Mix Examples - Surface Areas and Volumes Questions in English

(D) When a cone is cut by a plane parallel to its base,it is divided into two parts: a smaller cone at the top and a remaining part at the bottom. The remaining part,which has two circular bases of different radii,is known as the frustum of a cone.

(A) Let the radii of the two spheres be $r_1$ and $r_2$,respectively.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Given the ratio of volumes $V_1 : V_2 = 64 : 27$,we have:
$\frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \frac{64}{27}$
$\frac{r_1^3}{r_2^3} = \frac{64}{27} = \left(\frac{4}{3}\right)^3$
Taking the cube root on both sides,we get:
$\frac{r_1}{r_2} = \frac{4}{3}$
The surface area of a sphere is given by $A = 4 \pi r^2$.
The ratio of their surface areas is:
$\frac{A_1}{A_2} = \frac{4 \pi r_1^2}{4 \pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2$
Substituting the value of $\frac{r_1}{r_2}$:
$\frac{A_1}{A_2} = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$
Thus,the ratio of their surface areas is $16:9$.

(N/A) The outer rectangle $ABCD$ has length $L = 26 \, m$ and breadth $B = 12 \, m$.
Area of outer rectangle $= L \times B = 26 \times 12 = 312 \, m^2$.
The inner unshaded region consists of a rectangle and two semi-circles.
The breadth of the inner rectangle is $12 - (4 + 4) = 4 \, m$. This is also the diameter of the two semi-circles.
Radius of each semi-circle $r = 4 / 2 = 2 \, m$.
The length of the inner rectangle is $26 - (3 + 3 + 2 + 2) = 16 \, m$ (subtracting the $3 \, m$ gaps and the two radii of the semi-circles).
Area of inner rectangle $= 16 \times 4 = 64 \, m^2$.
Area of two semi-circles $= 2 \times (\frac{1}{2} \pi r^2) = \pi r^2 = \pi (2)^2 = 4\pi \, m^2$.
Total unshaded area $= 64 + 4\pi \, m^2$.
Area of shaded region $=$ Area of outer rectangle $-$ Total unshaded area
$= 312 - (64 + 4\pi) = (248 - 4\pi) \, m^2$.

(C) Let the volume of cone $A$ be $2V$ and the volume of cone $B$ be $V$. Let the height of cone $A$ be $h_1 \, cm$,then the height of cone $B$ is $(21 - h_1) \, cm$.
Given,the diameter of the cones $= 6 \, cm$.
Therefore,the radius of the cones $r = \frac{6}{2} = 3 \, cm$.
Volume of cone $A = 2V = \frac{1}{3} \pi r^2 h_1 = \frac{1}{3} \pi (3)^2 h_1 = 3 \pi h_1$.
Thus,$V = \frac{3}{2} \pi h_1$ ...... $(i)$
Volume of cone $B = V = \frac{1}{3} \pi r^2 (21 - h_1) = \frac{1}{3} \pi (3)^2 (21 - h_1) = 3 \pi (21 - h_1)$ ...... $(ii)$
Equating $(i)$ and $(ii)$:
$\frac{3}{2} \pi h_1 = 3 \pi (21 - h_1)$
$\frac{1}{2} h_1 = 21 - h_1$
$\frac{3}{2} h_1 = 21$
$h_1 = 14 \, cm$.
Height of cone $A = 14 \, cm$,Height of cone $B = 21 - 14 = 7 \, cm$.
Volume of cone $A = 3 \pi (14) = 3 \times \frac{22}{7} \times 14 = 132 \, cm^3$.
Volume of cone $B = 3 \pi (7) = 3 \times \frac{22}{7} \times 7 = 66 \, cm^3$.
Volume of the cylinder $= \pi r^2 H = \frac{22}{7} \times (3)^2 \times 21 = 22 \times 9 \times 3 = 594 \, cm^3$.
Volume of the remaining portion $=$ Volume of cylinder $-$ (Volume of cone $A$ $+$ Volume of cone $B$)
$= 594 - (132 + 66) = 594 - 198 = 396 \, cm^3$.

(N/A) Let $r$ be the radius of the hemisphere and the cone,and $h$ be the height of the cone. Given $h = 4\, cm$ and diameter $d = 8\, cm$,so $r = 4\, cm$.
Volume of the toy $=$ Volume of the hemisphere $+$ Volume of the cone
$= \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r + h)$
$= \frac{1}{3} \times \frac{22}{7} \times 4^2 \times (2 \times 4 + 4) = \frac{1}{3} \times \frac{22}{7} \times 16 \times 12 = \frac{1408}{7} \approx 201.14\, cm^3$.
$A$ cube circumscribes the toy. The height of the toy is $h + r = 4 + 4 = 8\, cm$ and the width is $8\, cm$. Thus,the edge of the cube is $a = 8\, cm$.
Volume of the cube $= a^3 = 8^3 = 512\, cm^3$.
Difference in volumes $= 512 - 201.14 = 310.86\, cm^3$.
Total surface area of the toy $=$ Curved surface area of cone $+$ Curved surface area of hemisphere
$= \pi r l + 2 \pi r^2$,where slant height $l = \sqrt{h^2 + r^2} = \sqrt{4^2 + 4^2} = 4\sqrt{2}\, cm$.
$= \pi r (l + 2r) = \frac{22}{7} \times 4 \times (4\sqrt{2} + 8) = \frac{88}{7} \times 4(\sqrt{2} + 2) = \frac{352}{7} (1.414 + 2) \approx 171.68\, cm^2$.

(N/A) Given that,a heap of rice is in the form of a cone.
Height of the cone $(h) = 3.5 \, m$.
Diameter of the cone $= 9 \, m$,so the radius $(r) = \frac{9}{2} = 4.5 \, m$.
Volume of the rice $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (4.5)^2 \times 3.5 = \frac{1}{3} \times \frac{22}{7} \times 20.25 \times 3.5 = 74.25 \, m^3$.
To cover the heap,we need the curved surface area of the cone,which is $\pi r l$,where $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{(4.5)^2 + (3.5)^2} = \sqrt{20.25 + 12.25} = \sqrt{32.5} \approx 5.70 \, m$.
Curved surface area $= \frac{22}{7} \times 4.5 \times 5.70 \approx 80.61 \, m^2$.
Thus,the volume is $74.25 \, m^3$ and the required canvas cloth is $80.61 \, m^2$.

(N/A) The rocket is a combination of a right circular cylinder and a cone.
Given: Diameter of the cylinder $= 6 \, cm$.
Radius of the cylinder $(r) = \frac{6}{2} = 3 \, cm$.
Height of the cylinder $(H) = 12 \, cm$.
Volume of the cylinder $= \pi r^2 H = 3.14 \times (3)^2 \times 12 = 3.14 \times 9 \times 12 = 339.12 \, cm^3$.
Curved surface area of the cylinder $= 2 \pi r H = 2 \times 3.14 \times 3 \times 12 = 226.08 \, cm^2$.
For the cone,slant height $(l) = 5 \, cm$ and radius $(r) = 3 \, cm$.
Height of the cone $(h) = \sqrt{l^2 - r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \, cm$.
Volume of the cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times (3)^2 \times 4 = 3.14 \times 3 \times 4 = 37.68 \, cm^3$.
Curved surface area of the cone $= \pi r l = 3.14 \times 3 \times 5 = 47.1 \, cm^2$.
Total volume of the rocket $= \text{Volume of cylinder} + \text{Volume of cone} = 339.12 + 37.68 = 376.8 \, cm^3$.
Total surface area of the rocket $= \text{CSA of cone} + \text{CSA of cylinder} + \text{Area of base of cylinder} = 47.1 + 226.08 + (\pi r^2) = 47.1 + 226.08 + (3.14 \times 3^2) = 47.1 + 226.08 + 28.26 = 301.44 \, cm^2$.

(C) Let the total height of the building be equal to the internal diameter of the dome,which is $2r \, m$.
Therefore,the radius of the building (or dome) is $\frac{2r}{2} = r \, m$.
The height of the cylindrical part is $2r - r = r \, m$.
Volume of the cylinder $= \pi r^2 h = \pi r^2 (r) = \pi r^3 \, m^3$.
Volume of the hemispherical dome $= \frac{2}{3} \pi r^3 \, m^3$.
Total volume of the building $=$ Volume of the cylinder $+$ Volume of the hemispherical dome.
Total volume $= \pi r^3 + \frac{2}{3} \pi r^3 = \frac{5}{3} \pi r^3 \, m^3$.
Given that the volume of air inside is $41 \frac{19}{21} \, m^3 = \frac{880}{21} \, m^3$.
Equating the volumes: $\frac{5}{3} \pi r^3 = \frac{880}{21}$.
Using $\pi = \frac{22}{7}$,we get $\frac{5}{3} \times \frac{22}{7} \times r^3 = \frac{880}{21}$.
$\frac{110}{21} r^3 = \frac{880}{21}$.
$r^3 = \frac{880}{110} = 8$.
$r = \sqrt[3]{8} = 2 \, m$.
Total height of the building $= 2r = 2 \times 2 = 4 \, m$.

(D) Given:
Length of the cuboid $(l) = 10 \, cm$,breadth $(b) = 5 \, cm$,and height $(h) = 4 \, cm$.
Volume of the cuboid $= l \times b \times h = 10 \times 5 \times 4 = 200 \, cm^3$.
Radius of each conical depression $(r) = 0.5 \, cm$ and depth $(h_1) = 2.1 \, cm$.
Volume of one conical depression $= \frac{1}{3} \pi r^2 h_1 = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 2.1 = \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 2.1 = 0.55 \, cm^3$.
Volume of $4$ conical depressions $= 4 \times 0.55 = 2.2 \, cm^3$.
Edge of the cubical depression $(a) = 3 \, cm$.
Volume of the cubical depression $= a^3 = 3^3 = 27 \, cm^3$.
Volume of wood in the stand $=$ Volume of cuboid $-$ (Volume of $4$ conical depressions $+$ Volume of cubical depression)
$= 200 - (2.2 + 27) = 200 - 29.2 = 170.8 \, cm^3$.

(N/A) For the given frustum of a cone,
bigger radius $r_{1} = 14 \,cm$,smaller radius $r_{2} = 7 \,cm$,and height $h = 24 \,cm$.
Slant height of the frustum of a cone $l = \sqrt{h^{2} + (r_{1} - r_{2})^{2}}$
$= \sqrt{24^{2} + (14 - 7)^{2}}$
$= \sqrt{576 + 49}$
$= \sqrt{625} = 25 \,cm$.
Curved Surface Area $(CSA)$ $= \pi l(r_{1} + r_{2})$
$= \frac{22}{7} \times 25 \times (14 + 7)$
$= \frac{22}{7} \times 25 \times 21 = 22 \times 25 \times 3 = 1650 \,cm^{2}$.
Total Surface Area $(TSA)$ $= \pi l(r_{1} + r_{2}) + \pi r_{1}^{2} + \pi r_{2}^{2}$
$= 1650 + \frac{22}{7} \times (14^{2} + 7^{2})$
$= 1650 + \frac{22}{7} \times (196 + 49)$
$= 1650 + \frac{22}{7} \times 245$
$= 1650 + 22 \times 35 = 1650 + 770 = 2420 \,cm^{2}$.
Volume $= \frac{1}{3} \pi h(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})$
$= \frac{1}{3} \times \frac{22}{7} \times 24 \times (196 + 49 + 98)$
$= \frac{22 \times 8}{7} \times 343$
$= 22 \times 8 \times 49 = 8624 \,cm^{3}$.
Thus,the $CSA$ is $1650 \,cm^{2}$,$TSA$ is $2420 \,cm^{2}$,and volume is $8624 \,cm^{3}$.

(N/A) Given: $R = 35 \, cm$,$r = 21 \, cm$,$h = 48 \, cm$.
First,calculate the slant height $l = \sqrt{h^2 + (R - r)^2} = \sqrt{48^2 + (35 - 21)^2} = \sqrt{2304 + 196} = \sqrt{2500} = 50 \, cm$.
Curved Surface Area $(CSA)$ $= \pi(R + r)l = \frac{22}{7} \times (35 + 21) \times 50 = \frac{22}{7} \times 56 \times 50 = 22 \times 8 \times 50 = 8800 \, cm^2$.
Total Surface Area $(TSA)$ $= \pi l(R + r) + \pi R^2 + \pi r^2 = 8800 + \frac{22}{7} \times (35^2 + 21^2) = 8800 + \frac{22}{7} \times (1225 + 441) = 8800 + \frac{22}{7} \times 1666 = 8800 + 22 \times 238 = 8800 + 5236 = 14036 \, cm^2$.
Volume $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 48 \times (35^2 + 21^2 + 35 \times 21) = \frac{22 \times 16}{7} \times (1225 + 441 + 735) = \frac{352}{7} \times 2401 = 352 \times 343 = 120736 \, cm^3$.

(A) Given: Radius of cylinder $(r)$ = $8 \, cm$,Height of cylinder $(h_1)$ = $9 \, cm$,Height of cone $(h_2)$ = $15 \, cm$.
Slant height of cone $(l)$ = $\sqrt{r^2 + h_2^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \, cm$.
Total Surface Area = Curved surface area of cylinder + Curved surface area of cone + Base area of cylinder.
$= 2\pi rh_1 + \pi rl + \pi r^2 = \pi r(2h_1 + l + r) = \frac{22}{7} \times 8 \times (2 \times 9 + 17 + 8) = \frac{176}{7} \times (18 + 17 + 8) = \frac{176}{7} \times 43 \approx 1081.14 \, cm^2$.
(Note: Using $\pi = 3.14$,Area $\approx 3.14 \times 8 \times 43 = 1079.68 \, cm^2$).
Volume = Volume of cylinder + Volume of cone = $\pi r^2 h_1 + \frac{1}{3} \pi r^2 h_2 = \pi r^2 (h_1 + \frac{h_2}{3}) = \frac{22}{7} \times 8^2 \times (9 + \frac{15}{3}) = \frac{22}{7} \times 64 \times 14 = 22 \times 64 \times 2 = 2816 \, cm^3$.

(N/A) $1$. Volume of the container: The container consists of a cylinder and a cone. The radius $r = 35 \,cm$,height of cylinder $h_1 = 52 \,cm$,and height of cone $h_2 = 12 \,cm$.
Volume of cylinder $V_1 = \pi r^2 h_1 = \frac{22}{7} \times 35 \times 35 \times 52 = 200,200 \,cm^3$.
Volume of cone $V_2 = \frac{1}{3} \pi r^2 h_2 = \frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 12 = 15,400 \,cm^3$.
Total volume $V = V_1 + V_2 = 200,200 + 15,400 = 215,600 \,cm^3$.
Since $1,000 \,cm^3 = 1 \,litre$,the volume is $215.6 \,litres$.
$2$. Total Surface Area: The surface area includes the base of the cylinder,the curved surface area of the cylinder,and the curved surface area of the cone.
Slant height of cone $l = \sqrt{r^2 + h_2^2} = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37 \,cm$.
Base area of cylinder $= \pi r^2 = \frac{22}{7} \times 35 \times 35 = 3,850 \,cm^2$.
Curved surface area of cylinder $= 2 \pi r h_1 = 2 \times \frac{22}{7} \times 35 \times 52 = 11,440 \,cm^2$.
Curved surface area of cone $= \pi r l = \frac{22}{7} \times 35 \times 37 = 4,070 \,cm^2$.
Total surface area $= 3,850 + 11,440 + 4,070 = 19,360 \,cm^2$.

(N/A) $1$. Slant height $(l)$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{21^2 + 20^2} = \sqrt{441 + 400} = \sqrt{841} = 29 \, cm$.
$2$. Curved Surface Area $(CSA)$ of the cone: $CSA_{cone} = \pi rl = \frac{22}{7} \times 21 \times 29 = 22 \times 3 \times 29 = 1914 \, cm^2$.
$3$. Curved Surface Area $(CSA)$ of the hemisphere: $CSA_{hemi} = 2\pi r^2 = 2 \times \frac{22}{7} \times 21 \times 21 = 2 \times 22 \times 3 \times 21 = 2772 \, cm^2$.
$4$. Total Curved Surface Area: $1914 + 2772 = 4686 \, cm^2$.
$5$. Volume of the cone $(V_{cone})$: $V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 21 \times 21 \times 20 = 22 \times 21 \times 20 = 9240 \, cm^3$.
$6$. Volume of the hemisphere $(V_{hemi})$: $V_{hemi} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 2 \times 22 \times 21 \times 21 = 19404 \, cm^3$.
$7$. Total Volume: $9240 + 19404 = 28644 \, cm^3$.

(B) To match the given parts,we identify the formulas for surface areas:
$1$. The Curved Surface Area $(CSA)$ of a cylinder is $2\pi rh$. Thus,$1-b$.
$2$. The Curved Surface Area $(CSA)$ of a sphere is $4\pi r^2$. Thus,$2-c$.
$3$. The Curved Surface Area $(CSA)$ of a cone is $\pi rl$. Thus,$3-d$.
$4$. The Total Surface Area $(TSA)$ of a hemisphere is $3\pi r^2$. Thus,$4-a$.
Therefore,the correct matching is $(1-b), (2-c), (3-d), (4-a)$.