Chapter 12, Problem 16RP

To determine

To calculate: The total volume of the castle, including the towers.

Answer to Problem 16RP

The total volume of the castle, including the towers is $153570.003{\text{ unit}}^{\text{3}}$ .

Explanation of Solution

Given information:

Main Castle: length $\left(l_1\right)=100\text{ units}$ , breadth $\left(b_1\right)=50\text{ units}$ and height $\left(h_1\right)=30\text{ units}$

Front-left Pillar: length of cuboid $\left(l_2\right)=3\text{ units}$ , breadth of cuboid $\left(b_2\right)=3\text{ units}$ , height of cuboid $\left(h_2\right)=50\text{ units}$ , length of square pyramid $\left(l_3\right)=3\text{ units}$ , breadth of square pyramid $\left(b_3\right)=3\text{ units}$ and height of square pyramid $\left(h_3\right)=3\text{ units}$

Front-right Pillar: radius of cylinder $\left(r_1\right)=3\text{ units}$ , height of cylinder $\left(h_4\right)=50\text{ units}$ , radius of cone $\left(r_2\right)=3\text{ units}$ and height of cone $\left(h_5\right)=3\text{ units}$

Rear-left Pillar: radius of cylinder $\left(r_3\right)=3\text{ units}$ , height of cylinder $\left(h_6\right)=50\text{ units}$ and radius of hemisphere $\left(r_4\right)=3\text{ units}$

Rear-right Pillar: base of the triangular pyramid’s base $\left(b_4\right)=3\text{ units}$ , altitude of the triangular pyramid $\left(h_7\right)=3\text{ units}$ , base of triangular prism’s base $\left(b_5\right)=3\text{ units}$ and altitude of triangular prism $\left(h_8\right)=50\text{ units}$

Formula used:

Volume of Cuboid $\left(V\right)=lbh$ …………………………………………. (i)

where l = length of cuboid

b = breadth of cuboid

h = height of cuboid

Volume of Square Pyramid $\left(V\right)=\left(\frac{1}{3}\right)lbh$ ……………………………. (ii)

where l = length of square pyramid

b = breadth of square pyramid

h = height of square pyramid

Volume of Cylinder $\left(V\right)=\pi r^{2}h$ ………………………………………. (iii)

where r = radius of cylinder

h = height of cylinder

Volume of Cone $\left(V\right)=\left(\frac{1}{3}\right)\pi r^{2}h$ ……………………………………… (iv)

where r = radius of cone

h = height of cone

Volume of Hemisphere $\left(V\right)=\left(\frac{2}{3}\right)\pi r^3$ ………………………………… (v)

where r = radius of hemisphere

Volume of the Triangular Pyramid $\left(V\right)=\left(\frac{1}{3}\right)\left(\frac{1}{2}bh\right)H$ ……………... (vi)

where b = base of the triangular pyramid

h = height of triangular base of pyramid

H = altitude of the triangular pyramid

Volume of Triangular Prism $\left(V\right)=\left(\frac{1}{2}bh\right)H$ ………………………... (vii)

where b = base of the triangular prism

h = height of triangular base of prism

H = altitude of the triangular prism

Pythagoras Theorem for any right-angled triangle,

  $h^2=b^2+p^2$ ………………………………………………………….. (viii)

where h = hypotenuse of the right-angled triangle

b = base of the right-angled triangle

p = perpendicular of the right-angled triangle

Calculation:

Total Volume of Castle $\left(V\right)$ = Volume of Main Castle $\left(V_1\right)$ + Volume of Front-left Pillar $\left(V_2\right)$ +

Volume of Front-right Pillar $\left(V_3\right)$ + Volume of Rear-left Pillar $\left(V_4\right)$ + Volume of Rear-right Pillar $\left(V_5\right)$

  $V=V_1+V_2+V_3+V_4+V_5$ …………………………………………….. (ix)

Now using equation (i) to calculate the volume of main castle,

  $V_1=l_1{b}_1{h}_1$

Substituting the given values of $l_1$ , $b_1$ and $h_1$ in above equation to get,

  $\begin{array}{l}{V}_1=100\times 50\times 30\\ V_1=150000{\text{ unit}}^{\text{3}}\end{array}$

Volume of Front-left Pillar = Volume of Cuboid + Volume of Square Pyramid

Using equations (i) and (ii) in above equation to get,

  $V_2=l_2{b}_2{h}_2+\left(\frac{1}{3}\right)l_3{b}_3{h}_3$

Substituting the given values in above equation,

  $\begin{array}{l}{V}_2=3\times 3\times 50+\left(\frac{1}{3}\right)\times 3\times 3\times 3\\ V_2=450+3\times 3\\ V_2=450+9\\ V_2=459{\text{ unit}}^{\text{3}}\end{array}$

Volume of Front-right Pillar = Volume of Cylinder + Volume of Cone

Using equations (iii) and (iv) in above equation to get,

  $V_3=\pi r_1{}^2{h}_1+\left(\frac{1}{3}\right)\pi r_2{}^2{h}_2$

Substituting the given values in the above equation,

  $\begin{array}{l}{V}_3=\pi \times 3^2\times 50+\left(\frac{1}{3}\right)\times \pi \times 3^2\times 3\\ V_3=\pi \times 9\times 50+\pi \times 3^2\\ V_3=450\pi +9\pi \\ V_3=459\pi \\ V_3=1441.991{\text{ unit}}^{\text{3}}\end{array}$

Volume of Rear-left Pillar = Volume of Cylinder + Volume of Hemisphere

Using equations (iii) and (v) in above equation to get,

  $V_4=\pi r_3{}^2{h}_3+\left(\frac{2}{3}\right)\pi r_4{}^3$

Substituting the given values in above equation,

  $\begin{array}{l}{V}_4=\pi \times 3^2\times 50+\left(\frac{2}{3}\right)\times \pi \times 3^3\\ V_4=450\pi +2\times \pi \times 3^2\\ V_4=450\pi +18\pi \\ V_4=468\pi \\ V_4=1470.265{\text{ unit}}^{\text{3}}\end{array}$

Volume of Rear-right Pillar = Volume of Triangular Prism + Volume of Triangular Pyramid

Using equations (vi) and (vii) in above equation to get,

  $V_5=\left(\frac{1}{2}{b}_{5}H\right)h_8+\left(\frac{1}{3}\right)\left(\frac{1}{2}{b}_{4}H\right)h_7$ ……………………………… (x)

Here we can see that height of triangular prism’s base is taken equal to the height of triangular pyramid’s base because the dimensions of bases of both the solids are same. This height of bases of both the solids i.e., His unknown. Let us calculate it by considering triangular pyramid’s base

We can calculate Hby bisecting any side whose dimension is known to us and then applying Pythagoras theorem to it.

In $△ADB$ as per Pythagoras Theorem,

  $\begin{array}{l}A{B}^2=B{D}^2+A{D}^2\\ A{D}^2=A{B}^2-B{D}^2\end{array}$

Taking square root of the terms on both sides of the above equation,

  $AD=\sqrt{A{B}^2-B{D}^2}$

Here $AD=H$ and substituting the known values in above equation

  $\begin{array}{l}H=\sqrt{3^2-{1.5}^2}\\ H=\sqrt{9-2.25}\\ H=\sqrt{6.75}\\ H=2.598\text{ units}\end{array}$

Substituting the given values along with the calculated value of Hin theequation (x),

  $\begin{array}{l}{V}_5=\left(\frac{1}{2}\times 3\times 2.598\right)\times 50+\left(\frac{1}{3}\right)\times \left(\frac{1}{2}\times 3\times 2.598\right)\times 3\\ V_5=\left(3\times 1.299\right)\times 50+\left(3\times 1.299\right)\\ V_5=194.85+3.897\\ V_5=198.747{\text{ unit}}^{\text{3}}\end{array}$

Now putting the obtained values of V_1,V_2,V_3,V₄and V₅in equation (ix) to get,

  $\begin{array}{l}V=150000+459+1441.991+1470.265+198.747\\ V=153570.003{\text{ unit}}^{\text{3}}\end{array}$

Hence, the total volume of the castle, including the towers is $153570.003{\text{ unit}}^{\text{3}}$ .