Mensuration II (Important results)


The following results are very important to solve various mensuration problems.

1.  The largest possible sphere that can be chiseled out from a cube of side "a" cm. 
 Diagonal of the sphere is a, so radius = a/2.
Remaining empty space in the cube = ${a^3} - \displaystyle\frac{{\pi {a^3}}}{6}$

2. The largest possible cube that can be chiseled out from a sphere of radius "a" cm
Here OA = radius of the sphere. So diameter of the sphere = 2a.
Let the side of the square = $x$, then the diagonal of the cube = $\sqrt 3 x$
$ \Rightarrow \sqrt 3 x = 2a$ $ \Rightarrow x = \dfrac{{2a}}{{\sqrt 3 }}$
Therefore side of the square = $\displaystyle\frac{{2a}}{{\sqrt 3 }}$ 

3.  The largest possible cube that can be chiseled out from a hemisphere of radius 'a' cm. 
Sol:

Given, the radius of the hemi sphere AC = $a$.  Let the side of the cube is $x$.
From the above diagram, $B{E^2} + E{D^2} = B{D^2}$
$ \Rightarrow {x^2} + {x^2} = B{D^2}$
$ \Rightarrow BD = \sqrt 2 x$
$ \Rightarrow BC = \dfrac{{\sqrt 2 x}}{2} = \dfrac{x}{{\sqrt 2 }}$
From $\Delta $ABC, $A{C^2} = A{B^2} + B{C^2}$
$ \Rightarrow {a^2} = {x^2} + {\left( {\dfrac{x}{{\sqrt 2 }}} \right)^2}$
$ \Rightarrow {a^2} = \dfrac{{3{x^2}}}{2}$
$ \Rightarrow x = \sqrt {\dfrac{2}{3}} a$
The edge of the cube = $a\sqrt {\displaystyle\frac{2}{3}} $

4.  The largest square that can be inscribed in a right angled triangle ABC when one of its vertices coincide with the vertex of right of the triangle.
Solution:

let the side of the square = $x$
DE // BC, therefore, $\Delta $ADE and $\Delta $ABC are similar.
$ \Rightarrow \dfrac{{AD}}{{AB}} = \dfrac{{DE}}{{BC}}$
$ \Rightarrow \dfrac{{b - x}}{b} = \dfrac{x}{a}$
$ \Rightarrow 1 - \dfrac{x}{b} = \dfrac{x}{a}$
$ \Rightarrow 1 = \dfrac{x}{b} + \dfrac{x}{a}$
$ \Rightarrow 1 = x\left( {\dfrac{{a + b}}{{ab}}} \right)$
$ \Rightarrow x = \left( {\dfrac{{ab}}{{a + b}}} \right)$
Side of the square = $\dfrac{{ab}}{{a + b}}$ and area of the square = ${\left( {\dfrac{{ab}}{{a + b}}} \right)^2}$

5.  The largest square that can be inscribed in a right angled triangle ABC when one of its vertices lies on the hypotenuse of the triangle
Solution 1:

From the above diagram,  $\Delta $ABC and $\Delta $AFD are similar.
$ \Rightarrow Tan\alpha  = \dfrac{{AD}}{{FD}} = \dfrac{{AB}}{{BC}}$
$ \Rightarrow \dfrac{{AD}}{x} = \dfrac{a}{b}$
$ \Rightarrow AD = \dfrac{{xa}}{b}$ - - - - - (1)
Also, $\Delta $ABC and $\Delta $EGC are similar.
$ \Rightarrow Tan\alpha  = \dfrac{{GE}}{{EC}} = \dfrac{a}{b}$
$ \Rightarrow \dfrac{x}{{EC}} = \dfrac{a}{b}$
$ \Rightarrow EC = \dfrac{{xb}}{a}$ - - - - - (2)
We know that c= AD + x + EC
$ \Rightarrow c = \dfrac{{xa}}{b} + x + \dfrac{{xb}}{a}$
$ \Rightarrow c = x\left( {\dfrac{a}{b} + 1 + \dfrac{b}{a}} \right)$
$ \Rightarrow c = x\left( {\dfrac{{{a^2} + ab + {b^2}}}{{ab}}} \right)$
$ \Rightarrow x = \left( {\dfrac{{abc}}{{{a^2} + ab + {b^2}}}} \right)$
Side of the square = ${\displaystyle\frac{{abc}}{{{a^2} + {b^2} + ab}}}$

Solution 2:

From the above diagram, drop a perpendicular to AC from vertex B. 
Area of $\Delta $ABC = $\dfrac{1}{2} \times a \times b = \dfrac{1}{2} \times BF \times c$
$\therefore BF = \dfrac{{ab}}{c}$ - - - - - (1)
Now $\Delta $BDE and $\Delta $ABC are similar. 
$ \Rightarrow \dfrac{{BG}}{{BF}} = \dfrac{{DE}}{{AC}}$
$ \Rightarrow \dfrac{{\dfrac{{ab}}{c} - x}}{{\dfrac{{ab}}{c}}} = \dfrac{x}{c}$
$ \Rightarrow 1 - \dfrac{{xc}}{{ab}} = \dfrac{x}{c}$
$ \Rightarrow 1 = \dfrac{{xc}}{{ab}} + \dfrac{x}{c}$
$ \Rightarrow 1 = x\left( {\dfrac{c}{{ab}} + \dfrac{1}{c}} \right)$
$ \Rightarrow 1 = x\left( {\dfrac{{{c^2} + ab}}{{abc}}} \right)$
$ \Rightarrow x = \dfrac{{abc}}{{{c^2} + ab}}$ = $\dfrac{{abc}}{{{a^2} + {b^2} + ab}}$

6.  The largest square that can be inscribed in a semi circle of radius 'r' units
Area of the square = $\displaystyle\frac{3}{5}{r^2}$

7.  The largest square that can be inscribed in a quadrant of radius 'r' cm. 
Side of the square = $\displaystyle\frac{r}{{\sqrt 2 }}$, and area of the square = $\displaystyle\frac{{{r^2}}}{2}$

8.  The largest circle that can be inscribed in the semi circle of radius 'r' cm is 

Inscribed circle area = $\displaystyle\frac{{\pi {r^2}}}{4}$
(Rememeber: Inscribed circle area is half of the semi circle area)

9.  The largest circle that can be inscribed in a quadrant of radius 'r' cm is 

Area of the circle = $\displaystyle\frac{{\pi {r^2}}}{{3 + 2\sqrt 2 }}$

Level - 2: 

10. The largest cube that can be chiseled out from a cone of height 'h' cm and radius of 'r' cm

Solution:

Let the side of the square = $x$
DHE is a right angle triangle.   Therefore, $D{E^2} = D{H^2} + H{E^2}$
$ \Rightarrow DE = \sqrt {{x^2} + {x^2}}  = \sqrt 2 x$
From the diagram, $\Delta ADE$ and $\Delta ABC$ are similar.
$ \Rightarrow \dfrac{{AF}}{{AG}} = \dfrac{{DE}}{{BC}}$
$ \Rightarrow \dfrac{{h - x}}{h} = \dfrac{{\sqrt 2 x}}{{2r}}$
$ \Rightarrow 1 - \dfrac{x}{h} = \dfrac{x}{{\sqrt 2 r}}$
$ \Rightarrow \dfrac{x}{{\sqrt 2 r}} + \dfrac{x}{h} = 1$
$ \Rightarrow x\left( {\dfrac{1}{{\sqrt 2 r}} + \dfrac{1}{h}} \right) = 1$
$\Rightarrow x\left( {\dfrac{{h + \sqrt 2 r}}{{\sqrt 2 rh}}} \right) = 1$
$ \Rightarrow x = \dfrac{{\sqrt 2 rh}}{{h + \sqrt 2 r}}$
Square side = $\dfrac{{\sqrt 2 rh}}{{h + \sqrt 2 r}}$

11.  Find the maximum volume of cylinder that can be made out of a cone of radius 'r' and height 'h'

Sol:
Let the radius of the cylinder = $x$ and height = $y$.  We have to maximize the volume of the cylinder V = $\pi {x^2}y$.  To maximize the volume we have to change one of the given variable in the equation into "r"and "h" and differentiate the equation with respect to the other variable.


From the above, $\Delta ADE$ and $\Delta ABC$ are similar. 
So, $ \Rightarrow \dfrac{{h - y}}{h} = \dfrac{{2x}}{{2r}}$ - - - - - (1)
$ \Rightarrow h - y = \dfrac{{xh}}{r}$
$ \Rightarrow h - \dfrac{{xh}}{r} = y$
Substituting the above result in the volume we get, 
$ V = \pi {x^2}\left( {h - \dfrac{{hx}}{r}} \right)$
$ V = \pi h{x^2} - \dfrac{{\pi h{x^3}}}{r}$
Differentiation the above equation w.r.t x, 
${V^1}(x) = 2\pi hx - \dfrac{{3\pi h{x^2}}}{r}$
By equating the above equation to zero, we can find the value of x where the above equation becomes maximum. 
$2\pi hx - \dfrac{{3\pi h{x^2}}}{r}$ = 0
$ \Rightarrow 2\pi hx\left( {2 - \dfrac{{3x}}{r}} \right) = 0$
h, x cannot be zero. So ${2 - \dfrac{{3x}}{r}}$ = 0
$ \Rightarrow x = \dfrac{{2r}}{3}$ - - - - - (2)
Substituting the above result in equation (1), 
$ \Rightarrow \dfrac{{h - y}}{h} = \dfrac{{2\left( {\dfrac{{2r}}{3}} \right)}}{{2r}}$
$ \Rightarrow \dfrac{{h - y}}{h} = \dfrac{2}{3}$
$ \Rightarrow h - y = \dfrac{2}{3}h$
$ \Rightarrow y = \dfrac{h}{3}$ - - - - - (3)
Therefore, maximum volume of the cylinder = $ \pi {\left( {\dfrac{{2r}}{3}} \right)^2}\left( {\dfrac{h}{3}} \right)$
$ \Rightarrow \dfrac{4}{{27}}\pi {r^2}h$

12. Find the maximum volume of sphere that can be inscribed in a cone.

Sol:
Let the cone height = $h$ and radius = $r$.  Also let the radius of the sphere inside the cone = $x$
We have to maximize the volume of the sphere = $V = \dfrac{4}{3}\pi {x^3}$
From the above diagram, $\Delta AFC$ and $\Delta ADE$ are similar.  ($\because$ AAA rule)
Therefore, $\dfrac{{CF}}{{DE}} = \dfrac{{AC}}{{AD}}$
From Pythagoras rule, $AC = \sqrt {A{F^2} + F{C^2}}  = \sqrt {{h^2} + {r^2}} $
Now, $\dfrac{r}{x} = \dfrac{{\sqrt {{r^2} + {h^2}} }}{{h - x}}$
$ \Rightarrow rh - rx = x\sqrt {{r^2} + {h^2}} $
$ \Rightarrow rh = rx + x\sqrt {{r^2} + {h^2}} $
$ \Rightarrow rh = x\left( {r + \sqrt {{r^2} + {h^2}} } \right)$
$ \Rightarrow x = \dfrac{{rh}}{{r + \sqrt {{r^2} + {h^2}} }}$

13. Find the maximum volume of cone that can be inscribed in a sphere.

Sol:

Let the radius of the sphere = $r$.
Let the radius of the cone = $x$, and height = $y$
We have to maximize the volume of the cone = V = $\dfrac{1}{3}\pi {x^2}y$ - - - - - (1)
From $\Delta $BCD, $B{D^2} = B{C^2} + C{D^2}$
$ \Rightarrow {r^2} = {x^2} + {\left( {y - r} \right)^2}$
$ \Rightarrow {r^2} = {x^2} + {y^2} + {r^2} - 2yr$
$ \Rightarrow {x^2} = 2yr - {y^2}$ - - - - (2)
Substituting the above result in equation (1), and differentiating w.r.t. y, and equating to zero, 
V= $\dfrac{1}{3}\pi \left( {2yr - {y^2}} \right)y$
${V^1}(y) = \dfrac{1}{3}\pi \left( {4yr - 3{y^2}} \right)$ = 0
$ \Rightarrow y = \dfrac{{4r}}{3}$
Now CD = $\dfrac{{4r}}{3} - r = \dfrac{r}{3}$
From $\Delta $BCD, 
${r^2} = {x^2} + {\left( {\dfrac{r}{3}} \right)^2}$
$ \Rightarrow x = \sqrt {\dfrac{{8{r^2}}}{9}}  = \dfrac{{2\sqrt 2 r}}{3}$ - - - - (3)
Volume of the cone = $\dfrac{1}{3}\pi {x^2}y$ = $\dfrac{1}{3}\pi {\left( {\dfrac{{2\sqrt 2 r}}{3}} \right)^2}\dfrac{{4r}}{3} = \dfrac{{32}}{{81}}\pi {r^3}$