Gyan Room - Geometry - Concepts & Shortcuts



  • Gyan Room - - There's always room to learn more!
    This thread is reserved for sharing concepts, short cuts and good questions from Geometry topic.

    0_1502944041457_MissionIIMpossible.png

    Happy Learning, Stay MBAtious!


  • Faculty and Content Developer at TathaGat


    (a) In a plane if there are n points of which no three are collinear, then

    1. The number of straight lines that can be formed by joining them is nC2.
    2. The number of triangles that can be formed by joining them is nC3.
    3. 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

    1. The number of straight lines that can be formed by joining them is nC2 – mC2 + 1.
    2. The number of triangles that can be formed by joining them is nC3 – mC3.
    3. 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 n-sided polygon which has,

    1. Exactly one side common with that of the polygon are n × (n – 4).
    2. Exactly two sides common with that of the polygon are n.
    3. 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,

    1. The number of rectangles of any size is ∑r³.
    2. The number of squares of any size is ∑r².

    (j) In a rectangle of p x q (p < q),

    1. The number of rectangles of any size is p x q x (p + 1) x (q + 1) / 4.
    2. 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 non-overlapping bounded or unbounded regions, then the maximum number of regions into which the plane can be divided is (3n + 1).


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    0_1507265144037_fd57711b-ed2e-465f-8377-9769baae759a-22279753_863840940456832_8436621857248364612_n.jpg

    Here r = Radius of circle
    a = Side of square
    n = Number of circles


  • Director at ElitesGrid | CAT 2017 - QA 100 Percentile / CAT 2016 - QA : 99.94, LR-DI - 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 = 7

    Example2: 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 =24-23=1
    so area=1/2*1=1/2


  • QA/DILR Mentor | Be Legend


    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 [(p-6)^2]/48 and for odd [(p-3)^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 + even

    because 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) + (101-2z) = 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


  • Converted IIM Indore call | Mentor for Banking/RBI/SSC exams


    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 * (N-2n)/n
    1 face painted 6[N-2n/n]^2
    0 face painted [(N-2n)/n]^3

    A 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 blue

    The cuboid is cut into mnp smaller cubes
    All faces coloured = 8
    two faces coloured = 4{(m-2)+(n-2)+(p-2)}
    1 face coloured = 2{(m-2)(n-2)+(n-2)(p-2)+(p-2)(m-2)}
    No face coloured = (m-2)(n-2) (p-2)


  • Faculty and Content Developer at TathaGat


    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/2

    Example : 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


  • Being MBAtious!


    0_1515040596307_51255019-13df-4d3b-afce-fb00caf58f77-image.png

    Let say radius of small circle (between the three circles) = r
    then 1/r = 1/a + 1/b + 1/c

    So if a = 3, b = 6, c = 2
    then 1/r = 1/3 + 1/2 + 1/6 = 1
    r = 1


  • Being MBAtious!


    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


  • Being MBAtious!


    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 : @Deekonda-Saikrishna)


  • Being MBAtious!


    Number of Triangles in the below figure = a * b * (b - 1) / 2, where a = number of horizontal lines, b = number of non-horizontal lines.

    0_1515062333851_ebcf8fc6-0799-42cd-b57a-eac5bf726909-image.png

    Here a = 4, and b = 4. So Total number of triangles = 4 * 4 * 3/2 = 24

    [credits : Apratim Roy]


  • Being MBAtious!


    Number of Triangles in the figures like the one given below = [n*(n+2)(2n+1)/8], where n is the number of levels.

    0_1515062819738_50fd6137-bdf0-40e3-a429-56d7abda3c67-image.png

    Here, n = 4 so number of triangles = 4 * 6 * 9/8 = [27] = 27 triangles

    0_1515063114160_c3f7e9a3-c8b6-45a2-bd0e-323cc76fae3c-image.png

    Here, n = 3 so number of triangles = 3 * 5 * 7 / 8 = [13.125] = 13 triangles

    [credits : Ratnesh Soni]


  • Being MBAtious!


    Ceva's Theorem

    0_1515063316695_3acb8fa5-2137-40bb-9b46-a1fbe61bdc3b-image.png

    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

    0_1515063516744_7bdcc0ca-ee11-4aa1-800c-964922200ef7-image.png

    x * 3 * 5 = 2 * 7 * 4
    x = 56/15


  • Being MBAtious!


    Mass point Geometry lecture - @gaurav_sharma


 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.