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
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzYRnVBEP6RZlohLC_AMUNj0u7ssGD8qyu5IuHFlAkyx7g4SQic4yhjG1ecihTRZYc639r5dZOqAKO_e8rwxxH84YU3VrUvR4zRXJIki2kZ3Ki86Bof8sYYrBYFkBwVvUWxusMW_70gwLG/s1600/2.png)
3. The largest possible cube that can be chiseled out from a hemisphere of radius 'a' cm.
Sol:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHqlBoTKjSUjWgxsLZZKITRjFy2euR-udTAmIrNblWRe0buw62aThhwIoocBk3FYYHrPQuFWP1Wb1g-sNdf_JwNaj_aByk5AHLqOTbSnFfVXtvrtxcR01ZDVr9o7ZoM_1Or1PmM3_TilA8/s1600/Graphic1.png)
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:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZaoVRVqKGZ3WUDNmaKAWtURffd8-bRbVoyggRCR5RefXXBaINAe2VOZ0o8tXGXJqCSpo0lhr-PVDb80DwoOQwlo7qu4wCD_bU2O3GVJzXIuIhUIfaY2Cb89w08yjSnPGNbX0jW1J2apdX/s1600/Graphic1.png)
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
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEikwAuLlCrsB8CeJxOvOhN7nNJKtfXbrqrfUH8vZsQAk1vuUc5o1e-OMUymWgKH3nad-hNs7Cwzqy-stgOUMjkLZ-BzMBkFvV9zNen-VTQn7D6X75SLZXSbU74uUHG_-2T1AFouzK83cQA_/s1600/8.png)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjuqk9bRH7fEvHC2a_M7ZY-0gUU_3c0UbwMwhlMWnUAsP8aKJuZgFi8qG9-NpcUHkdjZ3gyYZukWsfFYpzbizJS58MlCkTvr2O7E09otqAaQeyTfeAzxbRO-uGxijxDoE3PBcQ2SakTuMcr/s1600/Graphic15.png)
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:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiSVQpVXXNymsUpy03Kje_IG2tjzy7gdS0IOOXQpG7pIPLfuANWVPU_EDJCaSmBivki64NmiJSKwsAl1tC7rTx9zJirDeKO0atUu9p_pCjoGuSfmTUfMudsDbHhMwu93JGucgft_gMeEaFi/s1600/Graphic16.png)
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
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7pHNdVd6I1AvaLWLhDcv2lhu55J7rqK3qhYWrmHjq3MWNfRGiS3zujQ5VmX7QZWhKz9XyAoUM_Vg16V-rg4NKjVCEeE_rFFmcLUziVkmDOLgnw6KMZsKpxdVXRmlyyuieN-ae_HuJphrm/s200/12.png)
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
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgZSNHuvzcuNq-I-KZ78Ot1LK-s1dCNpzqg4H9tvhyphenhyphenN661HDExy8xg32UwO3AJ82toOcSkOHYoKNJCuJ-8xKxY_gmvDxsPqmMkEdaFloShIUlEdpAqNR1-WHZpcy_WTL3uu7bdl-B0xU6I7/s200/cube_in_cone.jpg)
Solution:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhrX2Nytlts4LHJ2ik63Wq3idfqxHo1gKwoo_MJJ9OXckW65ZECoSpd0FdzBEH3bpHJdRywC1H05lTT_oxpdHC7fVkAAgUUCGHVkkhyphenhyphenkYdSUUYDVmdkgn36bkH8C8Xd7V_JkP2mgaiU8Rxa/s320/Graphic1.JPG)
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'
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiBsuhYNbXwzSiHJ_xVuPeArqvwlCJ7h5ozGAP9CVgz1aWyXHmtLhSjqZqhOoMYrWUOUnII_iU-5lGSGCk85rpttZN8yhphYfBK-nXDBQyw86edh_3IU_mr6QKOYl_hOfLEY8npnEFMYlO3/s200/Graphic1.JPG)
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.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7eCDFnIqLIWKw6doCZQOOK_A4J6x6Z7X26WIUjXtFu3LJ551Ytf0tx8b1MuhU33LCZdNBqIP38xobbK_jQDi8f6SZ5cGbZBWAzE4BxB1ELVTXXn41bH1W6BiyP1I86-OpFB7GSWGGOmbN/s320/Graphic2.jpg)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEic2cD0uCh1A39aXeNQHR-yFwn6tdfayZbubeZ2t85gixeYYQcylZEYOmQLmiCCmoRsh7qqwpvSwOTfSDg7K3eTmt8c9IHZod3f0o0mowR9Cz4eX1A6n6NlRF3Akg2kdZvvDZltVq_QOZh2/s200/Graphic1.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhc-uS-ljmV5-76l-6fV297I7MyDyq_RdRR2jgDUYbkEauMA9FqDAXBmFJFvIvFilknG6yJQu-VCdCxTY0rTDTF6vMYc8cG-B9uVRd95XXC-Dn_c9xnsQztwW2AGQLktdD4-BM1M71D2roD/s320/Graphic2.jpg)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEit8maaFWORC0Y5UnJbVypyO56bc6SU6a4oTDngoFsi6m-tbJTzxEGJyrU22WSasM_5VaJyTtCS6xQqnPUC0jc0H2yx60P0pxLfRMfdmbm2IYtlR9WeSrhhyphenhyphen4gE4oAiRrxqzdYFk5FocTmc/s200/Graphic1.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJvWqJXHLxGKJVtg3tG2EeURpWgGe1uiVqDhF_k0hhnmHBP1Bx_sXy2Gt1Pb6zS2UtppYYMUH_ra8do_X50WRfvtQFO5mXCZ_AMa6LNB440nCDDLN5boJYcXhM2emd0eCREY49nmy6TcpJ/s320/Graphic2.jpg)
1. The largest possible sphere that can be chiseled out from a cube of side "a" cm.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_J5WQRxf4jIFkLkTIDVWqORDRN62B656mNtW0FNKdKriktW-qfRoqo4D7lu0lLOmPZCyH0obi2zvTQJRGGa70UoweTN4IUQ_ILbcHgyHSuZNa1lRUINE841B8-8Ua-m8cnTURDkdtFU11/s1600/1.png)
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
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzYRnVBEP6RZlohLC_AMUNj0u7ssGD8qyu5IuHFlAkyx7g4SQic4yhjG1ecihTRZYc639r5dZOqAKO_e8rwxxH84YU3VrUvR4zRXJIki2kZ3Ki86Bof8sYYrBYFkBwVvUWxusMW_70gwLG/s1600/2.png)
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 = $\displaystyle\frac{{2a}}{{\sqrt 3 }}$
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjEJS1Lie0MPTMGLzO7cdBXpLZakhJBdbv5VIXenayHZCsLkDQ5qMdLJOm6lw0lLr_xpnV0OW-3Dvb58jPF5yTWXbzvJpBwjBjJTszEfDjTaXvgcFJpAeLveMVucz-t6MKQ_umfu35g8C3s/s1600/6.png)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHqlBoTKjSUjWgxsLZZKITRjFy2euR-udTAmIrNblWRe0buw62aThhwIoocBk3FYYHrPQuFWP1Wb1g-sNdf_JwNaj_aByk5AHLqOTbSnFfVXtvrtxcR01ZDVr9o7ZoM_1Or1PmM3_TilA8/s1600/Graphic1.png)
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.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxVdAT9t4XqlKjpreSt0zzCnOtraPkI1qHc2N4LnJyTx4p0CS93-wY_ktOVDLtqn7n3chFOXQKx61MMB7AwDY-TxVXp_AwjACkn3dyqgxm33XmoijQjXUHzFHn_fYItvzy0eiVpIIp7VXn/s1600/7.png)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZaoVRVqKGZ3WUDNmaKAWtURffd8-bRbVoyggRCR5RefXXBaINAe2VOZ0o8tXGXJqCSpo0lhr-PVDb80DwoOQwlo7qu4wCD_bU2O3GVJzXIuIhUIfaY2Cb89w08yjSnPGNbX0jW1J2apdX/s1600/Graphic1.png)
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
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEikwAuLlCrsB8CeJxOvOhN7nNJKtfXbrqrfUH8vZsQAk1vuUc5o1e-OMUymWgKH3nad-hNs7Cwzqy-stgOUMjkLZ-BzMBkFvV9zNen-VTQn7D6X75SLZXSbU74uUHG_-2T1AFouzK83cQA_/s1600/8.png)
Solution 1:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjuqk9bRH7fEvHC2a_M7ZY-0gUU_3c0UbwMwhlMWnUAsP8aKJuZgFi8qG9-NpcUHkdjZ3gyYZukWsfFYpzbizJS58MlCkTvr2O7E09otqAaQeyTfeAzxbRO-uGxijxDoE3PBcQ2SakTuMcr/s1600/Graphic15.png)
$ \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:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiSVQpVXXNymsUpy03Kje_IG2tjzy7gdS0IOOXQpG7pIPLfuANWVPU_EDJCaSmBivki64NmiJSKwsAl1tC7rTx9zJirDeKO0atUu9p_pCjoGuSfmTUfMudsDbHhMwu93JGucgft_gMeEaFi/s1600/Graphic16.png)
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
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjslTztkoCyP9ayQ5txVJwNi5fKhZX0uLJtlFJjwP16_vksFQ-PFEn26dmbr3NSzF0mRERtlF73fTIdWtX6kOF5zPSMig-n9eHDX-En0tXjcKOy_teT-DXwqDeYCcaoj7PuoWBGpeejsP1w/s1600/9.png)
7. The largest square that can be inscribed in a quadrant of radius 'r' cm.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi-bMKFlFpDSoDm4eF5QahJqa7XtEdR1It2XfMXBahiJQKpgMbwjwR_-SkXfAdz3G4k9T6WZTxwaTMT1MjE2f8J4sxQp7Zanlhekz4hzop12P3_pBEYEEb3IDzZJjXSkpowDCD3c9w4wxpf/s200/10.png)
8. The largest circle that can be inscribed in the semi circle of radius 'r' cm is
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7pHNdVd6I1AvaLWLhDcv2lhu55J7rqK3qhYWrmHjq3MWNfRGiS3zujQ5VmX7QZWhKz9XyAoUM_Vg16V-rg4NKjVCEeE_rFFmcLUziVkmDOLgnw6KMZsKpxdVXRmlyyuieN-ae_HuJphrm/s200/12.png)
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
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgZSNHuvzcuNq-I-KZ78Ot1LK-s1dCNpzqg4H9tvhyphenhyphenN661HDExy8xg32UwO3AJ82toOcSkOHYoKNJCuJ-8xKxY_gmvDxsPqmMkEdaFloShIUlEdpAqNR1-WHZpcy_WTL3uu7bdl-B0xU6I7/s200/cube_in_cone.jpg)
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.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7eCDFnIqLIWKw6doCZQOOK_A4J6x6Z7X26WIUjXtFu3LJ551Ytf0tx8b1MuhU33LCZdNBqIP38xobbK_jQDi8f6SZ5cGbZBWAzE4BxB1ELVTXXn41bH1W6BiyP1I86-OpFB7GSWGGOmbN/s320/Graphic2.jpg)
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:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhc-uS-ljmV5-76l-6fV297I7MyDyq_RdRR2jgDUYbkEauMA9FqDAXBmFJFvIvFilknG6yJQu-VCdCxTY0rTDTF6vMYc8cG-B9uVRd95XXC-Dn_c9xnsQztwW2AGQLktdD4-BM1M71D2roD/s320/Graphic2.jpg)
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:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJvWqJXHLxGKJVtg3tG2EeURpWgGe1uiVqDhF_k0hhnmHBP1Bx_sXy2Gt1Pb6zS2UtppYYMUH_ra8do_X50WRfvtQFO5mXCZ_AMa6LNB440nCDDLN5boJYcXhM2emd0eCREY49nmy6TcpJ/s320/Graphic2.jpg)
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 = $\dfrac{1}{3}\pi {x^2}y$ = $\dfrac{1}{3}\pi {\left( {\dfrac{{2\sqrt 2 r}}{3}} \right)^2}\dfrac{{4r}}{3} = \dfrac{{32}}{{81}}\pi {r^3}$