# Non Verbal Reasoning - Analytical Reasoning - Discussion

Discussion Forum : Analytical Reasoning - Section 1 (Q.No. 1)

1.

Find the number of triangles in the given figure.

Answer: Option

Explanation:

The figure may be labelled as shown.

The simplest triangles are AHG, AIG, AIB, JFE, CJE and CED i.e. 6 in number.

The triangles composed of two components each are ABG, CFE, ACJ and EGI i.e. 4 in number.

The triangles composed of three components each are ACE, AGE and CFD i.e. 3 in number.

There is only one triangle i.e. AHE composed of four components.

Therefore, There are 6 + 4 + 3 + 1 = 14 triangles in the given figure.

Discussion:

52 comments Page 1 of 6.
Shalini said:
10 years ago

There is another method to solve this problem:

1. First check the possible Triangle by Assume each node or point as (A,B,C,D,E,F,G,I,J) .

2. Second check the possibility of Joining each node i.e (AHG,ABI,AGI,AGB,AEG).

3. Then check for B and see what possible joining are available i.e(BIA,BGA).

4. Again check for C (CED,CFE,CIE,CFD) D (DEC,DFC) E (ECJ,ECF) F(FJE,FCE,FCD) G (GAE,GAI,GAB,GAH) H(HAG,HAE).

5. NOW Combine all Elements.

AHG,ABI,AGI,AGB,AEG,BIA,BGA,CED,CFE,CIE,CFD,DEC,DFC,ECJ,ECF,FJE,FCE,FCD,GAE,GAI,GAB,GAH,HAG,HAE

6. Now Eliminate same letter (AHG -> HAG)i.e remove HAG.

7. If you eliminate you will get,(AHG,ABI,AGI,AGB,AEG,BGA,CED,CFE,CIE,CFD,DFC,ECJ,FJE,HAS).

TOTALLY 14 letter are remaining..

Hope you all Understand :).

1. First check the possible Triangle by Assume each node or point as (A,B,C,D,E,F,G,I,J) .

2. Second check the possibility of Joining each node i.e (AHG,ABI,AGI,AGB,AEG).

3. Then check for B and see what possible joining are available i.e(BIA,BGA).

4. Again check for C (CED,CFE,CIE,CFD) D (DEC,DFC) E (ECJ,ECF) F(FJE,FCE,FCD) G (GAE,GAI,GAB,GAH) H(HAG,HAE).

5. NOW Combine all Elements.

AHG,ABI,AGI,AGB,AEG,BIA,BGA,CED,CFE,CIE,CFD,DEC,DFC,ECJ,ECF,FJE,FCE,FCD,GAE,GAI,GAB,GAH,HAG,HAE

6. Now Eliminate same letter (AHG -> HAG)i.e remove HAG.

7. If you eliminate you will get,(AHG,ABI,AGI,AGB,AEG,BGA,CED,CFE,CIE,CFD,DFC,ECJ,FJE,HAS).

TOTALLY 14 letter are remaining..

Hope you all Understand :).

Madhuri said:
5 years ago

The easiest way is to pick an edge at a time from the boundary like AH and count the no. Of triangles having it as one of its sides. These are AHG and AHE.

Then take up the next edge, say AC and count the no of triangles formed with it. These are ACE and ACJ.

Similarly, for GE the triangles are GEA and GEI.

For AG, we have AGB, AGI

For CE, we have, CEJ, CEF

For CD, it is CDE and CDF

Besides these, there are ABI, EFJ.

No more triangles are there. And the counting has reached 14.

The way to confirm that no triangles are left uncounted is to pick the edges one by one in a cyclic manner that is to start from a vertex and keep on moving from it in a certain direction along the boundary of the given figure.

Then take up the next edge, say AC and count the no of triangles formed with it. These are ACE and ACJ.

Similarly, for GE the triangles are GEA and GEI.

For AG, we have AGB, AGI

For CE, we have, CEJ, CEF

For CD, it is CDE and CDF

Besides these, there are ABI, EFJ.

No more triangles are there. And the counting has reached 14.

The way to confirm that no triangles are left uncounted is to pick the edges one by one in a cyclic manner that is to start from a vertex and keep on moving from it in a certain direction along the boundary of the given figure.

(1)

Saiteja said:
7 months ago

@All.

Count the triangles by considering parts separately.

For the above figure consider the middle rectangle separately and count. Then left and right triangle along with the triangle included in the rectangle, total =14.

Count the triangles by considering parts separately.

For the above figure consider the middle rectangle separately and count. Then left and right triangle along with the triangle included in the rectangle, total =14.

(2)

Jay varshney said:
1 decade ago

The meaning of the component here is a triangle has made with how much points(a,b,c,d,e... j) in a straight line i.e AHE has made with 4 points in a straight line!

Rikesh karki said:
1 decade ago

Is there any shortcut or trick to solve above question because we have to solve these question in 1 minutes in our entrance exam and its seems very difficult?

(1)

Gajanan Konge said:
1 decade ago

@ Priyanka.

This formula doesn't work as, there is one more triangle present in the figure and it is AHE (That comprises of the four components).

This formula doesn't work as, there is one more triangle present in the figure and it is AHE (That comprises of the four components).

Chand said:
1 decade ago

@Priyanka.

Is there any standard such formula. I tried this formula for a structure a star with in a circle. It didn't work. Please tell me?

Is there any standard such formula. I tried this formula for a structure a star with in a circle. It didn't work. Please tell me?

B. Surya kumar peetha said:
6 years ago

CFD is covered by 4 triangles. This changes the calculation by triangles covered by three components are 2 and four triangle components 2.

(1)

Priyanka said:
1 decade ago

No. of total nodes + 2(no. of inner nodes)

Here,

Total no. of nodes=10

No. of inner nodes(i and j)=2

10+2(2)=14

Here,

Total no. of nodes=10

No. of inner nodes(i and j)=2

10+2(2)=14

(3)

Mamun Hasan said:
8 years ago

Nodes are vertices which are labeled as (A, B, . , I, J).

Such as: A is node similarly I is an internal node.

Such as: A is node similarly I is an internal node.

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