Measurement of Volume and Surface Area Questions in English

(C) The radius of the sphere $(r_s)$ is $6/2 = 3 \, cm$.
The volume of the sphere is given by $V = \frac{4}{3} \pi r_s^3 = \frac{4}{3} \pi (3)^3 = 36 \pi \, cm^3$.
The wire is in the shape of a cylinder. The radius of the wire $(r_w)$ is $0.2/2 = 0.1 \, cm$.
Let the length of the wire be $l$. The volume of the wire is $V = \pi r_w^2 l = \pi (0.1)^2 l = 0.01 \pi l \, cm^3$.
Since the sphere is melted to form the wire,their volumes are equal:
$36 \pi = 0.01 \pi l$
$l = \frac{36}{0.01} = 3600 \, cm$.
Converting the length to meters: $3600 \, cm = 36 \, m$.

(B) Let $r$ be the radius of the base of the cone.
Given,circumference of the base $= 2 \pi r = 44 \, m$.
$\therefore r = \frac{44 \times 7}{2 \times 22} = 7 \, m$.
The height of the cone is $h = 9 \, m$.
The volume of air contained in the tent is equal to the volume of the cone.
Volume $= \frac{1}{3} \pi r^2 h$.
Volume $= \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 9$.
Volume $= 22 \times 7 \times 3 = 462 \, m^3$.

(C) Let the original radius of the sphere be $r$.
Original surface area $A_1 = 4 \pi r^2$.
If the radius is doubled,the new radius $r' = 2r$.
New surface area $A_2 = 4 \pi (2r)^2 = 4 \pi (4r^2) = 16 \pi r^2$.
The increase in surface area is $A_2 - A_1 = 16 \pi r^2 - 4 \pi r^2 = 12 \pi r^2$.
The percentage increase is given by $\frac{\text{Increase}}{\text{Original Area}} \times 100 = \frac{12 \pi r^2}{4 \pi r^2} \times 100 = 3 \times 100 = 300 \%$.

(C) The volume $V$ of a cylinder is given by the formula $V = \pi r^2 h$,where $r$ is the radius and $h$ is the height.
Since the radius $r$ remains unchanged,the volume $V$ is directly proportional to the height $h$ $(V \propto h)$.
If the height $h$ is decreased by $8 \%$,the new height $h'$ becomes $h' = h - 0.08h = 0.92h$.
The new volume $V'$ will be $V' = \pi r^2 (0.92h) = 0.92 V$.
The percentage change in volume is given by $\frac{V' - V}{V} \times 100 \%$.
$= \frac{0.92V - V}{V} \times 100 \% = -0.08 \times 100 \% = -8 \%$.
The negative sign indicates a decrease.
Therefore,the volume of the cylinder decreases by $8 \%$.

(B) Let the original radius be $r$ and the original height be $h$. The original volume $V_1 = \pi r^2 h$.
The new radius $r' = r + 0.20r = 1.2r$. The height remains $h$.
The new volume $V_2 = \pi (1.2r)^2 h = \pi (1.44r^2) h = 1.44 \pi r^2 h$.
The increase in volume = $V_2 - V_1 = 1.44 \pi r^2 h - \pi r^2 h = 0.44 \pi r^2 h$.
Percentage increase = $\left( \frac{\text{Increase}}{\text{Original Volume}} \right) \times 100 = \left( \frac{0.44 \pi r^2 h}{\pi r^2 h} \right) \times 100 = 44 \%$.

(B) Let the total height of the cone be $H = 21\, cm$ and base radius be $R = 9\, cm$. The slant height $L = \sqrt{R^2 + H^2} = \sqrt{9^2 + 21^2} = \sqrt{81 + 441} = \sqrt{522} = 3\sqrt{58}\, cm$.
By similar triangles,the radius $r$ at height $h$ from the apex is given by $r/h = R/H$. The slant height $l$ at height $h$ from the apex is $l = h \cdot (L/H)$.
The curved surface area of a cone is $S = \pi r l = \pi (R/H) h \cdot (L/H) h = \pi (R L / H^2) h^2$.
Let the heights from the apex be $h_1, h_2, h_3$. The cuts are at $7\, cm$ and $14\, cm$ from the base,so heights from the apex are $h_1 = 21 - 14 = 7\, cm$,$h_2 = 21 - 7 = 14\, cm$,and $h_3 = 21\, cm$.
The curved surface area of the top part (small cone) is $S_1 = k h_1^2 = k(7^2) = 49k$,where $k = \pi R L / H^2$.
The curved surface area of the top two parts combined is $S_2 = k h_2^2 = k(14^2) = 196k$.
The curved surface area of the whole cone is $S_3 = k h_3^2 = k(21^2) = 441k$.
The curved surface area of the top part is $A_1 = S_1 = 49k$.
The curved surface area of the middle part is $A_2 = S_2 - S_1 = 196k - 49k = 147k$.
The curved surface area of the bottom part is $A_3 = S_3 - S_2 = 441k - 196k = 245k$.
The ratio $A_1 : A_2 : A_3 = 49 : 147 : 245$. Dividing by $49$,we get $1 : 3 : 5$.

(B) The total surface area of the remaining part consists of:
$1$. The curved surface area of the original cylinder: $2\pi Rh = 2 \times \frac{22}{7} \times 14 \times 15 = 1320\, cm^2$.
$2$. The area of the two bases of the original cylinder minus the area of the two circular cuts: $2 \times (\pi R^2 - \pi r^2) = 2 \times \pi \times (14^2 - 7^2) = 2 \times \frac{22}{7} \times (196 - 49) = 2 \times \frac{22}{7} \times 147 = 924\, cm^2$.
$3$. The curved surface area of the two small cylinders cut out: $2 \times (2\pi rh) = 2 \times (2 \times \frac{22}{7} \times 7 \times 5) = 880\, cm^2$.
$4$. The area of the inner circular base of the two small cylinders: $2 \times (\pi r^2) = 2 \times \frac{22}{7} \times 7^2 = 308\, cm^2$.
Total surface area = $1320 + 924 + 880 + 308 = 3432\, cm^2$.

(D) The diameter of the hemisphere is $d = 28\, cm$.
Therefore,the radius $r = d / 2 = 28 / 2 = 14\, cm$.
The curved surface area $(CSA)$ of a hemisphere is given by the formula $CSA = 2 \pi r^{2}$.
Substituting the values,we get $CSA = 2 \times (22 / 7) \times 14 \times 14$.
$CSA = 2 \times 22 \times 2 \times 14$.
$CSA = 44 \times 28 = 1232\, cm^{2}$.

(D) Let the radius be $r$ and the height be $h$.
Curved Surface Area $(CSA)$ $= 2\pi rh$,Volume $(V) = \pi r^2h$.
Given $\frac{2\pi rh}{\pi r^2h} = \frac{1}{7} \Rightarrow \frac{2}{r} = \frac{1}{7} \Rightarrow r = 14$.
Total Surface Area $(TSA)$ $= 2\pi rh + 2\pi r^2 = 2\pi r(h+r)$.
Given $\frac{2\pi r(h+r)}{\pi r^2h} = \frac{187}{770}$.
Substitute $r = 14$:
$\frac{2(h+14)}{14h} = \frac{187}{770} \Rightarrow \frac{h+14}{7h} = \frac{187}{770}$.
$\frac{h+14}{7h} = \frac{17}{70} \Rightarrow 10(h+14) = 17h$.
$10h + 140 = 17h \Rightarrow 7h = 140 \Rightarrow h = 20$.
The ratio of radius to height is $r:h = 14:20 = 7:10$.

(B) Let the side of the cube be $a$ and the radius of the hemisphere be $r$. Since the hemisphere is placed on the cube,the diameter of the hemisphere equals the side of the cube,so $2r = a$,or $r = a/2$.
The total height of the figure is $a + r = 221 \ cm$.
Substituting $r = a/2$,we get $a + a/2 = 221$,which implies $3a/2 = 221$,so $a = 442/3 \ cm$ and $r = 221/3 \ cm$.
The curved surface area of the hemisphere is $2\pi r^2$ and the total surface area of the cube is $6a^2$.
The ratio is given as $\frac{2\pi r^2}{6a^2} = \frac{11}{42}$.
Substituting $r = a/2$,we get $\frac{2\pi (a/2)^2}{6a^2} = \frac{2\pi (a^2/4)}{6a^2} = \frac{\pi}{12} = \frac{11}{42}$.
Using $\pi \approx 22/7$,we have $\frac{22/7}{12} = \frac{22}{84} = \frac{11}{42}$,which matches the given ratio.
The total volume $V$ is the sum of the volume of the cube and the hemisphere:
$V = a^3 + \frac{2}{3}\pi r^3 = a^3 + \frac{2}{3}\pi (a/2)^3 = a^3 + \frac{\pi a^3}{12} = a^3(1 + \frac{\pi}{12})$.
Substituting $a = 442/3 \approx 147.33$:
$V = (147.33)^3 \times (1 + \frac{22/7}{12}) = 3198545.6 \times (1 + 0.2619) \approx 4035955 \ cm^3$.
Wait,re-evaluating the height: If $a=14$,$r=7$,height $= 21$. Here height is $221$. If $a=147.33$,the volume is very large. Checking options,it seems the height was meant to be $21 \ cm$. If $a+r=21$,$1.5a=21 \implies a=14, r=7$.
$V = 14^3 + \frac{2}{3} \times \frac{22}{7} \times 7^3 = 2744 + 718.67 = 3462.67 \ cm^3$.

(D) The volume of one spherical ball is $V = \frac{4}{3} \pi r^3$,where $r = 3 \text{ cm}$.
Total volume of $10$ balls $= 10 \times \frac{4}{3} \pi (3)^3 = 10 \times \frac{4}{3} \pi (27) = 360 \pi \text{ cm}^3$.
Since $20 \%$ of the material is wasted,the volume of the new sphere $V_{new}$ is $80 \%$ of the total initial volume.
$V_{new} = 0.80 \times 360 \pi = 288 \pi \text{ cm}^3$.
Let the radius of the new sphere be $R$. Then,$\frac{4}{3} \pi R^3 = 288 \pi$.
Dividing both sides by $\pi$ and multiplying by $\frac{3}{4}$:
$R^3 = 288 \times \frac{3}{4} = 72 \times 3 = 216$.
Taking the cube root of both sides,$R = \sqrt[3]{216} = 6 \text{ cm}$.

(B) The total surface area $(TSA)$ of the original cylinder is given by $2 \pi r(r + h)$.
Substituting the values: $TSA_{original} = 2 \times \frac{22}{7} \times 7 \times (7 + 18) = 44 \times 25 = 1100 \, cm^2$.
When the cylinder is cut into $3$ equal parts by $2$ cuts parallel to the base,each new cylinder has a height of $h' = \frac{18}{3} = 6 \, cm$ and the same radius $r = 7 \, cm$.
The $TSA$ of one smaller cylinder is $2 \pi r(r + h') = 2 \times \frac{22}{7} \times 7 \times (7 + 6) = 44 \times 13 = 572 \, cm^2$.
The total surface area of the $3$ smaller cylinders is $3 \times 572 = 1716 \, cm^2$.
The increase in surface area is $1716 - 1100 = 616 \, cm^2$.
The percentage increase is $\frac{616}{1100} \times 100 = 56 \%$.

(A) Let the radius of the base of the cone be $R = 8\, cm$ and the height be $H = 15\, cm$.
The slant height $L$ of the cone is given by $L = \sqrt{R^2 + H^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\, cm$.
Let $r$ be the radius of the largest sphere that can be inscribed in the cone.
The formula for the radius of the inscribed sphere in a cone is $r = \frac{R \cdot H}{R + L}$.
Substituting the values: $r = \frac{8 \cdot 15}{8 + 17} = \frac{120}{25} = 4.8\, cm$.
We need to find the ratio of the radius of the base of the cone $(R)$ to the radius of the sphere $(r)$:
Ratio $= R : r = 8 : 4.8$.
To simplify,multiply both sides by $10$: $80 : 48$.
Divide by $16$: $80/16 : 48/16 = 5 : 3$.
Thus,the ratio is $5:3$.

(A) The total surface area $(TSA)$ of a hemisphere is given by the formula $TSA = 3\pi r^2$.
Given $TSA = 41.58 \text{ cm}^2$,we have $3\pi r^2 = 41.58$.
Dividing both sides by $3$,we get $\pi r^2 = \frac{41.58}{3} = 13.86 \text{ cm}^2$.
The curved surface area $(CSA)$ of a hemisphere is given by the formula $CSA = 2\pi r^2$.
Substituting the value of $\pi r^2$,we get $CSA = 2 \times 13.86 = 27.72 \text{ cm}^2$.

(B) The formula for the curved surface area $(CSA)$ of a cylinder is given by $CSA = 2 \pi rh$.
Given radius $r = 14 \, cm$ and height $h = 10 \, cm$.
Using $\pi = \frac{22}{7}$,we substitute the values:
$CSA = 2 \times \frac{22}{7} \times 14 \times 10$
$CSA = 2 \times 22 \times 2 \times 10$
$CSA = 880 \, cm^2$.

(B) The perimeter of the base of a right circular cone is given by $2 \pi r = 132 \, cm$.
Using $\pi = \frac{22}{7}$,we have $2 \times \frac{22}{7} \times r = 132$.
$r = \frac{132 \times 7}{44} = 21 \, cm$.
The slant height $l$ is given by $l = \sqrt{h^2 + r^2} = \sqrt{72^2 + 21^2} = \sqrt{5184 + 441} = \sqrt{5625} = 75 \, cm$.
The total surface area $(TSA)$ of the cone is $\pi r(l + r)$.
$TSA = \frac{22}{7} \times 21 \times (75 + 21) = 22 \times 3 \times 96 = 66 \times 96 = 6336 \, cm^2$.

(C) Volume of the cylindrical vessel $V_{cyl} = \pi r_1^2 h_1 = \pi \times 5^2 \times 8 = 200\pi\, cm^3$.
Volume of the solid cone $V_{cone} = \frac{1}{3} \pi r_2^2 h_2 = \frac{1}{3} \times \pi \times 4^2 \times 7 = \frac{112}{3}\pi\, cm^3$.
Volume of water required = $V_{cyl} - V_{cone} = 200\pi - \frac{112}{3}\pi = \frac{600\pi - 112\pi}{3} = \frac{488}{3}\pi\, cm^3$.
Using $\pi \approx 3.14159$,Volume $\approx \frac{488}{3} \times 3.14159 \approx 162.666 \times 3.14159 \approx 511.03\, cm^3$.
Given the options provided,the closest value is $511.24\, cm^3$.

(D) The maximum capacity of the canal is equal to the volume of the canal,which is the product of the area of the cross-section and the length of the canal.
The cross-section is an isosceles trapezium with parallel sides $a = 4 \ m$ and $b = 5 \ m$,and height (depth) $h = 2 \ m$.
Area of trapezium $= \frac{1}{2} \times (a + b) \times h$
Area $= \frac{1}{2} \times (4 + 5) \times 2 = 9 \ m^2$
Volume (Capacity) $= \text{Area} \times \text{Length} = 9 \times 120 = 1080 \ m^3$.

(B) The formula for the curved surface area $(CSA)$ of a hemisphere is $2 \pi r^2$.
Given $CSA = 27.72 \text{ cm}^2$.
$2 \times \frac{22}{7} \times r^2 = 27.72$
$r^2 = \frac{27.72 \times 7}{44} = 0.63 \times 7 = 4.41$
$r = \sqrt{4.41} = 2.1 \text{ cm}$.
To verify with volume: The formula for the volume $(V)$ of a hemisphere is $\frac{2}{3} \pi r^3$.
$V = \frac{2}{3} \times \frac{22}{7} \times (2.1)^3 = \frac{2}{3} \times \frac{22}{7} \times 9.261 = 44 \times 0.441 = 19.404 \text{ cm}^3$.
Since both conditions are satisfied,the radius is $2.1 \text{ cm}$.

(C) The diagonal of the square base is given as $d = 6\sqrt{2}\, cm$.
Since the diagonal of a square with side $a$ is $d = a\sqrt{2}$,we have $a\sqrt{2} = 6\sqrt{2}$,which gives $a = 6\, cm$.
The area of the square base is $A = a^{2} = 6^{2} = 36\, cm^{2}$.
The volume $V$ of a pyramid is given by the formula $V = \frac{1}{3} \times \text{Area of base} \times \text{height}$.
Substituting the given values,$V = \frac{1}{3} \times 36 \times 12$.
$V = 12 \times 12 = 144\, cm^{3}$.

(C) The volume of the solid cone is given by $V_{cone} = \frac{1}{3} \pi r_1^2 h_1 = \frac{1}{3} \times \pi \times 8^2 \times 24 = 512 \pi \, cm^3$.
The volume of the solid cylinder formed is $V_{cylinder} = \pi r_2^2 h_2 = \pi \times 6^2 \times 6 = 216 \pi \, cm^3$.
The volume of material wasted is $V_{wasted} = V_{cone} - V_{cylinder} = 512 \pi - 216 \pi = 296 \pi \, cm^3$.
The percentage of material wasted is $\frac{V_{wasted}}{V_{cone}} \times 100 = \frac{296 \pi}{512 \pi} \times 100 = \frac{296}{512} \times 100 = 57.8125 \% \approx 57.8 \%$.

(B) The total surface area $(TSA)$ of the original solid cylinder is given by the formula: $TSA = 2 \pi r(r + h)$.
Substituting the values $r = 7 \, cm$ and $h = 20 \, cm$:
$TSA_{original} = 2 \times \frac{22}{7} \times 7 \times (7 + 20) = 44 \times 27 = 1188 \, cm^2$.
When the cylinder is bisected along its height,we get two identical cylinders,each with height $h' = \frac{20}{2} = 10 \, cm$.
Each new cylinder has a $TSA$ of: $TSA_{new} = 2 \pi r(r + h') = 2 \times \frac{22}{7} \times 7 \times (7 + 10) = 44 \times 17 = 748 \, cm^2$.
The total surface area of the two new cylinders combined is: $2 \times 748 = 1496 \, cm^2$.
The increase in surface area is: $1496 - 1188 = 308 \, cm^2$.
The percentage increase is: $\frac{308}{1188} \times 100 \approx 25.93 \%$.

(A) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Total volume of the three spheres $= \frac{4}{3} \pi (2^3 + 4^3 + 6^3) = \frac{4}{3} \pi (8 + 64 + 216) = \frac{4}{3} \pi (288) = 384 \pi \, cm^3$.
Since $25 \%$ of the material is lost,the volume of the new sphere is $75 \%$ of the total initial volume.
Volume of the new sphere $= 0.75 \times 384 \pi = 288 \pi \, cm^3$.
Let the radius of the new sphere be $R$. Then,$\frac{4}{3} \pi R^3 = 288 \pi$.
$R^3 = 288 \times \frac{3}{4} = 72 \times 3 = 216$.
$R = \sqrt[3]{216} = 6 \, cm$.

(A) Volume of the cone $V_1 = \frac{1}{3} \pi r_1^2 h_1 = \frac{1}{3} \pi (9)^2 (36) = \frac{1}{3} \pi (81) (36) = 972 \pi \, cm^3$.
Volume of the cylinder $V_2 = \pi r_2^2 h_2 = \pi (9)^2 (9) = 729 \pi \, cm^3$.
Volume of material wasted $= V_1 - V_2 = 972 \pi - 729 \pi = 243 \pi \, cm^3$.
Percentage of material wasted $= \left( \frac{\text{Volume wasted}}{\text{Original volume}} \right) \times 100 = \left( \frac{243 \pi}{972 \pi} \right) \times 100$.
$= \frac{1}{4} \times 100 = 25 \%$.

(B) The circumference of a wheel is the distance it covers in one complete revolution,given by the formula $C = 2 \pi r$.
Given the radius $r = 3.5 \, cm$,the circumference is:
$C = 2 \times \frac{22}{7} \times 3.5 = 2 \times 22 \times 0.5 = 22 \, cm$.
To find the total distance covered in $20$ revolutions,we multiply the circumference by the number of revolutions:
$\text{Total Distance} = \text{Circumference} \times \text{Number of revolutions} = 22 \, cm \times 20 = 440 \, cm$.

(B) The longest rod that can be placed in a room is equal to the length of the space diagonal of the cuboid.
The formula for the space diagonal of a cuboid is $\sqrt{l^2 + b^2 + h^2}$,where $l$ is length,$b$ is breadth,and $h$ is height.
Given: $l = 2 \text{ m}$,$b = 2 \text{ m}$,$h = 6 \text{ m}$.
Longest Rod $= \sqrt{2^2 + 2^2 + 6^2}$
$= \sqrt{4 + 4 + 36}$
$= \sqrt{44}$
$= \sqrt{4 \times 11}$
$= 2 \sqrt{11} \text{ m}$.

(B) The distance covered by a wheel in one revolution is equal to its circumference.
Circumference of the wheel $= 2 \pi r$
Given,radius $r = 21 \, cm$.
Circumference $= 2 \times \frac{22}{7} \times 21 = 2 \times 22 \times 3 = 132 \, cm$.
Distance travelled in $10$ revolutions $= 10 \times (\text{Circumference})$
$= 10 \times 132 = 1320 \, cm$.

(B) The tunnel is a right circular cylinder. The area of the iron sheet required is equal to the curved surface area of the cylinder.
Given: Diameter $d = 5\, m$,so radius $r = d/2 = 2.5\, m$.
Length (height) $h = 10\, m$.
Curved Surface Area $= 2\pi rh$.
Substituting the values: Area $= 2 \times \pi \times 2.5 \times 10 = 50\pi\, m^2$.
Thus,the required area is $50\pi\, m^2$.

(D) $1$. Convert all units to meters and hours:
Area of the tank base = $80 \, m \times 40 \, m = 3200 \, m^2$.
Area of pipe opening = $40 \, cm^2 = 40 \times 10^{-4} \, m^2 = 0.004 \, m^2$.
Speed of water = $10 \, km/h = 10000 \, m/h$.
Time = $0.5 \, h$.
$2$. Volume of water entering the tank in $0.5 \, h$:
Volume = $\text{Area of pipe} \times \text{Speed} \times \text{Time} = 0.004 \, m^2 \times 10000 \, m/h \times 0.5 \, h = 20 \, m^3$.
$3$. Calculate the rise in water level $(h)$:
Volume = $\text{Base Area} \times h$
$20 \, m^3 = 3200 \, m^2 \times h$
$h = 20 / 3200 \, m = 1 / 160 \, m$.
$4$. Convert $h$ to $cm$:
$h = (1 / 160) \times 100 \, cm = 100 / 160 \, cm = 5/8 \, cm$.

(A) Volume of the cylinder $= \pi R^2 H = \pi \times (10.5)^2 \times 38 = \pi \times 110.25 \times 38 = 4189.5\pi\, cm^3$.
Volume of one ice cream cone (cone + hemisphere) $= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$.
Given $r = 3.5\, cm$ and $h = 12\, cm$.
Volume $= \frac{1}{3}\pi (3.5)^2 (12) + \frac{2}{3}\pi (3.5)^3 = \pi (12.25 \times 4) + \pi (2 \times 14.2916) \approx 49\pi + 28.583\pi = 77.583\pi\, cm^3$.
Alternatively,using fractions: Volume $= \frac{1}{3}\pi (\frac{7}{2})^2 (12) + \frac{2}{3}\pi (\frac{7}{2})^3 = \pi (49) + \pi (\frac{343}{12}) = \pi (\frac{588+343}{12}) = \frac{931}{12}\pi\, cm^3$.
Number of cones $= \frac{\text{Volume of cylinder}}{\text{Volume of one cone}} = \frac{4189.5\pi}{931\pi / 12} = \frac{4189.5 \times 12}{931} = \frac{50274}{931} = 54$.

(B) The total volume of the aquarium is equal to the product of the number of fishes and the volume of water required per fish.
Given:
Number of fishes $= 11$
Volume required per fish $= 1.54 \, m^3$
Total Volume $= 11 \times 1.54 \, m^3$
Total Volume $= 16.94 \, m^3$

(B) The volume of a prism is given by $V = \text{Area of base} \times \text{height}$.
Given $V = 100\, cm^3$ and height $h = 25\, cm$.
Therefore, $\text{Area of base} = \frac{V}{h} = \frac{100}{25} = 4\, cm^2$.
The base is a right-angled triangle. Let the shorter sides be $x$ and $2x$.
The area of the right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times 2x = x^2$.
Equating the area, $x^2 = 4$, which gives $x = 2\, cm$.
The sides of the triangle are $x = 2\, cm$ and $2x = 4\, cm$.
The longest side (hypotenuse) is $\sqrt{x^2 + (2x)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\, cm$.

(D) The volume of a hollow hemispherical bowl is given by the formula $V = \frac{2}{3} \pi (R^3 - r^3)$,where $R$ is the outer radius and $r$ is the inner radius.
Given $R = 8\, cm$ and $r = 4\, cm$,the volume is $V = \frac{2}{3} \pi (8^3 - 4^3) = \frac{2}{3} \pi (512 - 64) = \frac{2}{3} \pi (448) = \frac{896}{3} \pi\, cm^3$.
When the bowl is melted to form a solid right circular cone,the volume remains constant.
The volume of a cone is $V = \frac{1}{3} \pi r_c^2 h$,where $r_c = 8\, cm$ is the radius of the cone and $h$ is its height.
Equating the volumes: $\frac{896}{3} \pi = \frac{1}{3} \pi (8^2) h$.
$\frac{896}{3} = \frac{64}{3} h$.
$896 = 64h$.
$h = \frac{896}{64} = 14\, cm$.

(A) Let the radius of the solid sphere be $r_1$ and the radius of the solid hemisphere be $r_2$.
Total Surface Area $(TSA)$ of a sphere $= 4 \pi r_1^2$.
Total Surface Area $(TSA)$ of a hemisphere $= 3 \pi r_2^2$.
Given that their total surface areas are equal:
$4 \pi r_1^2 = 3 \pi r_2^2$
$\frac{r_1^2}{r_2^2} = \frac{3}{4}$
$\frac{r_1}{r_2} = \frac{\sqrt{3}}{2}$.
Now,the ratio of their volumes is:
$\frac{V_{\text{sphere}}}{V_{\text{hemisphere}}} = \frac{\frac{4}{3} \pi r_1^3}{\frac{2}{3} \pi r_2^3} = 2 \times \left(\frac{r_1}{r_2}\right)^3$.
Substituting the ratio $\frac{r_1}{r_2} = \frac{\sqrt{3}}{2}$:
Ratio $= 2 \times \left(\frac{\sqrt{3}}{2}\right)^3 = 2 \times \frac{3 \sqrt{3}}{8} = \frac{3 \sqrt{3}}{4}$.

(C) When objects are melted and recast,the total volume remains constant.
Let the radius of the new sphere be $R$.
The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^{3}$.
The sum of the volumes of the three smaller spheres is equal to the volume of the new larger sphere:
$\frac{4}{3} \pi r_{1}^{3} + \frac{4}{3} \pi r_{2}^{3} + \frac{4}{3} \pi r_{3}^{3} = \frac{4}{3} \pi R^{3}$
Dividing both sides by $\frac{4}{3} \pi$,we get:
$r_{1}^{3} + r_{2}^{3} + r_{3}^{3} = R^{3}$
Taking the cube root on both sides,the radius of the new sphere is:
$R = (r_{1}^{3} + r_{2}^{3} + r_{3}^{3})^{1/3}$

(A) The volume of the conical vessel is given by $V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times 5^2 \times 8 = \frac{200\pi}{3}\, cm^3$.
The volume of water that overflows is $25\%$ of the total volume of the cone,which is equal to the total volume of the spherical balls dropped into the vessel.
Volume of water overflowed $= \frac{1}{4} \times V_{cone} = \frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}\, cm^3$.
The volume of one spherical ball with radius $r_b = \frac{1}{2}\, cm$ is $V_{ball} = \frac{4}{3} \pi r_b^3 = \frac{4}{3} \times \pi \times (\frac{1}{2})^3 = \frac{4}{3} \times \pi \times \frac{1}{8} = \frac{\pi}{6}\, cm^3$.
Let $n$ be the number of balls. Then,$n \times V_{ball} = \text{Volume of water overflowed}$.
$n \times \frac{\pi}{6} = \frac{50\pi}{3}$.
$n = \frac{50\pi}{3} \times \frac{6}{\pi} = 50 \times 2 = 100$.
Therefore,the number of balls is $100$.

(B) Let the radius of the cylinder be $r$ and the height be $h$.
The curved surface area $(CSA)$ is $2 \pi r h$ and the total surface area $(TSA)$ is $2 \pi r h + 2 \pi r^2$.
The ratio is given as $CSA : TSA = 4 : 5$.
$\frac{2 \pi r h}{2 \pi r h + 2 \pi r^2} = \frac{4}{5}$
$\frac{h}{h + r} = \frac{4}{5}$
$5h = 4h + 4r \implies h = 4r$.
Given $CSA = 1232 \, cm^2$,we have $2 \pi r h = 1232$.
Substitute $h = 4r$ into the equation:
$2 \times \frac{22}{7} \times r \times (4r) = 1232$
$8 \times \frac{22}{7} \times r^2 = 1232$
$r^2 = 1232 \times \frac{7}{8 \times 22}$
$r^2 = 1232 \times \frac{7}{176} = 7 \times 7 = 49$
$r = 7 \, cm$.
Since $h = 4r$,then $h = 4 \times 7 = 28 \, cm$.

(A) Let the volume of the smaller cone (top part) be $V_1$ and the volume of the larger cone be $V_2$.
Since the plane divides the cone into two parts of equal volume,the volume of the smaller cone is half the volume of the larger cone.
Thus,$\frac{V_1}{V_2} = \frac{1}{2}$.
For a cone cut by a plane parallel to its base,the ratio of the volumes of the smaller cone and the larger cone is equal to the cube of the ratio of their corresponding heights:
$\frac{V_1}{V_2} = \left(\frac{h_1}{h_2}\right)^3 = \frac{1}{2}$.
Taking the cube root on both sides,we get $\frac{h_1}{h_2} = \frac{1}{\sqrt[3]{2}}$.
Here,$h_1$ is the height of the smaller cone and $h_2$ is the height of the larger cone.
The height of the frustum (the bottom part) is $h_2 - h_1$.
The ratio in which the height is divided is $h_1 : (h_2 - h_1)$.
Substituting the values,we get $1 : (\sqrt[3]{2} - 1)$.

(D) First,calculate the semi-perimeter $(s)$ of the triangular base:
$s = \frac{a + b + c}{2} = \frac{13 + 20 + 21}{2} = \frac{54}{2} = 27\, cm$
Next,calculate the area of the triangular base using Heron's formula:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
$= \sqrt{27(27 - 13)(27 - 20)(27 - 21)}$
$= \sqrt{27 \times 14 \times 7 \times 6}$
$= \sqrt{15876} = 126\, cm^2$
The volume of a right prism is given by the product of the base area and the altitude (height):
$\text{Volume} = \text{Base Area} \times \text{Height}$
$= 126\, cm^2 \times 9\, cm = 1134\, cm^3$

(C) When a solid is melted and recast into another shape,the volume remains constant.
Volume of cylinder = $\pi r_1^2 h_1 = \pi \times (8)^2 \times 2 = 128\pi\, cm^3$.
Volume of cone = $\frac{1}{3} \pi r_2^2 h_2 = \frac{1}{3} \times \pi \times r_2^2 \times 6 = 2\pi r_2^2$.
Since the volumes are equal: $128\pi = 2\pi r_2^2$.
Dividing both sides by $2\pi$,we get $r_2^2 = 64$.
Therefore,$r_2 = \sqrt{64} = 8\, cm$.

(C) Volume of the ditch $= \text{length} \times \text{width} \times \text{depth} = 48 \ m \times 16.5 \ m \times 4 \ m = 3168 \ m^3$.
The tunnel is cylindrical with diameter $d = 4 \ m$, so radius $r = 2 \ m$, and length $h = 56 \ m$.
Volume of earth and stones from the tunnel $= \pi r^2 h = \frac{22}{7} \times (2)^2 \times 56 = 22 \times 4 \times 8 = 704 \ m^3$.
The portion of the ditch filled $= \frac{\text{Volume of earth}}{\text{Volume of ditch}} = \frac{704}{3168}$.
Dividing both by $352$, we get $\frac{704 \div 352}{3168 \div 352} = \frac{2}{9}$.

(B) The volume of a right circular cone is given by $V = \frac{1}{3} \pi r^{2} h$.
The lateral surface area $(LSA)$ of a right circular cone is given by $LSA = \pi r l$,where $l = \sqrt{r^{2} + h^{2}}$ is the slant height.
According to the problem,the numerical values of the volume and the lateral surface area are equal:
$\frac{1}{3} \pi r^{2} h = \pi r \sqrt{r^{2} + h^{2}}$
Dividing both sides by $\pi r$ (assuming $r \neq 0$):
$\frac{1}{3} r h = \sqrt{r^{2} + h^{2}}$
Squaring both sides:
$\frac{r^{2} h^{2}}{9} = r^{2} + h^{2}$
Dividing both sides by $r^{2} h^{2}$:
$\frac{1}{9} = \frac{r^{2} + h^{2}}{r^{2} h^{2}}$
$\frac{1}{9} = \frac{r^{2}}{r^{2} h^{2}} + \frac{h^{2}}{r^{2} h^{2}}$
$\frac{1}{9} = \frac{1}{h^{2}} + \frac{1}{r^{2}}$
Therefore,the value of $\frac{1}{h^{2}} + \frac{1}{r^{2}}$ is $\frac{1}{9}$.

(D) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Total volume of the two solid spheres $= \frac{4}{3} \pi (1^3 + 6^3) = \frac{4}{3} \pi (1 + 216) = \frac{4}{3} \pi (217) \, cm^3$.
Let the internal radius of the hollow sphere be $r \, cm$. The volume of the material of the hollow sphere is given by $V = \frac{4}{3} \pi (R^3 - r^3)$,where $R = 9 \, cm$ is the external radius.
Equating the volumes: $\frac{4}{3} \pi (217) = \frac{4}{3} \pi (9^3 - r^3)$.
$217 = 729 - r^3$.
$r^3 = 729 - 217 = 512$.
$r = \sqrt[3]{512} = 8 \, cm$.
Thickness of the hollow sphere $=$ External radius $-$ Internal radius $= 9 \, cm - 8 \, cm = 1 \, cm$.

(A) Let $R$ be the radius of the hemisphere and $r$ be the radius of each small sphere.
The volume of the hemisphere is given by $V_h = \frac{2}{3} \pi R^3$.
The volume of one sphere is given by $V_s = \frac{4}{3} \pi r^3$.
According to the problem,the hemisphere is melted to form four spheres of equal volume:
$V_h = 4 \times V_s$
Substituting the formulas:
$\frac{2}{3} \pi R^3 = 4 \times (\frac{4}{3} \pi r^3)$
Simplifying the equation:
$\frac{2}{3} R^3 = \frac{16}{3} r^3$
$R^3 = 8 r^3$
Taking the cube root on both sides:
$R = 2r$
Therefore,$r = \frac{R}{2}$,which means the radius of each sphere is $1/2$ of the radius of the hemisphere.

(B) For the largest possible cube cut out from a sphere,the diagonal of the cube is equal to the diameter of the sphere.
Let the radius of the sphere be $r = 6 \sqrt{3} \text{ cm}$.
Diameter of the sphere $D = 2r = 2 \times 6 \sqrt{3} = 12 \sqrt{3} \text{ cm}$.
Let the side length of the cube be $a$. The diagonal of a cube is given by $\sqrt{3}a$.
Equating the diagonal of the cube to the diameter of the sphere:
$\sqrt{3}a = 12 \sqrt{3}$
$a = 12 \text{ cm}$.
The surface area of a cube is given by $6a^2$.
Surface area $= 6 \times (12)^2 = 6 \times 144 = 864 \text{ cm}^2$.

(B) The tank is initially full, and after removing water, $1/3$ of the tank remains filled. Therefore, the volume of water removed is $1 - 1/3 = 2/3$ of the total volume of the tank.
The volume of the cubical tank is $V = (\text{side})^3 = (1.2\, m)^3 = 1.728\, m^3$.
Since $1\, m^3 = 1000\, litres$, the total volume of the tank is $1.728 \times 1000 = 1728\, litres$.
The volume of water removed by $64$ buckets is $2/3$ of the total volume: $V_{\text{removed}} = 1728 \times (2/3) = 1152\, litres$.
The volume of each bucket is the total removed volume divided by the number of buckets: $V_{\text{bucket}} = 1152 / 64 = 18\, litres$.

(C) Area of base $= 10 \times 10 = 100 \text{ cm}^2$.
Perimeter of base $= 4 \times 10 = 40 \text{ cm}$.
The slant height $(l)$ of the pyramid is calculated using the height $(h = 12 \text{ cm})$ and half the side of the base $(a/2 = 5 \text{ cm})$:
$l = \sqrt{(a/2)^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}$.
Curved Surface Area $(C.S.A.)$ of the pyramid $= \frac{1}{2} \times \text{Perimeter of base} \times \text{slant height}$
$C.S.A. = \frac{1}{2} \times 40 \times 13 = 20 \times 13 = 260 \text{ cm}^2$.
Total Surface Area $(T.S.A.)$ $= C.S.A. + \text{Area of base}$
$T.S.A. = 260 \text{ cm}^2 + 100 \text{ cm}^2 = 360 \text{ cm}^2$.

(A) Given: Height of the cone $h = 14 \, m$,Floor area $= \pi r^{2} = 346.5 \, m^{2}$.
First,find the radius $r$:
$\frac{22}{7} \times r^{2} = 346.5$
$r^{2} = \frac{346.5 \times 7}{22} = 110.25$
$r = \sqrt{110.25} = 10.5 \, m$.
Now,find the slant height $l$:
$l = \sqrt{r^{2} + h^{2}} = \sqrt{10.5^{2} + 14^{2}} = \sqrt{110.25 + 196} = \sqrt{306.25} = 17.5 \, m$.
Curved surface area of the tent $= \pi r l = \frac{22}{7} \times 10.5 \times 17.5 = 22 \times 1.5 \times 17.5 = 577.5 \, m^{2}$.
Width of canvas $= 75 \, cm = 0.75 \, m$.
Length of canvas $= \frac{\text{Curved surface area}}{\text{Width}} = \frac{577.5}{0.75} = 770 \, m$.

(D) The volume of a prism is given by the formula: $\text{Volume} = \text{Area of base} \times \text{Height}$.
Given:
$\text{Area of base} = \frac{3 \sqrt{3}}{2} P^{2} \, \text{cm}^{2}$
$\text{Height} = 100 \sqrt{3} \, \text{cm}$
$\text{Volume} = 7200 \, \text{cm}^{3}$
Substituting these values into the formula:
$7200 = \left( \frac{3 \sqrt{3}}{2} P^{2} \right) \times (100 \sqrt{3})$
$7200 = \frac{3 \times 100 \times (\sqrt{3} \times \sqrt{3})}{2} \times P^{2}$
$7200 = \frac{300 \times 3}{2} \times P^{2}$
$7200 = 450 \times P^{2}$
$P^{2} = \frac{7200}{450}$
$P^{2} = 16$
$P = 4$.

(D) Let the radius of the sphere be $r \, cm$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
The surface area of a sphere is given by $A = 4 \pi r^2$.
According to the problem,the volume is numerically equal to the surface area:
$\frac{4}{3} \pi r^3 = 4 \pi r^2$
Dividing both sides by $4 \pi r^2$ (assuming $r \neq 0$):
$\frac{r}{3} = 1$
$r = 3 \, cm$.
The diameter of the sphere is $d = 2r = 2 \times 3 = 6 \, cm$.