Gyan Room  Geometry  Concepts & Shortcuts

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(a) In a plane if there are n points of which no three are collinear, then
 The number of straight lines that can be formed by joining them is nC2.
 The number of triangles that can be formed by joining them is nC3.
 The number of polygons with k sides that can be formed by joining them is nCk.
(b) In a plane if there are n points out of which m points are collinear, then
 The number of straight lines that can be formed by joining them is nC2 – mC2 + 1.
 The number of triangles that can be formed by joining them is nC3 – mC3.
 The number of polygons with k sides that can be formed by joining them is nCk – mCk.
(c) The number of diagonals of a n sided polygon are nC2 – n = n × (n – 3)/2.
(d) The number of triangles that can be formed by joining the vertices of a nsided polygon which has,
 Exactly one side common with that of the polygon are n × (n – 4).
 Exactly two sides common with that of the polygon are n.
 No side common with that of the polygon are n × (n – 4) × (n – 5)/6.
(e) The number of parallelograms formed if ‘x’ lines in a plane are intersected by ‘y’ parallel lines are x × y × (x – 1) × (y – 1)/4.
(f) If there are n lines drawn in a plane such that no two of them are parallel and no three of them are concurrent, then the number of different points at which these lines will intersect each other is nC2 = n × (n – 1)/2.
(g) If there are n straight lines in a plane and no two of them are parallel and no three pass through the same point. Their points of intersection are joined. Then the number of new lines obtained are n × (n – 1) × (n – 2) × (n – 3)/8.
(h) The sides of a triangle a, b and c are integers where a ≤ b ≤ c. If c is given then the number of different triangles is c × (c + 2)/4 or (c + 1)2/4, according to c as even or odd. Also, the number of isosceles triangles is (3c – 2)/2 or (3c – 1)/2, according to c as even or odd.
(i) In a square of n x n,
 The number of rectangles of any size is ∑r³.
 The number of squares of any size is ∑r².
(j) In a rectangle of p x q (p < q),
 The number of rectangles of any size is p x q x (p + 1) x (q + 1) / 4.
 The number of squares of any size is ∑ [(p + 1 – r) x (q + 1 – r)].
(k) If n straight lines are drawn in the plane such that no two lines are parallel and no three lines are concurrent. Then the number of parts into which these lines divide the plane is equal to [n x (n + 1)/2] + 1.
(l) If “n” parallel lines are passing through a circle, dividing the plane into distinct nonoverlapping bounded or unbounded regions, then the maximum number of regions into which the plane can be divided is (3n + 1).

hemant_malhotra last edited by hemant_malhotra
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Here r = Radius of circle
a = Side of square
n = Number of circles

hemant_malhotra last edited by hemant_malhotra
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Shoelace Theorem to find area of polygons with given vertices
Example 1: Find area of triangle (2,4) (3,8) (1,2)
2 .... 4
3 .... 8
1 .... 2
2 .... 4
then 2 * 8 + 3 * 2 + (1 * 4) = 6
and then 4 * 3 + (8) * 1 + (2 * 2) = 8
so take difference of these two
6  8 = 14 take mod 14
so area will be 1/2 * 14 = 7Example2: if we want to find area of (1,1) (2,3) (3,4) (5,6) (1,1)
1……..1
2 …….3
3……..4
5……...6
1…….1
1……...1 (this is repeated from first point)
1 * 3 + 2 * 4 + 3 * 6 5 * 1  1 = 23
now 1 * 2 + 3 * 3 + 4 * 5  6  1 = 24
now take difference =2423=1
so area=1/2*1=1/2

Number of Triangles with Integer sides for a given perimeter.
 If the perimeter p is even then, total triangles is [p^2]/48.
 If the perimeter p is odd then, total triangles is [(p+3)^2]/48
 If it asks for number of scalene triangle with a given perimeter P, then subtract 6 and apply the same formula . For even [(p6)^2]/48 and for odd [(p3)^2]/48.
Where [ x ] represents neatest integer function. For example [6.7] is 7 not 6 because its nearest integer.
Find the number of triangles with exactly one side odd and perimeter = 203
sum is odd so possibilities are
odd + odd + odd
or
odd + even + evenbecause one side HAS TO BE odd, its given in the question, so only these 2 possibilities.
But again question says that EXACTLY one side has to be odd .. so we remove the case when all sides are odd.
So total triangles possible  triangles with odd sides
total triangles = (n+3)^2/48 = 884
triangles with sides odd = (101  2x) + (101 2y) + (1012z) = 203
=> x + y + z = 50
unordered solutions = 234
so 234 triangles have all sides odd..
So total triangles with EXACTLY one side odd = 884  234 = 650

Cube problems  shortcuts
A cube is given with an edge of unit N. It is painted on all faces. It is cut into smaller cubes of edge of unit n. How many edges will have x faces painted?
No of smaller cubes (N/n)^3
No of smaller cubes 5 faces painted 0
4 faces painted 0
3 faces painted 8
2 faces painted 12 * (N2n)/n
1 face painted 6[N2n/n]^2
0 face painted [(N2n)/n]^3A wooden cuboid of dimensions m * n * p unit is painted in a fixed pattern.
(1) Two opposite faces are painted in red.
(2) Two other opposite faces are painted in green
(3) The remaining two faces are painted in blueThe cuboid is cut into mnp smaller cubes
All faces coloured = 8
two faces coloured = 4{(m2)+(n2)+(p2)}
1 face coloured = 2{(m2)(n2)+(n2)(p2)+(p2)(m2)}
No face coloured = (m2)(n2) (p2)

How many regular polygons can be formed by joining the vertices of a N  sided regular polygon ?
Number of distinct polygons possible = (Factors of N  2)
Number of indistinct polygons = Sum of factors of N barring N and N/2Example : There is a 36 sided regular polygon. How many regular polygons can be formed by joining the vertices of the regular polygon ?
N = 36 = 2^2 * 3^2
Number of Factors = 9
Number of distinct polygons possible = 9  2 = 7
Sum of factors = (2^0 + 2^1 + 2^2) (3^0 + 3^1 + 3^2) = 7 * 13 = 91
Number of indistinct polygons = Sum of factors of N barring N and N/2 = 91  (36 + 18) = 37

Let say radius of small circle (between the three circles) = r
then 1/r = 1/a + 1/b + 1/cSo if a = 3, b = 6, c = 2
then 1/r = 1/3 + 1/2 + 1/6 = 1
r = 1

Area of the largest rectangle that can be inscribed in a right triangle = 1/4 * Base * Height
Diameter of the circle inscribed in a right triangle of sides a, b and c (c is hypotenuse) is a + b  c

In a cyclic quadrilateral ABCD let say, AB = a, BC = b, CD = c and DA = d.
Ratio of the length of Diagonals (AC and BD) = (a * d + b * c) /(a * b + d * c)
Example : In a cyclic quadrilateral ABCD, let AB = 2 cm, BC = 3 cm, CD = 4 cm and DA = 5 cm.
Then ratio of the length of diagonals = (2 * 5 + 3 * 4)/(2 * 3 + 5 * 4) = 22/26 = 11/13(credits : @DeekondaSaikrishna)

Number of Triangles in the below figure = a * b * (b  1) / 2, where a = number of horizontal lines, b = number of nonhorizontal lines.
Here a = 4, and b = 4. So Total number of triangles = 4 * 4 * 3/2 = 24
[credits : Apratim Roy]

Number of Triangles in the figures like the one given below = [n*(n+2)(2n+1)/8], where n is the number of levels.
Here, n = 4 so number of triangles = 4 * 6 * 9/8 = [27] = 27 triangles
Here, n = 3 so number of triangles = 3 * 5 * 7 / 8 = [13.125] = 13 triangles
[credits : Ratnesh Soni]

Ceva's Theorem
Given a triangle ABC with a point P inside the triangle, continue lines AP, BP and CP to hit BC, CA and AB at D, E and F respectively. Then AF/FB * BD/DC * CE/EA = 1
Example :
Find x assuming blue lines are concurrent
x * 3 * 5 = 2 * 7 * 4
x = 56/15

Mass point Geometry lecture  @gaurav_sharma