{7} \text{ sq. cm} \end{array}\]2. Total Surface Area (TSA):
Formula for TSA of a closed cone = \( \pi r(l + r) \)
\[\begin{array}{l} \text{TSA} = \frac{22}{7} \times 15 \times (24 + 15) \\ = \frac{22}{7} \times 15 \times 39 \\ = \frac{22 \times 585}{7} \\ = \frac{12870}{7} \\ = 1838\frac{4}{7} \text{ sq. cm} \end{array}\]Therefore, the curved surface area is \( 1131\frac{3}{7} \) sq. cm and the total surface area is \( 1838\frac{4}{7} \) sq. cm.
(i) base area is 1.54 sq.m. and height is 2.4m.
(ii) the length of base diameter is 21m and slant height is 17.5m.
Explanation:
(i) Given: Base area = 1.54 sq.m, Height \( (h) \) = 2.4 m
Formula for Volume = \( \frac{1}{3} \times \text{Base Area} \times \text{height} \)
\[\begin{array}{l} \text{Volume} = \frac{1}{3} \times 1.54 \times 2.4 \\ = 1.54 \times 0.8 \\ = 1.232 \text{ cu. m} \end{array}\]Therefore, the volume is 1.232 cubic metres.
(ii) Given: Base diameter = 21 m, Slant height \( (l) \) = 17.5 m
Radius \( (r) \) = \( \frac{21}{2} \) = 10.5 m
First, we need to find the height \( (h) \):
\[\begin{array}{l} h = \sqrt{l^2 - r^2} \\ h = \sqrt{(17.5)^2 - (10.5)^2} \\ h = \sqrt{(17.5 + 10.5)(17.5 - 10.5)} \\ h = \sqrt{28 \times 7} \\ h = \sqrt{196} = 14 \text{ m} \end{array}\]Now, calculate the Volume = \( \frac{1}{3} \pi r^2 h \):
\[\begin{array}{l} \text{Volume} = \frac{1}{3} \times \frac{22}{7} \times (10.5)^2 \times 14 \\ = \frac{1}{3} \times \frac{22}{7} \times 110.25 \times 14 \\ = \frac{1}{3} \times 22 \times 110.25 \times 2 \\ = \frac{4851}{3} \\ = 1617 \text{ cu. m} \end{array}\]Therefore, the volume is 1617 cubic metres.
Given:
When the triangle is revolved around the 15 cm side, it forms a right circular cone where:
Height \( (h) \) = 15 cm
Base radius \( (r) \) = 20 cm
First, find the slant height \( (l) \):
\[\begin{array}{l} l = \sqrt{r^2 + h^2} \\ l = \sqrt{20^2 + 15^2} \\ l = \sqrt{400 + 225} = \sqrt{625} = 25 \text{ cm} \end{array}\]1. Curved Surface Area (CSA):
\[\begin{array}{l} \text{CSA} = \pi r l = \frac{22}{7} \times 20 \times 25 \\ = \frac{11000}{7} \\ = 1571\frac{3}{7} \text{ sq. cm} \end{array}\]2. Total Surface Area (TSA):
\[\begin{array}{l} \text{TSA} = \pi r(l + r) = \frac{22}{7} \times 20 \times (25 + 20) \\ = \frac{22}{7} \times 20 \times 45 \\ = \frac{19800}{7} \\ = 2828\frac{4}{7} \text{ sq. cm} \end{array}\]3. Volume:
\[\begin{array}{l} \text{Volume} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (20)^2 \times 15 \\ = \frac{22}{7} \times 400 \times 5 \\ = \frac{44000}{7} \\ = 6285\frac{5}{7} \text{ cu. cm} \end{array}\]Given:
Height \( (h) \) = 6 cm
Slant height \( (l) \) = 10 cm
First, we find the base radius \( (r) \):
\[\begin{array}{l} r = \sqrt{l^2 - h^2} \\ r = \sqrt{10^2 - 6^2} \\ r = \sqrt{100 - 36} = \sqrt{64} = 8 \text{ cm} \end{array}\]1. Total Surface Area (TSA):
\[\begin{array}{l} \text{TSA} = \pi r(l + r) = \frac{22}{7} \times 8 \times (10 + 8) \\ = \frac{22}{7} \times 8 \times 18 \\ = \frac{3168}{7} \\ = 452\frac{4}{7} \text{ sq. cm} \end{array}\]2. Volume:
\[\begin{array}{l} \text{Volume} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 8^2 \times 6 \\ = \frac{22}{7} \times 64 \times 2 \\ = \frac{2816}{7} \\ = 402\frac{2}{7} \text{ cu. cm} \end{array}\]Therefore, the TSA is \( 452\frac{4}{7} \) sq. cm and the Volume is \( 402\frac{2}{7} \) cu. cm.
Given:
Volume \( (V) \) = \( 100\pi \) cm³
Height \( (h) \) = 12 cm
We know that Volume of a cone = \( \frac{1}{3}\pi r^2 h \)
Equating the volume formula with the given value:
\[\begin{array}{l} \frac{1}{3} \pi r^2 (12) = 100\pi \\ \Rightarrow 4\pi r^2 = 100\pi \\ \Rightarrow 4r^2 = 100 \quad \text{[Cancelling } \pi \text{ from both sides]} \\ \Rightarrow r^2 = 25 \\ \Rightarrow r = 5 \text{ cm} \quad \text{[Radius cannot be negative]} \end{array}\]Now, calculate the slant height \( (l) \):
\[\begin{array}{l} l = \sqrt{r^2 + h^2} \\ l = \sqrt{5^2 + 12^2} \\ l = \sqrt{25 + 144} \\ l = \sqrt{169} \\ l = 13 \text{ cm} \end{array}\]Therefore, the slant height of the cone is 13 cm.
Explanation:
The tarpauline is used to make the curved surface of the tent. Therefore, the Curved Surface Area (CSA) is equal to the area of the tarpauline.
Given: CSA = 77 sq.m, Slant height \( (l) \) = 7 m
We know, CSA = \( \pi r l \)
\[\begin{array}{l} \pi r l = 77 \\ \Rightarrow \frac{22}{7} \times r \times 7 = 77 \\ \Rightarrow 22r = 77 \\ \Rightarrow r = \frac{77}{22} \\ \Rightarrow r = 3.5 \text{ m} \end{array}\]Now, calculate the base area of the tent:
\[\begin{array}{l} \text{Base Area} = \pi r^2 \\ = \frac{22}{7} \times (3.5)^2 \\ = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \\ = \frac{11 \times 7}{2} \\ = \frac{77}{2} = 38.5 \text{ sq.m.} \end{array}\]Therefore, the base area of the tent is 38.5 sq.m.
Given:
Base diameter = 21 m \( \Rightarrow \) Radius \( (r) \) = \( \frac{21}{2} \) = 10.5 m
Height \( (h) \) = 14 m
First, calculate the slant height \( (l) \):
\[\begin{array}{l} l = \sqrt{r^2 + h^2} \\ l = \sqrt{(10.5)^2 + (14)^2} \\ l = \sqrt{110.25 + 196} \\ l = \sqrt{306.25} = 17.5 \text{ m} \end{array}\]Now, calculate the Curved Surface Area (CSA):
\[\begin{array}{l} \text{CSA} = \pi r l \\ = \frac{22}{7} \times 10.5 \times 17.5 \\ = \frac{22}{7} \times \frac{21}{2} \times 17.5 \\ = 11 \times 3 \times 17.5 \\ = 33 \times 17.5 \\ = 577.5 \text{ sq.m.} \end{array}\]Calculate total expenditure:
\[\begin{array}{l} \text{Expenditure} = \text{CSA} \times \text{Rate per sq.m.} \\ = 577.5 \times 1.50 \\ = 866.25 \text{ ₹} \end{array}\]Therefore, the expenditure to colour the curved surface is ₹ 866.25.
Given:
Base diameter = 10 cm \(\Rightarrow\) Radius \( (r) \) = 5 cm
Total cost for polishing the whole surface = ₹ 429
Rate of polishing = ₹ 2.10 per cm²
1. Finding the height:
Total Surface Area (TSA) of the toy = \( \frac{\text{Total Cost}}{\text{Rate}} \)
\[\begin{array}{l} \text{TSA} = \frac{429}{2.10} = \frac{4290}{21} = \frac{1430}{7} \text{ cm}^2 \end{array}\]We know the formula for TSA of a solid cone is \( \pi r(r + l) \), where \( l \) is the slant height.
\[\begin{array}{l} \frac{22}{7} \times 5 \times (5 + l) = \frac{1430}{7} \\ \Rightarrow 110(5 + l) = 1430 \\ \Rightarrow 5 + l = \frac{1430}{110} = 13 \\ \Rightarrow l = 13 - 5 = 8 \text{ cm} \end{array}\]Now, calculate the height \( (h) \):
\[\begin{array}{l} h = \sqrt{l^2 - r^2} \\ h = \sqrt{8^2 - 5^2} \\ h = \sqrt{64 - 25} \\ h = \sqrt{39} \text{ cm} \end{array}\]2. Finding the quantity of wood (Volume):
\[\begin{array}{l} \text{Volume} = \frac{1}{3} \pi r^2 h \\ = \frac{1}{3} \times \frac{22}{7} \times (5)^2 \times \sqrt{39} \\ = \frac{1}{3} \times \frac{22}{7} \times 25 \times \sqrt{39} \\ = \frac{550\sqrt{39}}{21} \text{ cm}^3 \end{array}\]Therefore, the height of the toy is \( \sqrt{39} \) cm and the quantity of wood required is \( \frac{550\sqrt{39}}{21} \) cubic cm.
Explanation:
A "boya" (buoy) is a closed structure, so the quantity of iron sheet represents its Total Surface Area (TSA).
Given: TSA = \( 75\frac{3}{7} \) m² = \( \frac{528}{7} \) m²
Slant height \( (l) \) = 5 m
1. Finding radius and height:
\[\begin{array}{l} \text{TSA} = \pi r(r + l) = \frac{528}{7} \\ \Rightarrow \frac{22}{7} \times r(r + 5) = \frac{528}{7} \\ \Rightarrow 22r(r + 5) = 528 \\ \Rightarrow r(r + 5) = 24 \\ \Rightarrow r^2 + 5r - 24 = 0 \\ \Rightarrow r^2 + 8r - 3r - 24 = 0 \\ \Rightarrow r(r + 8) - 3(r + 8) = 0 \\ \Rightarrow (r - 3)(r + 8) = 0 \end{array}\]Since radius cannot be negative, \( r = 3 \) m.
Height \( (h) \) = \( \sqrt{l^2 - r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \) m.
2. Finding Volume of air:
\[\begin{array}{l} \text{Volume} = \frac{1}{3} \pi r^2 h \\ = \frac{1}{3} \times \frac{22}{7} \times (3)^2 \times 4 \\ = \frac{1}{3} \times \frac{22}{7} \times 9 \times 4 \\ = \frac{22 \times 3 \times 4}{7} = \frac{264}{7} = 37\frac{5}{7} \text{ m}^3 \end{array}\]3. Expenditure to colour the whole surface:
\[\begin{array}{l} \text{Cost} = \text{TSA} \times \text{Rate} \\ = \frac{528}{7} \times 2.80 \\ = 528 \times 0.40 \\ = 211.20 \text{ ₹} \end{array}\]Therefore, height is 4 m, volume is \( 37\frac{5}{7} \) m³, and expenditure is ₹ 211.20.
Explanation:
Number of persons = 11
1. Finding the Base Area:
Space required per person in the base = 4 m²
\[\begin{array}{l} \text{Total Base Area } (\pi r^2) = 11 \times 4 = 44 \text{ m}^2 \\ \end{array}\]2. Finding the Volume:
Air required per person = 20 m³
\[\begin{array}{l} \text{Total Volume } (\frac{1}{3} \pi r^2 h) = 11 \times 20 = 220 \text{ m}^3 \end{array}\]Now, substitute the value of the base area \( (\pi r^2 = 44) \) into the volume equation:
\[\begin{array}{l} \frac{1}{3} \times (\pi r^2) \times h = 220 \\ \Rightarrow \frac{1}{3} \times 44 \times h = 220 \\ \Rightarrow 44h = 220 \times 3 \\ \Rightarrow 44h = 660 \\ \Rightarrow h = \frac{660}{44} = 15 \text{ m} \end{array}\]Therefore, the height of the tent is 15 metres.
Given:
External diameter = 21 cm \(\Rightarrow\) Radius \( (r) \) = \( \frac{21}{2} = 10.5 \) cm
Total expenditure = ₹ 57.75
Rate = 10 paise per cm² = ₹ 0.10 per cm²
The "outer surface" refers to the Curved Surface Area (CSA) because a coronet (headpiece) is open at the base.
1. Finding the slant height (\( l \)):
\[\begin{array}{l} \text{CSA} = \frac{\text{Total Cost}}{\text{Rate}} \\ \Rightarrow \pi r l = \frac{57.75}{0.10} \\ \Rightarrow \frac{22}{7} \times 10.5 \times l = 577.5 \\ \Rightarrow \frac{22}{7} \times \frac{21}{2} \times l = 577.5 \\ \Rightarrow 11 \times 3 \times l = 577.5 \\ \Rightarrow 33l = 577.5 \\ \Rightarrow l = \frac{577.5}{33} = 17.5 \text{ cm} \end{array}\]2. Finding the height (\( h \)):
\[\begin{array}{l} h = \sqrt{l^2 - r^2} \\ h = \sqrt{(17.5)^2 - (10.5)^2} \\ h = \sqrt{(17.5 + 10.5)(17.5 - 10.5)} \\ h = \sqrt{28 \times 7} \\ h = \sqrt{196} = 14 \text{ cm} \end{array}\]Therefore, the height is 14 cm and the slant height is 17.5 cm.
Given:
Base diameter = 9 m \(\Rightarrow\) Radius \( (r) \) = 4.5 m
Height \( (h) \) = 3.5 m
Use \( \pi = 3.14 \)
1. Total Volume of wheat:
\[\begin{array}{l} \text{Volume} = \frac{1}{3} \pi r^2 h \\ = \frac{1}{3} \times 3.14 \times (4.5)^2 \times 3.5 \\ = \frac{1}{3} \times 3.14 \times 20.25 \times 3.5 \\ = 3.14 \times 6.75 \times 3.5 \\ = 74.1825 \text{ m}^3 \end{array}\]2. Minimum plastic sheet required (CSA):
First, find the slant height \( (l) \):
\[\begin{array}{l} l = \sqrt{r^2 + h^2} \\ l = \sqrt{(4.5)^2 + (3.5)^2} \\ l = \sqrt{20.25 + 12.25} \\ l = \sqrt{32.5} \end{array}\]We are given \( \sqrt{130} = 11.4 \). We can rewrite \( \sqrt{32.5} \) as:
\[\begin{array}{l} \sqrt{32.5} = \sqrt{\frac{130}{4}} = \frac{\sqrt{130}}{2} = \frac{11.4}{2} = 5.7 \text{ m} \end{array}\]Now calculate the Curved Surface Area:
\[\begin{array}{l} \text{CSA} = \pi r l \\ = 3.14 \times 4.5 \times 5.7 \\ = 3.14 \times 25.65 \\ = 80.541 \text{ m}^2 \end{array}\]Therefore, the total volume of wheat is 74.1825 m³ and the minimum quantity of plastic sheet required is 80.541 m².
📘 13. Very short Answer : (V.S.A.)
(A) M.C.Q.Explanation:
Slant height \( (l) = 15 \text{ cm} \)
Base diameter \( = 16 \text{ cm} \Rightarrow \) Radius \( (r) = \frac{16}{2} = 8 \text{ cm} \)
Lateral (curved) surface area of a cone \( = \pi r l \)
\[\begin{array}{l} \text{Lateral surface area} = \pi \times 8 \times 15 \\ = 120\pi \text{ cm}^2 \end{array}\]Correct Option: (c) \( 120\pi \text{ cm}^2 \)
Explanation:
Let the volumes be \( V_1 \) and \( V_2 \), radii be \( r_1 \) and \( r_2 \), and heights be \( h_1 \) and \( h_2 \).
We are given \( \frac{V_1}{V_2} = \frac{1}{4} \) and \( \frac{r_1}{r_2} = \frac{4}{5} \).
Volume formula: \( V = \frac{1}{3}\pi r^2 h \)
\[\begin{array}{l} \frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} \\ \Rightarrow \frac{1}{4} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) \\ \Rightarrow \frac{1}{4} = \left(\frac{4}{5}\right)^2 \times \left(\frac{h_1}{h_2}\right) \\ \Rightarrow \frac{1}{4} = \frac{16}{25} \times \left(\frac{h_1}{h_2}\right) \\ \Rightarrow \frac{h_1}{h_2} = \frac{1}{4} \times \frac{25}{16} = \frac{25}{64} \end{array}\]Correct Option: (d) 25:64
Explanation:
Let original radius be \( r \) and original height be \( h \).
Original volume \( V_1 = \frac{1}{3}\pi r^2 h \)
New radius \( = r \) and new height \( = 2h \) (doubled).
New volume \( V_2 = \frac{1}{3}\pi r^2 (2h) = 2 \left( \frac{1}{3}\pi r^2 h \right) = 2V_1 \)
Increase in volume \( = V_2 - V_1 = 2V_1 - V_1 = V_1 \)
\[\begin{array}{l} \text{Percentage increase} = \frac{\text{Increase}}{\text{Original Volume}} \times 100\% \\ = \frac{V_1}{V_1} \times 100\% = 100\% \end{array}\]Correct Option: (a) 100%
Explanation:
The standard interpretation of "increased by twice of its length" in this specific problem context (based on the given options) means it is doubled (multiplied by 2).
Let original radius be \( r \) and height be \( h \).
Original volume \( V_1 = \frac{1}{3}\pi r^2 h \)
New radius \( r' = 2r \) and new height \( h' = 2h \).
\[\begin{array}{l} \text{New Volume } V_2 = \frac{1}{3}\pi (2r)^2 (2h) \\ = \frac{1}{3}\pi (4r^2) (2h) \\ = 8 \left(\frac{1}{3}\pi r^2 h\right) \\ = 8 V_1 \end{array}\]The volume will be 8 times the previous one.
Correct Option: (d) 8 times
Explanation:
Given base radius \( R = \frac{r}{2} \)
Given slant height \( L = 2l \)
Total Surface Area of a cone \( = \pi R (L + R) \)
\[\begin{array}{l} \text{TSA} = \pi \left(\frac{r}{2}\right) \left(2l + \frac{r}{2}\right) \\ = \pi \left(\frac{r}{2}\right) \times 2 \left(l + \frac{r}{4}\right) \\ = \pi r \left(l + \frac{r}{4}\right) \text{ sq. unit} \end{array}\]Correct Option: (b) \( \pi r (l + \frac{r}{4}) \) sq. unit
📘 13. Very short Answer : (V.S.A.)
(B) True or FalseIf the length of base radius of a right circular cone is decreased by half and its height is increased by twice of it then the volume remains same.
Explanation:
Let original radius be \( r \) and original height be \( h \).
Original Volume, \( V_1 = \frac{1}{3}\pi r^2 h \)
New radius, \( r' = \frac{r}{2} \)
New height, \( h' = 2h \) (increased by twice implies it becomes \( 2h \)).
New Volume, \( V_2 \):
\[\begin{array}{l} V_2 = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 (2h) \\ = \frac{1}{3}\pi \left(\frac{r^2}{4}\right) (2h) \\ = \frac{1}{2} \left(\frac{1}{3}\pi r^2 h\right) \\ = \frac{V_1}{2} \end{array}\]The new volume is half of the original volume. It does not remain the same.
Answer: False
The height, radius and slant height of a right circular cone are always the three sides of a right-angled triangle.
Explanation:
By the geometric definition of a right circular cone, its vertical height (\( h \)) is perpendicular to the base radius (\( r \)). The slant height (\( l \)) connects the apex to the edge of the circular base.
These three lengths form a right-angled triangle where the slant height is the hypotenuse, satisfying Pythagoras' theorem: \( l^2 = h^2 + r^2 \).
Answer: True
📘 13. Very short Answer : (V.S.A.)
(C) Fill in the blanksExplanation:
If AC is the hypotenuse of the right-angled triangle ABC, then the right angle is at vertex B (\( \angle ABC = 90^\circ \)).
When the triangle is revolved around the side AB (taken as the central vertical axis), the side BC sweeps out the circular base of the cone.
Therefore, the side BC becomes the radius of the resulting cone.
Answer: BC
Explanation:
We know the formula for the volume of a cone is:
\[\begin{array}{l} \text{Volume } (V) = \frac{1}{3} \times \text{Base Area } (A) \times \text{height } (h) \\ \Rightarrow V = \frac{1}{3} A h \end{array}\]Solving for height (\( h \)):
\[\begin{array}{l} \Rightarrow 3V = A h \\ \Rightarrow h = \frac{3V}{A} \end{array}\]Answer: \( \frac{3V}{A} \)
Explanation:
Let the common base radius be \( r \) and the common height be \( h \).
Volume of the cylinder \( = \pi r^2 h \)
Volume of the cone \( = \frac{1}{3} \pi r^2 h \)
Ratio of their volumes (Cylinder : Cone):
\[\begin{array}{l} \text{Ratio} = (\pi r^2 h) : \left(\frac{1}{3} \pi r^2 h\right) \\ = 1 : \frac{1}{3} \\ = 3 : 1 \end{array}\]Answer: 3:1
📘 14. Short answer type question (S.A.)
Right Circular ConeExplanation:
Given, height \( (h) = 12 \text{ cm} \) and Volume \( (V) = 100\pi \text{ cm}^3 \).
We know the volume of a cone is \( V = \frac{1}{3}\pi r^2 h \).
\[\begin{array}{l} \frac{1}{3}\pi r^2 \times 12 = 100\pi \\ \Rightarrow 4\pi r^2 = 100\pi \\ \Rightarrow r^2 = \frac{100\pi}{4\pi} \\ \Rightarrow r^2 = 25 \\ \Rightarrow r = 5 \text{ cm} \quad [\text{Since radius cannot be negative}] \end{array}\]Answer: The length of the radius of the cone is \( 5\text{ cm} \).
Explanation:
Curved Surface Area (CSA) \(= \pi r l\)
Base Area \(= \pi r^2\)
According to the condition:
\[\begin{array}{l} \text{CSA} = \sqrt{5} \times \text{Base Area} \\ \pi r l = \sqrt{5} \pi r^2 \\ \Rightarrow l = \sqrt{5} r \quad \text{--- (1)} \end{array}\]We know the relationship between slant height (\(l\)), height (\(h\)), and radius (\(r\)):
\[\begin{array}{l} l^2 = h^2 + r^2 \end{array}\]Substitute the value of \( l \) from equation (1):
\[\begin{array}{l} (\sqrt{5}r)^2 = h^2 + r^2 \\ \Rightarrow 5r^2 = h^2 + r^2 \\ \Rightarrow 4r^2 = h^2 \\ \Rightarrow \frac{h^2}{r^2} = 4 \\ \Rightarrow \frac{h}{r} = 2 \end{array}\]Answer: The ratio of the height and the length of the radius is \( 2:1 \).
Explanation:
Base area \( A = \pi r^2 \)
Volume \( V = \frac{1}{3}\pi r^2 H \)
Substitute the value of the base area \( A \) into the volume formula:
\[\begin{array}{l} V = \frac{1}{3} A H \end{array}\]Now, calculate the value of \( \frac{AH}{V} \):
\[\begin{array}{l} \frac{AH}{V} = \frac{AH}{\frac{1}{3}AH} \\ = \frac{1}{\frac{1}{3}} \\ = 3 \end{array}\]Answer: The value of \( \frac{AH}{V} \) is \( 3 \).
Explanation:
Volume \(= \frac{1}{3}\pi r^2 h\)
Lateral Surface Area (LSA) \(= \pi r l\)
According to the given condition:
\[\begin{array}{l} \frac{1}{3}\pi r^2 h = \pi r l \\ \Rightarrow \frac{1}{3}rh = l \end{array}\]Squaring both sides:
\[\begin{array}{l} \frac{1}{9}r^2 h^2 = l^2 \end{array}\]We know that \( l^2 = h^2 + r^2 \), so substitute this in:
\[\begin{array}{l} \frac{1}{9}r^2 h^2 = h^2 + r^2 \end{array}\]Divide both sides by \( r^2 h^2 \):
\[\begin{array}{l} \frac{1}{9} = \frac{h^2 + r^2}{r^2 h^2} \\ \Rightarrow \frac{1}{9} = \frac{h^2}{r^2 h^2} + \frac{r^2}{r^2 h^2} \\ \Rightarrow \frac{1}{9} = \frac{1}{r^2} + \frac{1}{h^2} \end{array}\]Answer: The value of \( \frac{1}{h^2} + \frac{1}{r^2} \) is \( \frac{1}{9} \).
Explanation:
Let the radius of the cylinder be \( r_1 \) and the radius of the cone be \( r_2 \).
Ratio of radii: \( \frac{r_1}{r_2} = \frac{3}{4} \)
Let the height of the cylinder be \( h_1 \) and the height of the cone be \( h_2 \).
Ratio of heights: \( \frac{h_1}{h_2} = \frac{2}{3} \)
Volume of cylinder \( (V_{\text{cyl}}) = \pi r_1^2 h_1 \)
Volume of cone \( (V_{\text{cone}}) = \frac{1}{3}\pi r_2^2 h_2 \)
Now, calculate the ratio of their volumes:
\[\begin{array}{l} \frac{V_{\text{cyl}}}{V_{\text{cone}}} = \frac{\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} \\ = 3 \times \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) \\ = 3 \times \left(\frac{3}{4}\right)^2 \times \left(\frac{2}{3}\right) \\ = 3 \times \frac{9}{16} \times \frac{2}{3} \\ = \frac{18}{16} \\ = \frac{9}{8} \end{array}\]Answer: The ratio of the volumes of the cylinder and the cone is \( 9:8 \).
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