Maths - Lesson 3
Finding Line Equation from two points
(y - y₁)=m( x - x₁) Ex - We have two points (2,3) and (6,4) then Slope is = (4-3)/(6-2) = 1/4 Putting values in Equation (y - y₁)=m( x - x₁) (y - 3) = 1/4 (x - 2) 4y - 12 = x - 2 4y = x + 10 y = x/4 + 10/4 y = (1/4)x + 5/2 where m = 1/4 and y intercept is 5/2
Circle Arc Rule
1. Perpendicular from the center of a circle to a Chord of the circle bisects the chord . i.e OM⊥AB then , AM = MB 2. Conversely , the line joining the center of circle and the midpoint of a chord is perpendicular to the chord
Rectangles
All Parallelogram Properties , plus 1. All angles are equal = 90° 2. Diagonals are Congruent or Equal
Rhombus
All Parallelogram Properties , plus 1. All sides are equal 2. Diagonals are perpendicular Area of Rhombus = Base * Height
Inscribed Angle in Semicircle with vertices on Diameter is always 90°
As ∠HOJ = 180° the Angle HKJ will always be 90°
Inscribed Angle is half the inscribed Arc - Case II
Case II - Center Outside Inscribed Angle Lets have ∠QPR = x and ∠RPO = y As ∆POQ is Isosceles Triangle ∠OQP =∠OPQ = x +y and ∠QOP = 180 - 2(x +y) In ∆POR ∠OPR = ∠ORP = y (Isosceles Triangle) and ∠POR = 180 - 2y also ∠ROQ = ∠POR - ∠QOP ∠ROQ = 180 -2y - ( 180 -2x -2y) ∠ROQ = 2x Hence ∠ROQ = 2x which is twice as big as ∠RPQ = x
Q. Attached
FAQ: The lines are clearly parallel! Why can't we assume that they are? Would the GRE include such an obvious trap? This is a very tricky convention that takes some time to get used to. In short, we cannot assume that lines are parallel on the GRE unless we are explicitly told they are parallel or there's a mathematical reason that they must be parallel. There's a very, very good chance you will see a question on your GRE that hinges on an assumption like this, so it's important to learn what we can and can't assume to be true about a figure.
Cards
Four French Suites 1. Red Diamonds ♦ - 13 Cards 2. Red Hearts ♥ - - 13 Cards 3. Black Spades ♠ - 13 Cards 4. Black Clubs ♣ - 13 Cards Each Sets has Two through 10 , A King , Queen , Jack and Ace cards .
Inscribed Angles in the same Circle intercept the same Arc or Chord are Equal
Here ∠LNM = ∠LPM
Transversal Lines
Line cutting two parallel lines
Binomial Formula
P = nCr * p∧r *[(1-p)∧n-r] where p = Probability of Success in One Trial n = Number of Trials r = Number of Success Can only Use Binomial in case "Exactly" is mentioned .
Q.When a certain coin is flipped, the probability of heads is 0.5. If the coin is flipped 6 times, what is the probability that there are exactly 3 heads?
Probability = Chances of Event/Total Chances In 6 times you will have exactly 2⁶ Output = 64 Now getting 3 heads = 6C3 = 20 Ways Probability = 20/64 = 5/16 Using Binomial Theorem = 6C3 * (1/2)³ * (1/2)³ = 6*5*4/6 * 1/64 = 20/64 = 5/16
Quadrilaterals
Relations
Polygons
The N sided polygon will have (n-3) Diagonals (n-2) Triangles and (n-2)*180° Total 5 Sides - Pentagon - 540° 6 Sides - Hexagon - 720° 7 Sides - Heptagon/Septagon - 900° 8 Sides - Octagon - 1080°
Counting with Identical terms
The Number of ways in which n things may be arranged among themselves , taking them all at a time , when p of things are alike , q of them are alike and r of them are alike and rest of them are different . n!/(p!q!r!)
Reflection of point over x axis
The reflection point will have Same X and ±Y Ex (5,3) will have (5,-3) as reflection point
Reflection of point over y axis
The reflection point will have same Y and ±X Ex (5,3) will have (-5,3) as reflection point
Slope for Parallel Lines are Equal
True
Right Angle Triangle - Sin , Tan and Cos
Use attached Diagram
Cube
Volume = a³ Surface Area = 6a² Space Diagonal² = a² + a² + a² Space Diagonal = a√3
Two Circles intersecting
When two circles intersect each other , the line joining the 1. Center Bisects the common chord 2. Perpendicular to the common chord .
Congruent Angles
When two lines cross , four angles are formed here ∠1 = ∠4 = ∠5 = ∠8 and ∠2 = ∠3 = ∠6 = ∠7
Slope Line Equation
y = mx +c where m = Slope of line = (y₂ - y₁)/(x₂ -x₁) and c = Intercept of y
Probability Guessing Answers
1. Use your Instincts from Real World. For example you know that Probability of 8 heads coming straight is very less and will not be 0.8 or 80% so choose answers which is 20% or 0.2 or lower numbers. 2. If the Complementary Rule is given in the problem , look for Complementary Answers . Ex - Attached So In the example you can see 0.4 and 0.6 are complementary to each there , so choose these two option in Guessing .
Mutually Exclusive Events or Disjoints
Events A and B are Mutually Exclusive means 1. A Alone Happens 2. B Alone Happens 3. Neither A nor B Happens 4. Both A and B cant Happen together Ex - Consider a six-sided die as before, only in addition to the numbers 1 through 6 on each face, we have the property that the even-numbered faces are colored red, and the odd-numbered faces are colored green. Let event A be rolling a green face, and event B be rolling a 6. Then Probability of Green face(Odd Numbered) = 3/6 = 1/2 Probability of rolling 6 = 1/6 It is obvious that events A and B cannot simultaneously occur, since rolling a 6 means the face is red, and rolling a green face means the number showing is odd. Therefore P(A and B) = 0 and are mutually Exclusive .
Q. Events A and B are independent . If P(A) = 0.6 and P(B) = 0.8 , what does P(A or B) equal
Events A and B are independent and hence chances of event A and B happening together is True , hence these are NOT Mutually Exclusive Events . P(A) = 0.6 P(B) = 0.8 P(A or B) = P(A) + P(B) - P(A and B) P(A and B) ≠ 0 as these are not Mutually Exclusive events . P(A and B) = P(A)*P(B) P(A and B) = 0.6 * 0.8 = 0.48 P(A or B) = 0.6 + 0.8 - 0.48 = 0.92
Q. Six children will give presentations , one after the other in same order . Luke's presentation makes a reference to something mentioned in Ruth's , so Ruth's presentation must come before Luke's ( not necessarily immediately before ). How many orders are possible.
For Every Order in which R Comes before L { C , R , B , D , A , L} there's a corresponding order in which L Comes before R { C , L , B , D , A , R} Therefore in exactly half R comes first and in other half L Comes first . If we divide the total arrangements by 2 ,we can get arrangements where R comes before L . 6!/2 = 720/2 = 360 Another Way - Lets say R is the first one to give presentation then R - 1 Options , L - 5 Options,A - 4 Options, B - 3 Options , C - 2 Options , D - 1 Options i.e total of 5*4! Lets say R is the Second one to give presentation then R - 1 Options , L - 4 Options, A - 4 Options, B - 3 Options , C - 2 Options , D - 1 Options i.e total of 4*4! If we continue we will get 5*4! + 4*4! + 3*4! + 2*4! + 1*4 = 4!(5+4+3+2+1) = 4!*15 = 360
Tangent rule 1
From an External Point P , a secant P -A - B intersecting the circle at A and B and a Tangent PC are Drawn then PA*PB = PC²
Independent Events
If two events A and B are independent then they are NOT mutually Exclusive . i.e chances of two even happening together is Possible . Consider a fair coin and a fair six-sided die. Let event A be obtaining heads, and event B be rolling a 6. Then we can reasonably assume that events A and B are independent, because the outcome of one does not affect the outcome of the other. The probability that both A and B occur is P(A and B) = P(A)P(B) = (1/2)(1/6) = 1/12.
Similar Triangles - Sharing Same Vertex - Ex 1
In the attached image ∆ABD ~ ∆CBA as 1. They share ∠B together 2. ∠BAD and ∠ACB is 90°. Hence ∆CBA/∆ABD = CB/AB = CA/AD = BA/BD = k where k is a scale factor Also ∆ADB ~ ∆CDA as 1. They share ∠D together 2. ∠BAD and ∠ACB is 90°. Hence ∆CDA /∆ADB = CD/AD = CA/AB = DA/DB = k where k is a scale factor if the Sides are in proportion and scaled to k(i.e multiplied by k) then the Area will be multiplied by k². Area of ∆CBA /∆ABD = k² Area of ∆CDA /∆ADB = k²
Equilateral Triangle
Is an extension of Isosceles triangle with all side equal and angles as 60° Area of Equilateral Triangle = (√3/4)Side²
Q. Points J(5,2) and K(-2,-5) are two vertices of an Isosceles triangle . If L is a third Vertex and has a Y co-ordinate of 4 . What is the x co-ordinate of L.
J(5,2) and K(-2,-5) are reflections over line y = -x. And we know that L is on Line Y = -x , hence if y is 4 , X is -4.
Q. Attached
Let's call the third side L. From triangle inequality theorem, we know (3x + 4) - (2x + 1) < L < (3x + 4) + (2x + 1) x + 3 < L < 5x + 5 Clearly (x + 2) is always less than (x + 3), so choices B is wrong. Clearly (5x + 6) is always more than (5x + 5), so choice D is wrong. For the other choices, we have to be careful. Everything here is in variable form, and we are asked to find all POSSIBLE lengths, so if any value of x, any value at all, makes the expression work in the inequality above, then the expression works as a "possible" length. The algebraic expression does not need to work at all values of x; rather, it only needs to work at some or even just one value of x in order for that expression to be a "possible" length. The only restriction we have is Would (4x + 5) be possible? Let's try different values. x = 1 gives us 4 < 9 < 10, which is possible, so A is a possible length. Would (6x + 1) be possible? Let's try different values. x = 1 gives us 4 < 7 < 10, which is possible, so C is a possible length. Would (2x + 17) be possible? Let's try different values. x = 1 gives us 4 < 19 > 10 ---- no good. x = 2 gives us 5 < 21 > 15 ---- no good. x = 3 gives us 6 < 23 > 20 ---- no good. x = 4 gives us 7 < 25 = 25 ---- no good. x = 5 gives us 8 < 27 < 30, which is possible, so E is a possible length. All we needed was one possible value of x that makes the inequality true. Choices A & C & E are correct.
Q. Question is Attached
Lets Say , Science will have : 4 Options Science Lab will have only 1 option as it has to follow Science English Can have 3 Options Maths can have 2 options and History will have 1 option Total Arrangements = 4*1*3*2*1 = 24 Another Way of solving this - E M H |S SL| Lets say Science and SL is in one box and cant be separated . Then number of ways to arrange E M H and that box is 4! ways = 24
In the figure , the line represents one way roads allowing travel only northwards or only Eastwards . Along how many distinct routes can a Car reach point B from A .
Lets an eastward step be denoted by E and Northward step be denoted by N . There are 5 Eastward Steps and 3 Northwards . Therefore we are trying to find number of ways of arranging EEEEENNN. Hence total number of ways to reach B is = 8!/(5!*3!)
Q. If a right triangle has area 28 and hypotenuse 12, what is its perimeter?
Lets say sides are a , b and hypotenuse h Area = 1/2 a*b 28 = 1/2 a*b ab = 56 Also h² = a² + b² 144 = a² + b² (a + b)² = a² + b² + 2ab (a + b)² = 144 + 112 a + b = 16 a + b +c = 16 + 12 = 28
Slope for Perpendiculars are reverse
Line 1 has slope m₁ then Line 2 which is perpendicular to Line 1 will be = - 1/m₁
Isosceles Triangle
Median = Angle Bisector = Perpendicular Bisector are all same . So if a question says that any of these two are equal then we should conclude that the triangle given is an Isosceles Triangle .
Two sides of triangle DEF are equal to 3. Which of the following, taken alone, would be sufficient in finding the area of triangle DEF? Indicate all such statements 1.The ratio of DE to EF = 1 : √2 2.The sum of angles DEF and EFD is 135 degrees 3.The sum of angles DEF and FDE is 90 degrees
Need Clarity With the information provided in [A], there are still two possible scenarios: DE: EF: DF = 1:1:√2 DE: EF: DF = 1:√2:√2 Notice that two of the sides are equal for each of the scenarios, since the question states two sides equal 3. Yet, the two different proportions above result in two different areas. Thus, [A] is not sufficient. With the information provided in [B], there are still two possible scenarios: 45 degrees, 45 degrees, and 90 degrees 67.5 degrees, 67.5 degrees, and 45 degrees [B] is not sufficient. For [C], DEF and FDE must be the two sides that equal Since the remaining angle, EFD, must equal 90. And you can't have two angles in a triangle that both equal 90 degrees. Therefore, given the information in [C], we have a 45:45:90 triangle with sides 3:3:3√2. Since, such a triangle has only one possible area, [C] is sufficient
Q.Events A and B are independent.The probability that events A and B both occur is 0.6 Column A The probability that event A occurs Column B 0.3
P( A and B) = P(A) * P(B) 0.6 = P(A) * P(B) P(B) =0.6/P(A) P(B) = 6/10*P(A) Now P(A) > 1 so that P(B) is less than 1. Hence Column A > Column B
Generalized AND Rule
P(A and B) = P(A) * P(B|A) P(A and B) = P(A|B) * P(B) For Independent Events P(B|A) = P(B) and P(A|B) = P(A) P(A and B) = P(A) * P(B) P(A and B) = P(B) * P(A) For Dependent Events P(A and B) = P(A) * P(B|A) P(A and B) = P(B) * P(A|B)
Without Replacement
Pick Cards and put them aside ( for example , cards dealt to a hand ) Each new choice from a smaller and smaller deck .
Q. A Committee of three will be selected from a group of Eight Employees , Including Alice and Bob . What is the probability that the chosen committee of three includes Alice and not BOB.
Probability = No. of Success Events / Total Number of Events Total Number of Selection = 8C3 = 8!/3!*5! = 8*7*6/3*2 = 56 For no. of Success Events = Alice should be there , so 2 seats left = Probability of Selecting 2 from remaining 6( Leaving BOB ) is = 6C2 = 15 Ways OR For no. of Success Events = Alice should be there , so 2 seats left = 6 * 5 = 30 Ways /2 ( to remove dups) = 15 Probability = 15/56
Q. Ten Dice , each fair with six sides , are rolled simultaneously . What is the probability of getting exactly two fives among them .
Probability of Getting 5 on one dice = 1/6 P = 10C2 * (1/6)² * ( 1- 1/6)¹⁰⁻² = 10C2 * (1/6)² * (5/6)⁸ = 10C2 * (5⁸/6¹⁰)
Pythagorean Triplets
Pythagorean Triplets are number sets which satisfy below conditions a² + b² = c² where c is hypotenuse and a, b are sides Below are few triplets and all multiples should also follow Pythagoras laws. {1,1,√2} where √2 is hypotenuse {1,2,√5} where √5 is hypotenuse {3,4,5} where 5 is hypotenuse {5,12,13} where 13 is hypotenuse {7,24,25} where 25 is hypotenuse {8,15,17} where 17 is hypotenuse {9,40,41} where 41 is hypotenuse
Regular Polygons
Regular Polygons 1. All the sides have same Size 2. All the sides have same Angle Regular Pentagon has 540/5 , 108° each angle Regular Hexagon has 720/6 , 120° each angle Regular Heptagon has 900/7 , 128.57° each angle Regular Octagon has 1080/8 , 135° each angle
Q. Six Children will sit in a row of six chairs, but jackie and Marilyn cannot be seated next to each other . How many arrangements are possible .
Six children can be arranged in 6! Ways Lets Say Jackie and Marlyn is in a Box then A B C D | J M | can be arranged in 5! * 2! Ways = 240 Ways So Jackie and Marlyn not sitting together is (6! - 240) = 720 - 240 = 480 Ways
Slope
Slope of a Line is Rise/Run or Vertical/Horizontal or y/x or (y₂ - y₁)/(x₂ -x₁) Lets think about a slope of -2/3 and lets take a point (5,6) i.e Rise(y) = -2 and Run(x) = 3 or Rise(y) = 2 and run(x) = -3 Next point on the right side : (8,4) , (11,2) , (14,0) Next point on the left side : (2,8) , (-1,10), (-4,12)
Sqaure
Square has the Maximum Area out of all the rectangles when inscribed in Circle . Area of Square = 2* (Radius of Circle)² So if a Circle with radius 10 is given then Maximum Area will be of Sqaure which will be equal to 200 .
Tangent Rule 2
The Angle that a Tangent to a Circle makes with a chord drawn from the point of Contact is Equal to the Angle Subtended by that Chord in the Alternate Segment of the Circle .
Tangents
The Straight Line Drawn at Right Angles to a diameter of a circle from its extremity is a Tangent . Conversely if a Straight LIne is Tangent to a Circle , then the radius drawn to the point of Contact will be perpendicular to the Tangent . Rule 1 - If two tangents are drawn from a Exterior Point a. the length of two tangent Segments are Equal b. The Line joining the exterior point to the center of Circle bisects the angle between the tangents c. Chord joining two points on Circle is Perpendicular.
Triangle Inequality Theorem
The Sum of Two sides must be greater than the third side . In other words Difference of other two side < Length of a Side < Sum of other two side
Q. A librarian has a set of 10 books , including four different books on Abraham Lincoln . The Librarian wants to put the ten books on a shelf with the four Lincoln books next to each other , somewhere on the shelf . How many different arrangement of the 10 books are possible .
Think set of 4 books as one Big Book. Then there are 6 other books and arranging 7 books ( 6 + 1 big book ) can be done in 7! ways . 4 Books can arrange themselves in 4 ways . Hence total number of ways = 7!*4!
Trapezoids
Trapezoids has exactly one pair of Parallel sides and the opposite angles are supplementary ∠K + ∠J = 180° and ∠L + ∠M = 180° For Isosceles Trapezoids or Symmetrical Trapezoids If KL || JM and if KJ = LM then ∠K = ∠L and ∠J = ∠M and the diagonals have equal Lengths .
If the perimeter of two Regular figures ( Figures with all equal sides) then the figure with greater number of sides has larger Area
True
Similar Triangles
Two Triangles are similar if Two angles are equal or Share Two Angles . In that case the sides will be proportional
Q. Three cube-shaped aquariums that are five inches on each side are filled with water to capacity. All of the water from those three aquariums is to be transferred into a larger cube aquarium so that it must be filled to at least 50% of its total capacity without overflowing. Which of the following could be the length, in inches, of a side of the larger aquarium? Indicate all possible values?
Volume of Three Aquarium = 3*125 =375 Now the New Aquarium should be filled at least 50% from this . That means the new Aquarium should have maximum volume of 750 (375*2) and minimum volume of 375 ( to avoid overflowing) Lets say side of new Aquarium is x then 375 < x³ < 750 ~7.3 < x³ < ~9.2 Only option suits this is 8.4
Q. A box contains 4 red chips and 2 blue chips. If two chips are selected at random without replacement, what is the probability that the chips are different colors?
Way 1 - Total Chances = 6C2 Getting different color chips = 4C1*2C1 Probability = 4C1*2C1 / 6C2 = 8/15 Way 1 - Probability = P(1st Red and 2nd Blue) Or P(1st Blue and 2nd Red) = (4/6*2/5) + (2/6*4/5) = 8/15
GRE Assumption
We *Cant* Assume 1. Equal Lengths 2. horizontal or Vertical 3. Parallel or Perpendicular 4. Right Angles
Combination nCr
We are concerned with an End group of r individuals chosen from a pool of n and Order of Selection doesnt Matter . nCr = n!/r! (n-r)! nC1 = n i.e 10C1 = 10 nCr = nC(n-r) i.e 10C1 = 10C9 Use combination only when Order doesn't Matter .
Q. Two sides of triangle are 8 and 13 , whats the possible length of third side .
We know that sum of two sides must be greater than the third side 8+13 > x -> x < 21 also x+8 > 13 -> x> 5 Hence x should be 5<x<21
With Replacement
Whatever Selection is drawn , put it back into the deck and make the next choice from a full and newly shuffled deck . Each and every choice comes from a full and shuffled deck .
Vertical Angles
When two lines intersect , four angles are formed . Each opposite pair are called Vertical Angles and are always Congruent or Equal . The red angles ∠JQM and ∠LQK are equal , as are the blue angles ∠JQL and ∠MQK.
Reflection of point over y = x
Y and X will be switched Ex (5,3) will have (3,5) as reflection point Ex (5,-3) will have (-3,5) as reflection point
Reflection of point over y=-x
Y and X will be switched and sign will be changed Ex (5,3) will have (-3,-5) as reflection point Ex (5,-3) will have (3,-5) as reflection point
Q. Given that ∠PQS = 40, find ∠TPS
∠PQS = ∠PRS = 40 As ∠POR = 180 then ∠PSR = 90 Hence ∠OPS = 180 - 90 - 40 = 50 And ∠TPS = 90 - 50 = 40
Supplementary Angles
∠x + ∠y = 180° called Supplementary Angles
In Counting
1. Always Start with most restrictive Stage 2. If asked to count how many arrangements obey a restriction , it may be easier to count that does not , and Subtract them from the total . 3. If a Question is asking about two terms to be in one order and not another , think about whether the two groups are related by Symmetry .
Q. ABCD is a trapezoid with lengths shown in figure . Find Diagonal AC
1. Draw a perpendicular from B to AD, name the point E , draw another perpendicular from C to AD name it F Now EF = 11 and AE = FD = (21-11)/2 = 5 In ∆BEA -> AE = 5 , AB = 13 hence BE = 12 ( Pythagorean Triplets) In ∆CFD -> DF = 5 , DC = 13 hence CF = 12 ( Pythagorean Triplets). In ∆CFA -> CF = 12 , AF = 16 , CA =? Taking GCF of 12 and 16 is 4 , hence in smaller triangle the sides will be 3 , 4 and hypotenuse will be 5( Pythagorean Triplets). Scaling back CA will be 5*4 = 20 .
Chord Rule
1. Equal Cords are equidistant from Center . Conversely if two chords are equidistant from Center then they are equal . 2. In the attached figure if two chords AB and CD intersect at point P then AP*PB = CP*PD
Parallelogram
1. Opposite sides are parallel . AB is parallel to CD, and AD is parallel to BC. 2. Opposite sides are equal. AB equals CD, and BC equals AD. 3. the opposite angles are equal. 4. the diagonals bisect each other
Q. In a bag are 6 Identical Green marbles , 4 Identical blue Marbles and 2 Identical Red marbles . If 3 marbles are picked from Random from the twelve in the bag , how many distinct sets of 3 can be Selected .
Three of Kind - { G , G , G} , { B , B , B} , { R , R, R} Two of Kind - {2G, B} , {2G, R} , {2R , G} , {2R, B }, {2B, G} , {2B, R} One of kind - {R, G, B } Total - 10
Q.Set A: {1, 3, 4, 6, 9, 12, 15} If three numbers are randomly selected from set A without replacement, what is the probability that the sum of the three numbers is divisible by 3?
Total Chances = 7C3 Set of number divisible by 3 : {3,6,9,12,15} No. Of ways = 5C3 Probability = 5C3/7C3 = 2/7
Arc
Measure of the Central Angle equals the measure of the Arc If ∠AOB = ∠COD then Arc(AEB) = Arc(CFD) Similarly Equal Length chords in the same circle intersect equal length Arcs . If JK = LM then arc (JNK) = arc(LPM)
Q. Question is Attached
Lets Say Dahlia has 1 option George has 2 Options Al - 5 Options B - 4 Options C - 3 Options E - 2 Options F - 1 Options Total Arrangements - 5!*2 = 120*2 = 240
Probability of A or B
P(A or B) = P(A) + P(B) - P( A and B) for mutually Exclusive Events P(A and B) = 0 hence P(A or B) = P(A) + P(B)
Graph for Quadratic Equation
1. The Graph of Quadratic Equation is Parabola . 2. y = ax² + bx + c if a > 0 means the parabola opens upwards if a < 0 means the parabola opens downwards if |a| > 1 the parabola is skinny if |a| < 1 the parabola is wide 3. ax² + bx + c = 0 , Solution of this equation gives the X intercept of parabola . It can be one, two or not defined . 4. All parabola has vertical Symmetry . The line of Symmetry always passes through the vertex which is the lowest or highest point of Parabola .
Reflection of point
1. The Mirror Line is always perpendicular bisector of the segment between the original and its reflected images . 2. Any point in the mirror line is Equidistant from the original and its reflected image . i.e Creating Isosceles Triangle .
Inscribed Angle is half the inscribed Arc - Case I
Angle Created by two points at Circle will always be Twice the Angle created by them at third point on the circumference . Case I - Center Inside Inscribed Angle. Lets have ∠BAO = x and ∠CAO = y as ∆BAO and ∆CAO are Isosceles triangles . ∠BAO = ∠ABO = x and ∠CAO = ∠ACO = y As we know that Exterior Angle is equal to the sum of remote Interior angles . Therefore ∠BOD = 2x ∠COD = 2y Hence ∠BOC = 2(x+y) which is twice as big as ∠BAC = (x +y)
Circle
Area = πr² Circumference = 2πr
Q. An integer is randomly selected from the integers from 200 to 900 inclusive. Column A Probability that the number is either even or prime. Column B 14/13 Which one is greater Column A or B
As probability is Equal to or less than 1 . Column B is greater than column A
Q. ABCDEFGH is a regular Octagon , with two diagonals drawn . Find x .
Each angle in Regular Octagon is 135° Now AE divides Octagon in two equal parts and hence we can say that its dividing ∠A and ∠E in two equal parts . Therefore ∠MAH = 135/2 = 67.5° Now in Isosceles trapezoid ABCH ∠A = ∠B = 135° ∠AHM = ∠BCM = (360 - 135 - 135)/2 = 45° In ∆AMH , ∠MAH = 62.5 , ∠AHM = 45 hence ∠AMH = 180 - 67.5 - 45 = 67.5° ∠AMH = x = 67.5 as these are vertical angles and are equal ,
Q. Attached
Lets Say the Triangle is right triangle the . 11² = 6² + 8² 121 = 35 + 64 121 = 100 Clearly 121 > 100 i.e ∠x > 90°
Dependent Events
P(A|B) Assume that for whatever reason , we know that event B is TRUE , then given this condition , what is the probability that A happens ?What is the Probability of A, given B . Example - A = Randomly Chosen person is Male B = Selection Pool is the US Senate P(A) = 0.5 considering there are 50% Male overall P(A|B) = ? , i.e Whats the Probability of selecting a Male from US Senate or Probability of A given B is ? As we know in US Senate there are 80% of Male . so P(A|B) = 0.8
Q.A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what is the probability that they are matching shoes?
P(Matching Shoes) = P(Selecting 1st Shoe And 2nd Shoe Matches 1) = P(Selecting 1st Shoe) * P( 2nd Shoe matching 1 ) = 20/20 * 1/19 = 1 * 1/19 = 1/19
Q.The probability is 0.6 that an "unfair" coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?
P(tail) = 0.6 P(head) = 1 - 0.6 = 0.4 P(At Least One tail) = 1 - P(All heads) = 1 - [ P(H)₁ * P(H)₂ * P(H)₃ ] = 1 - [ 0.4 * 0.4 * 0.4] = 1 - 0.064 = 0.936
Q. Find the length of RS
To find the length of third side , first get the GCF of the two sides given and then scale down it to a smaller Triangle . Use it to find the third side and then scale it up . GCF of 36 and 72 is 36 Hence create a Triangle with side 1 and 2 and find its Hypotenuse . Scale it up to give the answer of bigger side.
Q. Attached
Total Number of Vertex = 12 Number of Vertex on which diagonal can be drawn = 9 So total you will have 12*9 Diagonals , however the diagonals are counted twice ex from K to A , and A to K ∴ Total Number of Diagonals = 12*9/2 = 54 Another way - Total Number of Diagonal for any Polygon is n - 3 . So this figure will have 12 - 3 = 9 Diagonals for a vertex. Total number of Diagonals = 9*6 = 54 as
Two Circles touching Internally
Two Circles touch internally at P , The common chord AD of the larger circle intersects the smaller circle in B and C , as shown in the figure . Shows that APB = CPD
Q.When a coin is flipped, the probability of getting heads is 0.5, and the probability of getting tails is 0.5. A coin is flipped 5 times Column A Probability of getting exactly 2 heads Column B Probability of getting exactly 3 heads
Using Binomial Theorem - Probability of one event P(H) =1/2 Total Number of flips = 5 So for Exactly 2 heads Probability = 5C2 * (1/2)² * (1/2)³ And for Exactly 3 heads Probability = 5C3 * (1/2)³ * (1/2)² both are same , hence Column A = Column B
Rectangular Solid
Volume = Height * width * depth Surface Area = 2hw + 2 wd + 2dh Space Diagonal² = height² + Width² + Depth²
Cylinder
Volume = πr² * h Surface Area = 2πr*h + 2πr²
Q. An Amusement Park has 12 different major rides . A Coupon gives its holder access to any 3 of these rides for free . How many set of these rides are possible.
first place will have 12 Options Second Place will have 11 Options Third Place will have 10 Options So total count = 12*11*10 but it will count different arrangements of the three item which should not matter at all . {R₁ , R₂ , R₃ } ,{R₁ , R₃ , R₂} ,{R₂,R₃ ,R₁} ,{R₂ ,R₁ ,R₃}, {R₃ ,R₁ ,R₂} ,{R₃ ,R₂ , R₁ } So all 6 above are not required .(3!) Hence total count = 12*11*10/6 = 220 Or 12C3 = 12!/(3!*7!) = 12*11*10*9*8/3*2*1 = 12*11*10/6 = 220
Q. The diagram is not drawn to scale . Circle O has radius of 8 and sector JOK has an area of 16π . Find the Length of arc JLK .
r = 8 so Area of Circle = 64π Area of Sector = 16π Hence Area of Sector / Area of Circle = 16π/64π = 1/4 Length of JK = 1/4 of Length of Circle = (1/4) * (2*8*π) = 4π Hence Length of JLK = 16π - 4π = 12π
