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Visual Mensuration

Under this section of mensuration, we will be required to visualize some of the unknown dimensions of any structure with the help of the given dimension. Some of the typical examples of Visual Mensuration are given below:
 
Given is a cube ABCDEFGH of side length ‘a’ units. Its top face is ABCD and its bottom face is EFGH. Since the side length of this cube is ‘a’ units, so, AB = BC = CD = AD = AE = EF = FD = FG = GH = EH = BH = CG = ‘a’
 
Description: n-16-1.tif

Minimum length between vertex A and vertex G We can find the minimum distance between these two vertices in the following two ways:
  • Aerial distance Aerial distance can be understood by assuming that there is a fly at vertex A and it has to reach vertex G through the minimum possible distance. This distance will be the diagonal distance between the vertices A and G
     
    Description: 2370.png
  • Physical distance Physical distance can be understood by assuming that there is an ant at vertex A and it has to reach vertex G through the minimum possible distance. Since it cannot fly, so ant will cover this distance along the two faces viz., face ABEH and face EFGH or face ADFE and face CDFG or face ABCD and face CDFG or face ABCD and face BCGH.
Description: n-16-2.tif

The shortest possible distance can be the diagonal only. Let us assume that the ant is going via face ABEH and face EFGH. So, the ant will cover first the diagonal distance between A and P, where P is the mid point of side EH, and then from P to vertex G.
 
Now AP2 = EP2 + GP2 = Description: 2379.png
 
And GP2 = HP2 + GH2 = Description: 2388.png
 
Hence, the minimum possible physical distance Description: 2392.png
 
Let us find out why the distance AE + EG cannot be the shortest?
 
First calculate AE and EG.
AE = a
EG = Description: 2402.png 
So, AE + EG = a + Description: 2411.pnga = a (1+ √2)
And obviously, Description: 2415.pnga(1 + √2)[For a>0]
 
 
We can understand this phenomenon by having a bit of mental mapping. As we have seen earlier that the minimum possible distance can be the diagonal distance only. Now let us cut open the face ABHE by making a cut mark at EH so that the faces ABHE and EFGH are in a plane, lying horizontal on the ground. Now the minimum possible distance between A and G will be the diagonal of the newly formed rectangle AFGB. This diagonal will pass through the mid-point of EH.
 
So, AG2 = AF2 + FG2 = (2a)2 + (a)2 = 5a2
AG = Description: 2424.png
 
Minimum length between vertex A and vertex O, where O is the mid-point of FG
 
Description: n-16-3.tif
 
  • Aerial distance We can see this situation vis-à-vis a cuboid of side lengths ‘a’ units, ‘a’ units and Description: 2428.png units.
     
    Minimum aerial distance between A and O
     
    Description: 2437.png
  • Physical distance Let us assume that the ant is moving through the faces ABHE and EFGH. Ant will first go the point Q, where Q is the mid-point of E and P. And then the ant will cover QO.
Description: n-16-4.tif
AQ2 = AE2 + EQ2 Description: 2446.png
So, AQ = QO = Description: 2455.png
So, the minimum distance between the vertex A and point O = AQ + QO = Description: 2464.png.





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