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.

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    Happy Learning, Stay MBAtious!


  • QA/DILR Mentor


    (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 n-sided 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) There are three co-planar parallel lines. If p points are taken on each of the lines, the maximum number of triangles with vertices at these points is p3 + 3p2(p – 1).
    (i) 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.

    (j) 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.

    (k) 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)].

    (l) 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.

    (m) 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).


  • QA/DILR Mentor


    In an isosceles right angled triangle , both the acute angles are equal to 45°
    The hypotenuse = √2 × perpendicular side.
    For a right angled triangle of given perimeter, an isosceles right angled triangle has maximum area.
    A square is broken into two isosceles right angled triangles by its diagonal.
    Ratio of sides : 1:1 :√2.

    Right angled triangle with angles 30°- 60°- 90°
    The side opposite to 30° = 1/2 × hypotenuse
    The side opposite to 60° = √3/2 × hypotenuse
    In an 30°- 60°- 90° triangle , the measure of the sides are in the ratio : 1:√3:2
    One half of an equilateral triangle is a 30°- 60°- 90° triangle.


  • Director at ElitesGrid | 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 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)


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