What is the minimum number of colour pencils required to fill the spaces in the below figure with no two adjacent spaces have the same colour?
The given figure can be labelled as shown :
The spaces P, Q and R have to be shaded by three different colours definitely (since each of these three spaces lies adjacent to the other two).
Now, in order that no two adjacent spaces be shaded by the same colour, the spaces T, U and S must be shaded with the colours of the spaces P, Q and R respectively.
Also the spaces X, V and W must be shaded with the colours of the spaces S, T and U respectively i.e. with the colours of the spaces R, P and Q respectively. Thus, minimum three colour pencils are required.
Figures 2 and 4 are identical.
NA
The simplest triangles are AKI, AIL, EKD, LFB, DJC, BJC, DHC and BCG i.e. 8 in number.
The triangles composed of two components each are AKL, ADJ, AJB and DBC i.e. 4 in number.
The triangles composed of the three components each are ADC and ABC i.e. 2 in number.
There is only one triangle i.e. ADB composed of four components.
Thus, there are 8+ 4 + 2 + 1= 15 triangles in the figure.
As per the given figure in above question, we can say that
From first figure to second figure the circle and the rectangle interchange positions and the upper shaded square moves to the lower side.
Thus , figure ( 3 ) will come on the place of ? in question figure .As shown in answer figures .
NA
NA
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.