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# 7/28/2016 MensurationII(Importantresults)

MensurationII(Importantresults)
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Thefollowingresultsareveryimportanttosolvevariousmensurationproblems.

1.Thelargestpossiblespherethatcanbechiseledoutfromacubeofside"a"cm.

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3
a
Remainingemptyspaceinthecube=a 3

6
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Letthesideofthesquare=x ,thenthediagonalofthecube=3 x
2a
3 x = 2a x =
3

2a
Thereforesideofthesquare=
3

Sol:

Fromtheabovediagram,BE 2
+ ED
2
= BD
2

x
2 2
+ x = BD
2

BD = 2 x
2 x x
BC = =
2 2

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7/28/2016 MensurationII(Importantresults)

From ABC,AC 2
= AB
2
+ BC
2

2
x
2 2
a = x + ( )
2
2
3x
a
2
=
2

2
x = a
3

2
Theedgeofthecube=a
3

4.ThelargestsquarethatcanbeinscribedinarightangledtriangleABCwhenoneofitsverticescoincidewiththe
vertexofrightofthetriangle.

Solution:

letthesideofthesquare=x
=
AB BC
b x x
=
b a
x x
1 =
b a
x x
1 = +
b a
a + b
1 = x( )
ab

ab
x = ( )
a + b
2
ab ab
Sideofthesquare= andareaofthesquare=( )
a + b a + b

5.ThelargestsquarethatcanbeinscribedinarightangledtriangleABCwhenoneofitsverticesliesonthe
hypotenuseofthetriangle

Solution1:

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7/28/2016 MensurationII(Importantresults)

T an = =
FD BC
=
x b
xa
b

## Also, ABCand EGCaresimilar.

GE a
T an = =
EC b
x a
=
EC b
xb
EC = (2)
a

xa xb
c = + x +
b a
a b
c = x( + 1 + )
b a
2 2
a + ab + b
c = x( )
ab

abc
x = ( )
2 2
a + ab + b

abc
Sideofthesquare=
2 2
a + b + ab

Solution2:

Fromtheabovediagram,dropaperpendiculartoACfromvertexB.
1 1
Areaof ABC= a b = BF c
2 2
ab
BF = (1)
c

## Now BDEand ABCaresimilar.

BG DE
=
BF AC
ab
x
c x
=
ab c

c
xc x
1 =
ab c
xc x
1 = +
ab c
c 1
1 = x( + )
ab c

2
+
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( ) 3/8
7/28/2016 MensurationII(Importantresults)
2
c + ab
1 = x( )
abc

abc abc
x = =
2 2 2
c + ab a + b + ab

3
Areaofthesquare= r
2

5

2
r r
Sideofthesquare= ,andareaofthesquare=
2 2

2
r
Inscribedcirclearea=
4

(Rememeber:Inscribedcircleareaishalfofthesemicirclearea)

2
r
Areaofthecircle=
3 + 22

Level2:

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7/28/2016 MensurationII(Importantresults)

Solution:

Letthesideofthesquare=x
DHEisarightangletriangle.Therefore,DE 2
= DH
2
+ HE
2

2 2
DE = x + x = 2 x
AF DE
=
AG BC
h x 2 x
=
h 2r
x x
1 =
h 2 r
x x
+ = 1
2 r h

1 1
x( + ) = 1
2 r h

h + 2 r
x( ) = 1
2 rh

2 rh
x =
h + 2 r

2 rh
Squareside=
h + 2 r

Sol:
y .To
maximizethevolumewehavetochangeoneofthegivenvariableintheequationinto"r"and"h"anddifferentiate
theequationwithrespecttotheothervariable.

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7/28/2016 MensurationII(Importantresults)

h y 2x
So, = (1)
h 2r
xh
h y =
r
xh
h = y
r

Substitutingtheaboveresultinthevolumeweget,
hx
2
V = x (h )
r
3
hx
2
V = hx
r

Differentiationtheaboveequationw.r.tx,
2
3hx
1
V (x) = 2hx
r

Byequatingtheaboveequationtozero,wecanfindthevalueofxwheretheaboveequationbecomesmaximum.
2
3hx
2hx =0
r
3x
2hx (2 ) = 0
r

3x
h,xcannotbezero.So2 =0
r
2r
x = (2)
3

Substitutingtheaboveresultinequation(1),
2r
2( )
h y 3
=
h 2r
h y 2
=
h 3
2
h y = h
3
h
y = (3)
3
2
2r h
Therefore,maximumvolumeofthecylinder=( ) ( )
3 3

4
2
r h
27

12.Findthemaximumvolumeofspherethatcanbeinscribedinacone.

Sol:

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7/28/2016 MensurationII(Importantresults)

4
Wehavetomaximizethevolumeofthesphere=V = x
3

## Fromtheabovediagram,AF C andADE aresimilar.( AAArule)

CF AC
Therefore, =

FromPythagorasrule,AC = AF
2
+ FC
2 2
= h + r
2

r r 2 + h2
Now, =
xh x

2 2
rh rx = x r + h

2 2
rh = rx + x r + h

2 2
rh = x (r + r + h )

rh
x =

2 2
r + r + h

13.Findthemaximumvolumeofconethatcanbeinscribedinasphere.

Sol:

1
Wehavetomaximizethevolumeofthecone=V= x y
2
(1)
3

From BCD,BD 2
= BC
2
+ CD
2

2 2 2
r = x + (y r)

2 2 2 2
r = x + y + r 2yr

x
2
= 2yr y
2
(2)
Substitutingtheaboveresultinequation(1),anddifferentiatingw.r.t.y,andequatingtozero,
1
V= (2yr y ) y
2

1
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7/28/2016 MensurationII(Importantresults)
1
V
1
(y) = (4yr 3y )
2
=0
3
4r
y =
3
4r r
NowCD= r =
3 3

From BCD,
r 2
2 2
r = x ) + (
3

2
8r 22 r
x = = (3)
9 3
2
1 1 22 r 4r 32
Volumeofthecone= x y
2
= ( ) = r
3

3 3 3 3 81

Aptitude
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