# Counting Figures Model Question Set 1 Practice Questions Answers Test With Solutions & More Shortcuts

#### COUNTING FIGURES PRACTICE TEST [3 - EXERCISES]

Counting Figures Model Question Set 1

Counting Figures Model Question Set 2

Counting Figures Model Question Set 3

**Question : 1**

a) a

b) b

c) c

d) d

Answer »Answer: (c)

We may the label the figure as shown

The Simplest triangles are AGH, GFO, LFO, DJK, EKP, PEL and IMN i.e., 7 in number.

The triangles composed of two components each are GFL, KEL, AMO, NDP, BHN, CMJ, NEJ and HFM i.e., 8 in number.

The triangles composed of three components each are IOE, IFP, BIF and CEI i.e., 4 in number.

The triangles composed of four components each are ANE and DMF i.e., 2 in number.

The triangles composed of five components each are FCK, BGE and ADL i.e., 3 in number.

The triangles composed of six components each are BPF, COE, DHF and AJE i.e., 4 in number.

Thus, there are 7 + 8 + 4 + 2 + 3 + 4 = 28. triangles in the given figure.

**Question : 3**

a) a

b) b

c) c

d) d

Answer »Answer: (a)

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.

Thus, there are 6 + 4 + 3 + 1 = 14 triangles in the given figure.

**Question : 4**

a) a

b) b

c) c

d) d

Answer »Answer: (a)

Smallest Triangles = ΔBOC, ΔCOD, ΔFOG, ΔGOH, ΔHOA, ΔAOB = 6

Triangles formed with two small triangles = ΔAOG, ΔAOC = 2

Largest Triangles = ΔACG, ΔEGC = 2

∴ Total Triangles = 6 + 2 + 2 = 10

**Question : 5**

a) a

b) b

c) c

d) d

Answer »Answer: (b)

The figure may be labelled as shown

**Triangles**

The Simplest triangles are KJN, KJO, CNB, OEF, JIL, JIM, BLA and MFG i.e.,8 in number.

The triangles composed of two components each are CDJ, EDJ, NKO, JLM, JAH and JGH i.e.,6 in number.

The triangles composed of three components each are BKI, FKI, CJA and EJG i.e.,4 in number.

The triangles composed of four components each are CDE and AJG i.e.,2 in number.

The only triangle composed of six components is BKF.

Thus, there are 8 + 6 + 4 + 2 + 1 = 21. triangles in the given figure.

**Parallelograms:**

The Simplest ||gms are NJLB and JOFM i.e.,2 in number.

The ||gms composed of two components each are CDKB, DEFK, BIHA and IFGH i.e.,4 in number.

The ||gms composed of three components each are BKJA, KFGJ, CJIB and JEFI i.e.,4 in number.

There is only one ||gm i.e., BFGA composed of four components.

The ||gms composed of five components each are CDJA, DEGJ, CJHA and JEGH i.e.,4 in number.

The only ||gm composed of six components is CEFB.

The only ||gm composed of ten components is CEGA.

Thus, there are 2 + 4 + 4 + 1+ 4 + 1 + 1 = 17. ||gms in the given figure.

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