Geometrical Figures, Chessboard and Grid Questions with Solutions — Set 4 (Q31-Q40)

This set contains Geometrical Figures, Chessboard and Grid Questions with Solutions — Set 4 (Q31-Q40) 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.

31. Find the number of squares in a chess board of 10×12 Grid instead of 8×8 Grid.

32. In how many ways 2 rooks can be placed on a chessboard such that they are not in attacking position?

33. Consider 37 points in a plane such that no three of them are collinear. A student is told to 1st draw all the possible polyg0ns of same number of sides & then draw all the possible diagonals. What is the maximum number of such diagonals possible? ( if a particular line is diagonal of two polygon then count both of them)

34. Consider a polygon of k sides with n points on each side (no point on the vertices)
(i) How many straight lines can be drawn from these ‘kn’ points such that each line passes through exactly 2 of the given points.
(ii) if ‘T’ is the maximum number of triangles that can be drawn from these kn points as vertices then find the value of ‘T’.
(iii) if ‘Q’ is the maximum number of quadrilaterals that can be drawn from these ‘kn’ points as vertices then find the value of Q.

35. Consider a polygon of n sides. what is the number of triangles that can be drawn taking vertices of these polygons as vertices of triangles and no side of triangles is common with any side of the polygon?

36. Consider ‘n’ straight lines in a plane such that no two of which are parallel and no three of which pass through the same point. How many new straight lines can be drawn from the point of intersection of these straight lines?

37. 150 circles are drawn on a plane. What is the maximum number of regions made by this system of 150 circles

38. The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior points respectively on them.The number of triangles that can be constructed using these interior points as vertices will be?

39. A rectangle with sides 2n-1 and 2m-1 is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side length is?

40. Let Tₙ be the number of all possible triangles formed by joining vertices of an n – sided regular polygon. If Tₙ₊₁ – Tₙ = 10, then what will be the value of n.

31. Find the number of squares in a chess board of 10×12 Grid instead of 8×8 Grid.
Solution:-
total number of squares that can be found within an m×n chessboard (m>n) is given by this formula:

  \(\frac{{2{n^3} + 3{n^2} + n}}{6} + \frac{{(m – n)(n)(n + 1)}}{2}\)
Here m = 12 & n = 10
Hence required number of square is:
\(\frac{{2 \times 1000 + 3 \times 100 + 10}}{6} + \frac{{(12 – 10)(10)(10 + 1)}}{2}\)
= 495          Answer

32. In how many ways 2 rooks can be placed on a chessboard such that they are not in attacking position?
Solution:-
We know that rooks can attack only in same row or in same column.
The first rook has 64 options to be placed & since the second rook can not be placed in the same row or the same column with first rook.
So it has 7 rows & 7 columns left for it & it can be placed in 7×7 = 49 ways.
But the order in which the rooks are placed is not important. So, it will be divided by 2!
∴ total number of ways
=  \(\frac{{64 \times 49}}{2}\)
= 1568           Answer

33. Consider 37 points in a plane such that no three of them are collinear. A student is told to 1st draw all the possible polyg0ns of same number of sides & then draw all the possible diagonals. What is the maximum number of such diagonals possible? ( if a particular line is diagonal of two polygon then count both of them)
Solution:-
we know that ³⁷Cᵣ will have maximum value when
r =  \(\frac{{37 – 1}}{2}\) = 18
or
r =  \(\frac{{37 + 1}}{2}\) = 19
and from one polygon we will get  \(\frac{{18.(18 – 3)}}{2}\)  = 135 diagonals.
Hence maximum number of diagonals is = 135.(³⁷C₁₈)        Answer

34. Consider a polygon of k sides with n points on each side (no point on the vertices)
(i) How many straight lines can be drawn from these ‘kn’ points such that each line passes through exactly 2 of the given points.
(ii) if ‘T’ is the maximum number of triangles that can be drawn from these kn points as vertices then find the value of ‘T’.
(iii) if ‘Q’ is the maximum number of quadrilaterals that can be drawn from these ‘kn’ points as vertices then find the value of Q.
Solution:-
(i). From k sides we have to select 2 sides & this can be done in ᵏC₂ ways.
Now from each selected side we have to select 1 point & that can be done in ⁿC₁×ⁿC₁ ways.
So total number of such straight lines

(ii) From k.n points without any restriction we can get maximum ᵏⁿC₃ triangles.
From this we have to reduce the number of triangles that we can not get due to collinear points which is equal to k.ⁿC₃
So total number of such triangle is 
= ᵏⁿC₃ – k.ⁿC₃               Answer

Alternately: In order to draw a triangle we need 3 non-collinear points. So here we have two cases:-
case(i):  If three selected points from three different sides:

case(ii):  if two points are selected from one side while one point is selected from different sides:-

= 2.ᵏC₂(ⁿC₂×ⁿC₁)
So total number of such triangle is 
= ᵏC₃×ⁿC₁×ⁿC₁×ⁿC₁ + 2.ᵏC₂.(ⁿC₂×ⁿC₁)                Answer

(iii). From kn points without any restriction we can get maximum ᵏⁿC₄ quadrilaterals.
From this we have to reduce the number of quadrilaterals that we can not get if three points are collinear & this can happen in two cases:-
case(i):  when all 4 selected points are collinear
 = k × ⁿC₄
case(ii): when three points are collinear
= k × ⁿC₃ ×(k-1) × n
So total number of such quadrilaterals is :

= ᵏⁿC₄ – k.ⁿC₄ – k. ⁿC₃.(k-1).n           Answer

Alternately: In order to draw a quadrilateral we need 4 points such that no three of them are collinear. Here we have three cases:
case(i): if 4 points selected from 4 different sides: 

case(ii): if two points are selected from one side while 1 point is selected from two different sides:

case(iii): if two points are selected from one side while two points are selected from different side:

ᵏC₂×ⁿC₂×ⁿC₂

So total number of such quadrilateral is 
= ᵏC₄×(ⁿC₁×ⁿC₁×ⁿC₁×ⁿC₁) + ᵏC₃×(ⁿC₂×ⁿC₁×ⁿC₁×3) + ᵏC₂×ⁿC₂×ⁿC₂              Answer

35. Consider a polygon of n sides. what is the number of triangles that can be drawn taking vertices of these polygons as vertices of triangles and no side of triangles is common with any side of the polygon?
Solution:-
Without any restriction number of tiangle formed is ⁿC₃
Now from ths value reduce the number of triangles which are being counted in this value but actually they are not desired, now consider the following cases:
case(i): Number of triangles having three sides common with the sides of polygon is 0.
case(ii):  Number of triangles having two sides common with the sides of polygon is n.
case(iii): Number of triangles having one side common with the sides of polygon

36. Consider ‘n’ straight lines in a plane such that no two of which are parallel and no three of which pass through the same point. How many new straight lines can be drawn from the point of intersection of these straight lines?
Solution:-
Since each line is intersected by every other line at a point.
So number of intersection point = number of ways to select 2 straight lines out of ‘n’ lines
= ⁿC₂

∴ Number of straight line from x points is 
= ˣC₂

Since in the question we are asked only th new lines & we can not here one thing that on particular line is interesected by remaining (n-1) straight lines a (n-1) points & from these points we will not get a new line. Hence from one line we will not get ⁿ⁻¹C₂ lines. 

37. 150 circles are drawn on a plane. What is the maximum number of regions made by this system of 150 circles?
Solution:-
we know the formula that if number of circle is n then number of regions is n(n-1) + 2
Hence when number of circles is 150 thn maximum regions is
= 150.(150 – 1) + 2
= 22352 Answer

38. The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior points respectively on them.The number of triangles that can be constructed using these interior points as vertices will be?
Solution:-
we have following cases in this question:-
case (i): if one point is selected from each side then number of triangle is = 3×4×5 = 60
case (ii):  if one point is selected from one side and 2 points from another side then number of ways is
= ³C₂×(4 + 5) + ⁴C₂×(5+3) + ⁵C₂×(3+4)
= 3×9 + 6×8 + 10×7
= 27 + 48 + 70
= 145
So total number of triangle is
= 60 + 145
= 205 Answer

39. A rectangle with sides 2n-1 and 2m-1 is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side length is?
Solution:-

total length of horizontal side = 2m – 1 → odd

total length of vertical side = 2n – 1 → odd

A rectangle is formed by two horizontal & two vertical lines.
& total number of hrizontal lines = 2m → Even
& total number of vertical lines = (2n – 1) + 1 = 2n → Even
Since total number of lines are even
Hence half lines will be at odd places & other half will be at even places
Horizontal ⇒ odd lines → m          even lines → m
Vertical ⇒ odd lines → n                even lines → n
 & we know that
                                 even – even = even
                                 odd – odd = even
                                 even – odd = odd
                                 odd – even = odd
Since 

& as per question side length of the selected rectangle should be odd.
So as per above formula one side should be odd & the other one should be even to get the length of sides of desired rectangle as odd.

⟹ A vertical line of odd place of the rectangle can be selected in ᵐC₁ = m ways
& other vertical line of odd place can be selected in ᵐC₁ = m ways.
By combining above four we will get a rectangle whose side length is odd
∴ total number of required rectangle is
= n.n.m.m
= n².m²        Answer

40. Let Tₙ be the number of all possible triangles formed by joining vertices of an n – sided regular polygon. if Tₙ₊₁ – Tₙ = 10, then what will be the value of n?
Solution:- 

As per the given condition
ⁿ⁺¹C₃ – ⁿC₃ = 10
we know that  ⁿCᵣ + ⁿCᵣ ₋ ₁ = ⁿ ⁺ ¹Cᵣ
i.e. ⁿ ⁺ ¹Cᵣ – ⁿCᵣ = ⁿCᵣ ₋ ₁
Hence ⁿC₃ ₋ ₁ = 10
ⁿC₂ = 10
n(n-1) = 20
∴ n = 5    Answer

Continue practising with Geometrical Figures, Chessboard and Grid Questions 41 to 50 → Set 5 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.