A regular Pentagon have 5 sides and 5 lines of symmetry.
Find the number of triangles in the given figure?
The simplest triangles are AKI, AIL, EKD, LFB, DJC, DKJ, KIJ, ILJ, JLB, BJC, DHC and BCG i.e. 12 in number.
The triangles composed of two components each are AKJ, ALJ, AKL, ADJ, AJB and DBC i.e. 6 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 12 + 6 + 2 + 1 = 21 triangles in the figure.
8(6+5) - 10 = ?
? = 8(11) - 10
? = 88 - 10
? = 78.
'&' is a Logical Symbol and is called as Ampersand.
^ is called Caret
- is called Bar
v is called Reversed Caret.
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.
1. Contains an infinite number of points
2. can be used to create other geometric shapes
3. is a term that does not have a formal definition
In Geometry, unless it's stated, a line will extend in one dimension and goes on forever in both ways.
This would not mean that K and L will always be together. It just implies that, if K is there, then L will also be there.
At the same time, it can happen that L is there but K isn't.
Remember, the condition is on K, not on L.
Find the minimum number of straight lines in the below figure?
The given figure can be labelled as :
Straight lines :
The number of straight lines are 19
i.e. BC, CD, BD, AF, FE, AE, AB, GH, IJ, KL, DE, AG, BH, HI, GJ, IL, JK, KE and DL.
This question concerns a committee's decision about which five of eight areas of expenditure to reduce. The question requires you to suppose that K and N are among the areas that are to be reduced, and then to determine which pair of areas could not also be among the five areas that are reduced.
The fourth condition given in the passage on which this question is based requires that exactly two of K, N, and J are reduced. Since the question asks us to suppose that both K and N are reduced, we know that J must not be reduced:
Reduced :: K, N
Not reduced :: J
The second condition requires that if L is reduced, neither N nor O is reduced. So L and N cannot both be reduced. Here, since N is reduced, we know that L cannot be. Thus, adding this to what we've determined so far, we know that J and L are a pair of areas that cannot both be reduced if both K and N are reduced:
Reduced :: K, N
Not reduced :: J, L
Answer choice (B) is therefore the correct answer.
We may label the figure as shown.
The Simplest triangles are AFB, FEB, EBC, DEC, DFB and AFD i.e 6 in number.
The triangles composed of two components each are AEB, FBC, DFC, ADE, DBE and ABD i.e 6 in number.
The triangles composed of three components each are ADC and ABC i.e 2 in number.
There is only one triangle i.e DBC which is composed of four components.
Thus, there are 6 + 6 + 2 + 1 = 15 triangles in the figure
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.