Geometrical Figures, Chessboard and Grid Questions with Solutions — Set 2 (Q11-Q20)

This set contains Geometrical Figures, Chessboard and Grid Questions with Solutions — Set 2 (Q11-Q20) covering a mix of question types and difficulty levels — from basic to advanced — exactly as asked in real competitive exams.

Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.

✏️ Attempt each question on your own first — then check the solution below.

11. consider a set of two octagons, what is the maximum number of quadrilaterals that can be drawn from these 16 points ( vertices of two octagons ) as vertices such that at least one point is selected from each octagon.

12. There are two parallel lines. First line has two points A & B while second line has 10 points. How many triangle can be formed taking these 12 points as vertices of triangle?

13. Consider a polygon of ‘n’ sides. If the number of diagonal of polygon is 65 then how many triangles can be drawn from taking vertices of polygon as vertices of triangle?

14. Consider a set of 8 non-overlapping triange in a plane such that no three points in the plane is collinear then
(i). if all the possible triangle are drawn taking vertices of these triangles such that not nore than 1 point is selected from a triangle then find the total number triangle hence drawn?

(ii). Find the total number of new triangles that can be drawn from this system of triangles?

15. Consider a square with 5 points on each side (no point on vertices)
(i). How many straight lines can be drawn from these 20 points such that each line passes through exactly 2 of the given points?
(ii). How many triangle can be drawn from these 20 points as vertices?
(iii). How many quadrilateral can be drawn from these 20 points as vertices?

16. Consider vertices of an octagon & mid point of its sides (total 16 points). How many triangles can be drawn taking these 16 points as vertices of the triangle?

17. If number of triangles that can be drawn from the given set of points as vertices is 56 then find the number of straight lines that can be drawn from these points.

18. Consider a set of ‘m’ parallel lines and another set of ‘n’ parallel lines then what is the number of vertices of parallelogram thus formed by these parallel lines?

19. Consider three set of parallel lines, having a, b, c points respectively & no three apart from the given three are collinear
(i). if p represents number of straight lines that can pass through this system of a + b + c points excluding the three original lines then how many of the following represents correct value of p.
(1) ᵃ⁺ᵇ⁺ᶜC₂ – (ᵃC₂ + ᵇC₂ + ᶜC₂)
(2) ab + bc + ca
(3) ᵃC₁×ᵇC₁ + ᵇC₁×ᶜC1 + ᶜC1×ᵃC₁
(4) 2(ab + bc + ca)
(ii) what is the number of triangle that can be formed from the system of a + b + c points?
(iii) what is the number of quadrilaterals that can be formed from the system of a + b + c points?

20. Consider a set of ‘m’ parallel lines and another set of ‘n’ parallel lines then what is the number of parallelograms thus formed by these parallel lines?

11. consider a set of two octagons, what is the maximum number of quadrilaterals that can be drawn from these 16 points ( vertices of two octagons ) as vertices such that at least one point is selected from each octagon.
Solution:-
⁸C₁×⁸C₃ + ⁸C₂×⁸C₂ + ⁸C₃×⁸C₁
= 8×56 + 28×28 + 5×8
=  1680         Answer

12. There are two parallel lines. First line has two points  A & B while second line has 10 points. How many triangle can be formed taking these 12 points as vertices of triangle?
Solution:-
Required number of triangles
= ¹²C₃ – ¹⁰C₃
= 220 – 120
= 100         Answer

13. Consider a polygon of ‘n’ sides. If the number of diagonal of polygon is 65 then how many triangles can be drawn from taking vertices of polygon as vertices of triangle?
Solution:-
number of diagonal of polygon = 65
∴ ⁿC₂ – n = 65
\(\frac{{n(n – 3)}}{2}\;\)  = 6
n(n-3) = 2×5×13

∴ n = 13

So polygon has 13 vertices
now taking theser vertices as vertices of a traingle, number of triangle = ¹³C₃
= 286         Answer

14. Consider a set of 8 non-overlapping triangle in a plane such that no three points in the plane is collinear then
(i). if all the possible triangle are drawn taking vertices of these triangles such that not more than 1 point is selected from a triangle then find the total number triangle hence drawn?
(ii). Find the total number of new triangles that can be drawn from this system of triangles?
Solution:-
(i).

3 trignle out of 8 triangle can be selected in ⁸C₃ ways
& now one vertices from each of these 3 selected triangle can be selected in ³C₁ ways
Hence required number of triangle
= ⁸C₃׳C₁׳C₁׳C₁          Answer

(ii).

total points in plain = 8×3 = 24
total nubmer of triangle drawn is ²⁴C₃ but out of these 8 are original triangles. Hence number of new triangle is
= ²⁴C₃ – 8     Answer

15. Consider a square with 5 points on each side (no point on vertices)
(i). How many straight lines can be drawn from these 20 points such that each line passes through exactly 2 of the given points?
(ii). How many triangle can be drawn from these 20 points as vertices?
(iii). How many quadrilateral can be drawn from these 20 points as vertices?
Solution:-
(i).
From 4 sides we have to select 2 sides that can be done in ⁴C₂ ways.
Now from each of these selected sides we have to select 1 point out of 5 points and that can be done in ⁵C₁×⁵C₁ ways.
So total number of such straight lines
= ⁴C₂×⁵C₁×⁵C₁          Answer

(ii).
From 20 points without any restriction we can get maximum ²⁰C₃ triangles.
From this we have to reduce the number of triangles that we can not get since 4 sides has 5 collinear points each, which is equal to 4.(⁵C₃)
So total number of required triangle is
= ²⁰C₃ – 4.(⁵C₃)                  Answer

(iii).
From 20 points without any restriction we can get maximum ²⁰C₄ quadrilaterals.
From this we have to reduce the number of quadrilateral that we can not get if three points are collinear & this can happen in two cases:
case(i): when all the four selected points are collinear
⁴C₁×⁵C₄ = 4×5 = 20
case(ii): when three points are collinear
⁴C₁×⁵C₃׳C₁×⁵C₁ = 4×10×3×5 = 600
So total number of such quadrilateral = ²⁰C₄ – 20 – 600
= ²⁰C₄ – 620                  Answer

16. Consider vertices of an octagon & mid point of its sides (total 16 points). How many triangles can be drawn taking these 16 points as vertices of the triangle?
Solution:-
Without any restriction we can get maximum ¹⁶C₃ triangles.
From this we have to reduce the number of triangles that we can not get if three points are collinear this can happen in ⁸C₁׳C₃ = 8 ways.
So total number of such triangles = ¹⁶C₃ – 8               Answer

17. If number of triangles that can be drawn from the given set of points as vertices is 56 then find the number of straight lines that can be drawn from these points.
Solution:-
Here it is given that ⁿC₃ = 56
⟹\(\;\frac{{(n – 2)(n – 1)n}}{6} = 56\)
⟹ n(n-1)(n-2) = 8×7×6
∴ n = 8
Hence number of straight lines = ⁸C₂ = 28     Answer

18. Consider a set of ‘m’ parallel lines and another set of ‘n’ parallel lines then what is the number of vertices of parallelogram thus formed by these parallel lines?
Solution:-
Total number of vertices of parallelogram
= total number of intersection point of m & n parallel lines
= ᵐC₁×ⁿC₁
= mn            Answer

19. Consider three set of parallel lines, having a, b, c points respectively & no three apart from the given three are collinear
(i). if p represents number of straight lines that can pass through this system of a + b + c points excluding the three original lines then how many of the following represents correct value of p.
(1) ᵃ⁺ᵇ⁺ᶜC₂ – (ᵃC₂ + ᵇC₂ + ᶜC₂)
(2) ab + bc + ca
(3) ᵃC₁×ᵇC₁ + ᵇC₁×ᶜC1 + ᶜC1×ᵃC₁
(4) 2(ab + bc + ca)
(ii) what is the number of triangle that can be formed from the system of a + b + c points?
(iii) what is the number of  quadrilaterals that can be formed from the system of a + b + c points?
Solution:-

(i). Without any restriction we can select 2 points from a + b + c points in ᵃ⁺ᵇ⁺ᶜC₂ ways but since points are collinear
∴ we will not get ᵃC₂ + ᵇC₂ + ᶜC₂ these straight lines excluding the original three lines.
Hence required number of straight lines is given by
ᵃ⁺ᵇ⁺ᶜC₂ – (ᵃC₂ + ᵇC₂ + ᶜC₂)         Answer
alternatively if we select 1 point from a points and one point from b points then number of straight lines is ᵃC₁× ᵇC₁ = ab
Similarly from other two pairs we will get ᵇC₁×ᵃC₁ = ca
so total number of straight lines = ab + bc + ca        Answer

(ii) Without any restriction we can select 3 points from a + b + c points in ᵃ⁺ᵇ⁺ᶜC₃ ways but since points are collinear hence we will not get ᵃC₃ + ᵇC₃ + ᶜC₃ these triangles.
Hence required number of triangle is given by
 ᵃ⁺ᵇ⁺ᶜC₃ – (ᵃC₃ + ᵇC₃ + ᶜC₃)         Answer

(iii). In order to from a quadrilateral we need 4 points & no three of them should be collinear so we have two cases:-
case(1):- 2 point from one straight line then 1 from each of the remaining two straight lines. In this case the number of qudrilateral is
ᵃC₂×ᵇC₁×ᶜC₁ + ᵃC₁×ᵇC₂×ᶜC₁ + ᵃC₁×ᵇC₁×ᶜC₂

case(2):- when 2 points from any of the two straight lines
ᵃC₂×ᵇC₂ + ᵇC₂×ᶜC₂ + ᶜC₂×ᵃC₂
Hence total number of such quadrilateral is :
(ᵃC₂×ᵇC₁×ᶜC₁ + ᵃC₁×ᵇC₂×ᶜC₁ + ᵃC₁×ᵇC₁×ᶜC₂) + (ᵃC₂×ᵇC₂ + ᵇC₂×ᶜC₂ + ᶜC₂×ᵃC₂)          Answer

20. Consider a set of ‘m’ parallel lines and another set of ‘n’ parallel lines then what is the number of parallelograms thus formed by these parallel lines?
Solution:-
To get a parallelogram we need to select 2 lines from ‘m’ lines that can be done in ᵐC₂ ways, at the same time 2 lines from ‘n’ lines and that can be done in ⁿC₂ ways.
Hence total number of parallelogram is
=  ᵐC₂×ⁿC₂        Answer

Continue practising with Geometrical Figures, Chessboard and Grid Questions 21 to 30 → Set 3 or revisit the Geometrical Figures, Chessboard and Grid Concept Page to strengthen your formulas and tricks before moving ahead.

Consistent practice is the key to mastering Geometrical Figures, Chessboard and Grid for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and international exams including GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests. Want to understand the concept better? Read about Shortest Path Problem on Wikipedia before attempting the next set.

This page is part of our complete series of Geometrical Figures, Chessboard and Grid questions with solutions for competitive exams — covering every question type from basic to advanced so you can build speed, accuracy and confidence. Practising these questions regularly will also strengthen your core Geometrical Figures, Chessboard and Grid concept before your exam day.