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 AIR LINES EXAM PREPARATION MATERIALS


Mathematics

Choose the best alternative among the given choices for the following 25 questions.

1.    Which number(s) is(are) equal to the quarter of its (their) square?

A.   0 and 1     B. 0 and 2      C. 0 and 4     D. 1 and 2

2.    Tom travels 60 miles per hour going to a neighboring city and 50 miles per hour coming back using the same road. He drove a total of 5 hours away and back. What is the distance from Tom's house to the city he visited?(round your answer to the nearest mile).

A.   136miles     B. 163miles     C. 123miles     D. 130miles

3.    Find an equation of the line containing (- 4,5) and perpendicular to the line 5x - 3y = 4.

A.      B.    C. 5y + 3x = 13    D. 5y + 5x = 13

4.    A rectangle field has an area of 300 square meters and a perimeter of 80 meters. What are the length and width of the field respectively?

A.   30 and 10   B. 20and 15    C. 50m and 6m    d. 25 and 12

5.    A rectangular garden in Mrs Dorothy's house has a length of 100 meters and a width of 50 meters. A square swimming pool is to be constructed inside the garden. Find the length of one side of the swimming pool if the remaining area (not occupied by the pool) is equal to one half the area of the rectangular garden.

A.   30meters    B. 25meters    C. 20meters    D. 50meters

6.    The numbers 2 , 3 , 5 and x have an average equal to 4. What is x?

A.   2             B. 6               C. 4              D. 5

7.    The numbers x , y , z and w have an average equal to 25. The average of x , y and z is equal to 27. Find w.

A.   23           B. 19             C. 25           D. 29

8.    A is a constant. Find A such that the equation 2x + 1 = 2A + 3(x + A) has a solution at x = 2.

A.   1/5           B. -1/5            C. 5           D. -5

9.    Find the length and width of a rectangle that has a length 3 meters more than its width and a perimeter equal in value to its area ?

A.   L = 2, W= 3   B. L =6, W= 3    C. L = 3, W = 2  D. L = 5, W = 3

10.  Given the symmetric shape below with a known perimeter of 77, find the area of the shape.

A.   289.84  B. 238 .45     C. 345.5    D. 200

11. A real estate agent received a 6% commission on the selling price of a house. If his commission was $8,880, what was the selling price of the house?

A.   $53280    B. $1480     C. $148,000    D. $8880

12. An electric motor makes 3,000 revolutions per minutes. How many degrees does it rotate in one second?

A.   180,000 degrees / second          B. 18,000 degrees / second                                              

                  C . 50 degrees / second                    D. 500 degrees / second

13. If a tire rotates at 400 revolutions per minute when the car is traveling 72km/h, what is the circumference of the tire?

A. 0.3meters     B. 333.3meters    C. 3meters    D. 33.3meters

14. In a shop, the cost of 4 shirts, 4 pairs of trousers and 2 hats is $560. The cost of 9 shirts, 9 pairs of trousers and 6 hats is $1,290. What is the total cost of 1 shirt, 1 pair of trousers and 1 hat?

A. $150      B. $170       C. $200    D. $250

15. Four children have small toys. The first child has 1/10 of the toys, the second child has 12 more toys than the first, the third child has one more toy of what the first child has and the fourth child has double the third child. How many toys are there?

A. 20 toys     B. 25 toys     C  40 toys     D.30 toys

16. A class average mark in an exam is 70. The average of students who scored below 60 is 50. The average of students who scored 60 or more is 75. If the total number of students in this class is 20, how many students scored below 60?

A. 5          B. 4          C. 6          D. 3

17. An airplane flies against the wind from A to B in 8 hours. The same airplane returns from B to A, in the same direction as the wind, in 7 hours. Find the ratio of the speed of the airplane (in still air) to the speed of the wind.

A. 0.06     B. 15    C. 8/7     D. none

18. Find the point(s) of intersection of the parabola with equation y = x2 - 5x + 4 and the line with equation y = 2x – 2

A.   (1 , 0) and (6 , 20)    B. (1 , 1) and (6 , 10)  

C. (2 , 0) and (6 , 10)     D. (1 , 0) and (6 , 10)

19. Find the constant k so that : -x2 - (k + 7)x - 8 = -(x - 2)(x - 4)

A.   1        B   -1          C. 2          D. -2

20. Factor the expression 6x2 - 13x + 5

A. (3x - 4)(2x - 1)    B. (2x - 5)(2x - 1)     C. (x - 5)(2x - 1)    D. (3x - 5)(2x - 1)

21.  Find all zeros of the polynomial P(x) = x3 - 3x2 - 10x + 24 knowing that x = 2 is a zero of the polynomial.

A. 4 , -3 and -2    B. 4 , 3 and 2    C. 4 , -3 and 2  D. -4 , -3 and 2

22.  If x is an integer, what is the greatest value of x which satisfies 5 < 2x + 2 < 9?

A. 3            B. 2            C. 5          D. 4

23.  Sets A and B are given by: A = {2 , 3 , 6 , 8, 10} , B = {3 , 5 , 7 , 9}.  Find the intersection of sets A and B.   

A. {3,5}         B.  {3}            C. {2,5,9}             D. {7}

24.  Simplify | - x2 + 4x - 4 |.

A. -(x - 2)2          B (x - 2)2          C. (x - 3)2          D. -(x - 3)2 

25.  Simplify 8 x3 / 2 x-3

A. 4 x6     B. 4    C. 16X6     D. None

                           TO GET FULL ANSWER 


Mathematics

 

  1. Which real numbers are equal to their cubes?

           0 , 1 , -1 are all equal to their respective cubes.

  1. Write 4*10-2 as a decimal.

         0.02

  1. Write 0.12*10-3 as a decimal.

0.00012

  1. Write 2 log3 x + log3 5 as a single logarithmic expression.

log3 x2 + log3 5 = log3(5x2)

  1. Factor the algebraic expression 6x2 - 21xy + 8xz - 28yz.

3x(2x - 7y) + 4z(2x - 7y) = (2x - 7y)(3x + 4z)

  1. Factor the algebraic expression (x - 1)2 - (y - 2)2.

[(x - 1)- (y - 2)][(x - 1)+ (y - 2)] = (x - y + 1)(x + y - 3)

  1. Factor the algebraic expression x2 - z4.

(x + z2)(x - z2) = (x + z2)(x + z)(x - z)

  1. Evaluate the algebraic expression |-2x - y + 3| for x = 3 and y = 5

|-2(3) - (5) + 3| = |-8| = 8

  1. Simplify the algebraic expression -2(x - 3) + 4(-2x + 8)

-2x + 6 -8x + 32 = -10x + 38

  1. Expand and simplify the algebraic expression (x + 3)(x - 3) - (-x - 9)

x2 - 9 + x + 9 = x2 + x

  1. Which property is used to write a(x + y) = ax + ay

distributivity

  1. Simplify 8 x3 / 2 x-3

4 x6

  1. Simplify (-a2b3)2(c2)0

a4b6

  1. For what value of k is the point (-2, k) on the line with equation -3x + 3y = 4?

k = -2/3

  1. For what value of a will the system given below have no solutions?

    2x + 6y = -2
    -3x + ay = 4

a = -9

  1. Which equation best describes the relationship between x and y in this table?

x

y

0

-4

4

-20

-4

12

8

-36
    1. y = - x/4 - 4
    2. y = - x/4 + 4
    3. y = - 4x - 4
    4. y = - 4x + 4

C. y = - 4x - 4

 

  1. Which equation best represents the area of the rectangle below?



                       x+1
         




                                                        

x-1    




    1. area = 2(x+1) + 2(x-1)
    2. area = 4(x+1)(x-1)
    3. area = 2x2
    4. area = x2 - 1

D. area = (x + 1)(x - 1) = x2 - 1

 

  1. Which line given by its equation below contains the points (1, -1) and (3, 5)?
    1. -2y -6x = 0
    2. 2y = 6x - 8
    3. y = 3x + 4
    4. y = -3x + 4

B. 2y = 6x - 8

  1. Solve the equation 2|3x - 2| - 3 = 7.

solution set : {7/3 , -1}

  1. Solve for x the equation (1/2)x2 + mx - 2 = 0.

solution set : {-m + sqrt(m2 + 4) , -m - sqrt(m2 + 4)}

  1. For what values of k the equation -x2 + 2kx - 4 = 0 has one real solution?

k = 2 , k = -2

  1. For what values of b the equation x2 - 4x + 4b = 0 has two real solutions?

all values of b less than 1

  1. Function f is described by the equation f (x) = -x2 + 7. What is the set of values of f(x) corresponding to the set for the independent variable x given by {1, 5, 7, 12}?

{6 , -18 , -42 , -137}

  1. Find the length and width of a rectangle whose perimeter is equal to 160 cm and its length is equal to triple its width.

width = 20 cm and length = 60 cm

  1. Simplify: |-x| + |3x| - |-2x| + 3|x|

      |-x| + |3x| - |-2x| + 3|x| = 5|x|

 

  1. If (x2 - y2) = 10 and (x + y) = 2, find x and y.

 

Grade 12 Math Word Problems with Solutions and Answers

Grade 12 math word problems with detailed solutions are presented.



  1. Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming pool in 4 hours. How many hours will it take 4 large and 2 small pumps to fill the swimming pool.(We assume that all large pumps are similar and all small pumps are also similar.)
  1. Find all sides of a right triangle whose perimeter is equal to 60 cm and its area is equal to 150 square cm.
  1. A circle of center (-3 , -2) passes through the points (0 , -6) and (a , 0). Find a.
  1. Find the equation of the tangent at ( 0 , 2) to the circle with equation

(x + 2)2 + (y + 1)2 = 13

  1. An examination consists of three parts. In part A, a student must answer 2 of 3 questions. In part B, a student must answer 6 of 8 questions and in part C, a student must answer all questions. How many choices of questions does the student have?
  1. Solve for x

x2 - 3|x - 2| - 4x = - 6

 

  1. The right triangle ABC shown below is inscribed inside a parabola. Point B is also the maximum point of the parabola (vertex) and point C is the x intercept of the parabola. If the equation of the parabola is given by y = -x2 + 4x + C, find C so that the area of the triangle ABC is equal to 32 square units.

problem 3.

  1. The triangle bounded by the lines y = 0, y = 2x and y = -0.5x + k, with k positive, is equal to 80 square units. Find k.
  1. A parabola has two x intercepts at (-2 , 0) and (3 , 0) and passes through the point (5 , 10). Find the equation of this parabola.
  1. When the polynomial P(x) = x3 + 3x2 -2Ax + 3, where A is a constant, is divided by x2 + 1 we get a remainder equal to -5x. Find A.
  1. When divided by x - 1, the polynomial P(x) = x5 + 2x3 +Ax + B, where A and B are constants, the remainder is equal to 2. When P(x) is divided by x + 3, the remainder is equal -314. Find A and B.
  1. Find all points of intersections of the 2 circles defined by the equations

(x - 2)2 + (y - 2)2 = 4

(x - 1)2 + (y - 1)2 = 4

  1. If 200 is added to a positive integer I, the result is a square number. If 276 is added to to the same integer I, another square number is obtained. Find I.
  1. The sum of the first three terms of a geometric sequence is equal to 42. The sum of the squares of the same terms is equal to 1092. Find the three terms of the sequence.
  1. A rock is dropped into a water well and it travels approximately 16t2 in t seconds. If the splash is heard 3.5 seconds later and the speed of sound is 1087 feet/second, what is the height of the well?
  1. Two boats on opposite banks of a river start moving towards each other. They first pass each other 1400 meters from one bank. They each continue to the opposite bank, immediately turn around and start back to the other bank. When they pass each other a second time, they are 600 meters from the other bank. We assume that each boat travels at a constant speed all along the journey. Find the width of the river?
  1. Find the constants a and b so that all the 4 lines whose equation are given by

x + y = -1

-x + 3y = -11

ax + by = 4

2ax - by = 2
  1. pass through the same point.
  3. Find the area of the right triangle shown below.

problem 17.

  1. It takes pump A 2 hours less time that pump B to empty a swimming pool. Pump A is started at 8:00 a.m. and pump B is started at 10:00 a.m. At 12:00 p.m. 60% of the pool is empty when pump B broke down. How much time after 12:00 p.m. would it take pump A to empty the pool?
  1. The number of pupils in school A is equal to half the number of pupils in school B. The ratio of the boys in school A and the boys in school B is 1:3 and the ratio of the girls in school A and the girls in school B is 3:5. The number of boys in school B is 200 higher than the number of boys in school A. Find the number of boys and girls in each school.

Solutions to the Above Questions

  1.  
    1. Let R and r the rate of work of the large and the small pumps respectively
    2. 4(2R + r) = 1 : 2 large and 1 small work for 4 hours to do 1 job
    3. 4(R + 3r) = 1 : 1 large and 3 small work for 4 hours to do 1 job
    4. T(4R + 4r) = 1 : Find time T if 4 large and 4 small are to do one job.
    5. Solve for R and r the system of first two equations then sustitute in the third and solve for T to find the time. T = 5/3 hours = 1 hour 40 minutes.
  1.  
    1. x + y + H = 60 : perimeter , x, y and H be the two legs and the hypotenuse of the right triangle
    2. (1/2)xy = 150 : area
    3. x2 + y2 = H2: Pythagora's theorem.
    4. 3 equations with 3 unknowns.
    5. (x + y)2 - 2xy = H2 : completing the square in the third equation.
    6. x + y = 60 - H : express x + y using the first equation and use the second equation to find xy = 300 and substitute in equation 5.
    7. (60 - H)2 - 600 = H2 : one equation with one unknown.
    8. Solve for H to find H = 25 cm. Substitute and solve for x and y to find x = 15 cm and y = 20 cm.
  1.  
    1. sqrt((-6 + 2)2 + (0 + 3)2) = (a + 3)2 + (0 + 2)2 : distances from center to any point on the circle are equal to the radius.
    2. a = -3 + sqrt(21) , a = -3 - sqrt(21) : solve for a and find two solutions.
  1.  
    1. (-2 , -1) : center of circle
    2. m = (2 - -1) / (0 - -2) = 3 / 2 : slope of line through the center and the point of tangency (0 , 2)
    3. The line through the center and the point of tangency (0 , 2) is perpendicular to the tangent.
    4. M = -2 / 3 : slope of tangent
    5. y = -(2/3)x + 2 : equation of tangent given its slope and point (0 , 2).
  1.  
    1. 3C2 * 8C6 * 1 = 84 : Use of fundamental theorem of counting
  1.  
    1. x2 - 3|x - 2| - 4x = - 6 : given
    2. Let Y = x - 2 which gives x = Y + 2
    3. (Y + 2)2 - 3|Y| - 4(Y + 2) = - 6 : substitute in above equation
    4. Y2 - 3|Y| + 2 = 0
    5. Y2 = |Y|2 : note
    6. |Y|2 - 3|Y| + 2 = 0 : rewrite equation as
    7. (|Y| - 2)(|Y| - 1) = 0
    8. |Y| = 2 , |Y| = 1 : solve for |Y|
    9. Y = 2, -2 , 1 , -1 : solve for Y
    10. x = 4 , 0 , 3 , 1 : solve for x using x = Y + 2.
  1.  
    1. h = -b / 2a = 2 : x coordinate of the vertex of the parabola
    2. k = -(2)2 + 4(2) + C = 4 + C : y coordinate of vertex
    3. x = (2 + sqrt(4 + C)) , x = (2 - sqrt(4 + C)) : the two x intercepts of the parabola.
    4. length of BA = k = 4 + C
    5. length of AC = 2 + sqrt(4 + C) - 2 = sqrt(4 + C)
    6. area = (1/2)BA * AC = (1/2) (4 + C) * sqrt(4 + C)
    7. (1/2) (4 + C) * sqrt(4 + C) = 32 : area is equal to 32
    8. C = 12 : solve above for C.
  1.  

problem 8

    1. A(0,0) , B(2k/5 , 4k/5) , C(2k ,0) : points of intersection of the 3 points of intersection of the 3 lines
    2. (1/2) * (4k/5) * (2k) = 80 : area given
    3. k = 10 : solve the above equation for k , k positive is a given condition.
  1.  
    1. y = a(x + 2)(x - 3) : equation of the parbola in factored form
    2. 10 = a(5 + 2)(5 - 2) : (5 , 10) is a point on the graph of the parabola and therefore satisfies the equation of the parabola.
    3. a = 5/7: solve the above equation for a.
  1.  
    1. Divide x(3 + 3x2 -2Ax + 3) by (x2 + 1) to obtain a remainder = -x(1 + 2A)
    2. -x(1 + 2A) = 5x : remainder given
    3. -(1 + 2A) = 5 : polynomials are equal if they corresponding coefficient area equal.
    4. A = -3
  1.  
    1. P(1) = 15 + 2(13) +A*(1) + B = 2 : remainder theorem
    2. P(-3) = (-3)5 + 2(-3)3 +A*(-3) + B = -314
    3. A = 4 and B = -5 : solve the above systems of equations.
  1.  
    1. x2 - 4x + 2 + y2 - 4y + 2 = 4 : expand equation of first circle
    2. x2 - 2x + 1 + y2 - 2y + 1 = 4 : expand equation of second circle
    3. -2x - 2y - 6 = 0 : subtract the left and right terms of the above equations
    4. y = 3 - x : solve the above for y.
    5. 2x2 - 6x + 1 = 0 : substitute y by 3 - x in the first equation, expand and group like terms.
    6. (3/2 + sqrt(7)/2 , 3/2 - sqrt(7)/2) , (3/2 - sqrt(7)/2 , 3/2 + sqrt(7)/2) : solve the above for x and use y = 3 - x to find y.
  1.  
    1. I + 200 = A2 : 200 added to I (unknown integer) gives a square.
    2. I + 276 = B2 : 276 added to I (unknown integer) gives another square.
    3. B2 = A2 + 76 : eliminate I from the two equations.
    4. add squares A2 (0, 1, 4, 9, 16, 25,...) to 76 till you obtain another square B2.
    5. 76 + 182 = 400 = 202
    6. A2 = 182 and B2 = 202
    7. I = A2 - 200 = 124
  1.  
    1. sum1 = a + a r + a r2 = 42: the sum of the three terms given, r is the common ratio.
    2. sum2 = a2 + a2r2 + ar2r4 = 1092: the sum of the squares of the three terms given .
    3. sum1 = a + ar + ar2 = a(r3 - 1) / (r - 1) = 42 : apply formula for a finite sum of geometric series.
    4. sum2 = a2 + a2r2 + ar2r4 = a2(r6 - 1) / (r2 - 1) = 1092: the sum of squares is a also a sum of geometric series.
    5. sum2/sum12 = 1092 / 422 = [ a2(r6 - 1)/(r2 - 1)] / [a2(r3 - 1)2 / (r - 1)2]
    6. (r2 - r + 1) / (r2 + r + 1) = 1092 / 422
    7. r = 4 , r = 1/4 : solve for r
    8. a = 2 : substitute r = 4 and solve for a
    9. a = 32 : substitute r = 1/4 and solve for a
    10. a = 2 , ar = 8 , ar2 = 32 : find the three terms for r = 4
    11. a = 32 , ar = 8 , ar2 = 2 : find the three terms for r = 1/4
  1.  
    1. T1 + T2 = 3.5 : T1 time for the rock to reach bottom of well and T2 time for the sound to reach the top of the well.
    2. 16 * T12 = 1087 * T2 : same distance which the height of the well.
    3. T2 = 3.5 - T1 : solve for T2
    4. 16 * T12 = 1087 * (3.5 - T1)
    5. T1 = 3.34 seconds
    6. Height = 16 * (3.34)2 = 178 feet (to the nearest unit)
  1.  

problem 16.

    1. S1*t1 = 1400 : S1 speed of boat 1, t1 : time to do 1400 meters(boat 1)
    2. 1400 + S2*t1 = X : S2 speed of boat 2
    3. S1*t2 = X + 600 : t2 time to do X + 600 (boat 2)
    4. S2*t2 = 2X - 600
    5. S1 = 1400/t1
    6. S2 = (X-1400)/t1
    7. T = t2/t1 : definition
    8. substitute S1, S2 and t2/t1 using the above expressions in equations 3 and 4 to obtain
    9. 1400*T = X + 600
    10. X*T - 1400*T = 2X - 600 : 2 equations 2 unknowns
    11. Eliminate T and solve for X to obtain X = 3600 meters.
  1.  
    1. solve the system of the first two equations to obtain the solution (2 , -3)
    2. The above solution is also a solution to the last two equations.
    3. a(2) + b(-3) = 4
    4. 2a(2) - b(-3) = 2
    5. a = 1 and b = -2/3 : solution to the above system of equations.

Geometry Problems with Solutions and Answers for Grade 12

Grade 12 geometry problems with detailed solutions are presented.



  1. In the triangle ABC sides AB and CB have equal lengths and the measure of angle ABC is equal to 36 degrees. What is the measure of angle BOC where O is the center of the circle?

problem 1.

  1. Circles C1 and C2 have equal radii and are tangent to that same line L. Circle C3 is tangent to C1 and C2. x is the distance between the between the centers of C1 and C2. Find the distance h, from the center of C3 to line L, in terms of x and the radii of the three circles.

problem 2.

  1. All three circles are tangent to the same line and to each other. Circles C2 and C3 have equal radii. Find the radius of C2 if the radius of C1 is equal to 10 cm.

problem 2.

  1. CD is parallel to AB and the measure of angle t is equal to 90 degrees. Find the area of the circle in terms of x.

problem 3.

  1. The shaded region below is the common area to four semicircles whose diamters are the sides of the square with side length 4x. Find the area of the shaded region in terms of x.

problem 4.

  1. The two circles below are concentric (have same circle). The length of the chord tangent to the smaller circle is equal to 20 mm. What is the area of the ring (shaded area) between the two circles?

problem 6.

  1. Find a, b and c so that the quadrilateral is a parallelogram with area equal to 80 square units.

problem 7.

  1. A right triangle is shown below. Find the lengths x, y and z.

problem 8.

  1. A rectangle is shown below. Find the length x.

problem 9.

  1. The two circles below have equal radii of 4 units each and the distance between their centers is 6 units. Find the area of the shaded region.

problem 10.

Solutions to the Above Problems

  1.  
    1. measure of angle BAC = (180 - 36)/2 = 72 degres : isosceles triangle
    2. measure of angle BOC = 2 * measure of angle BAC = 144 degrees : inscribed angle and central angle intercepting the same arc.
  1.  
    1. Let R1, R2 and R3 be the radii of circles C1, C2 and C3 respectively with R1 = R2 = R
    2. h = C3O + R
    3. C3O2 + (x/2)2 = (R + R3)2 : Pythagora's theorem applied to triangle C3OC1.
    4. h = R + C3O = R + sqrt[ (R + R3)2 - (x/2)2 ]

solution to problem 2.

  2.  
    1. Let r, R2 and R3 be the radii of circles C1, C2 and C3 respectively with R2 = R3 = R
    2. (r + R)2 = R2 + (R - r)2 : Pythagora's theorem applied to triangle MC1C3
    3. R = 4r = 40 cm : expand and solve for R.

solution to problem 2.

  2.  
    1. The measure of angle BtA is equal to 90 degrees : vertical angles
    2. AB is the diameter of the circle : converse of Thales theorem
    3. triangles BtA and CtD are similar : CD parallel to AB
    4. 3/5 = AB/x : corresponding sides proportional.
    5. AB = 3x/5 : solve for AB
    6. radius = AB/2 = 3x/10
    7. area = Pi(3x/10)2 = 0.09 Pi x2
  1.  
    1. The square below has side length 2x, half of the given square. Part of this is shaded and the other part is not shaded. Let us find the area of the non shaded part (white). The shaded part is a quarter of a disk (circle).
    2. area of non shaded area = (2x)2 - (1/4) Pi (2x)2
    3. If we go back to the given shape in problem 5, the area of the non shaded part is 8 times the non shaded area in the present shape which was calculated above.
    4. area of shaded part in shape of problem 5 = total area of square - total non shaded area = (4x)2 - 8*[ (2x)2 - (1/4) Pi (2x)2 ]
    5. = 16x2(Pi/2 - 1)

solution to problem 5.

  2.  
    1. R2 = r2 + 102 : Pythagora's theorem
    2. Pi(R2 - r2) = 100 Pi : area of ring

solution to problem 7.

  2.  
    1. The area of a parallelogram may be calculated using the cross product of vectors AB and AD as follows.
    2. area = |AB x AD| where AB and AD are 3-dimensional vectors.
    3. vector AB = <4 , b + 2 , 0> : set third component to zero since the given shape is 2-dimensional.
    4. vector AD = <6 , 4 , 0>
    5. |<4 , b + 2 , 0> x <6 , 4 , 0>| = 80 : modulus of cross product is equal to area
    6. |4 - 6b| = 80
    7. b = 14 and b = -38/3 : we select the solution b = 14 since point B(2,b) is in quadrant I.
    8. Since ABCD is a parallelogram, vector AB = vector DC
    9. Vector AB = <4 , 16 , 0>
    10. vector DC =
    11. c - 4 = 4 and d - 2 = 16 : If two vectors are equal their corresponding components area equal.
    12. c = 8 and d = 18 : solve the above equations.
  1.  
    1. y2 + z2 = 122 : Pythagora's theorem
    2. x2 + z2 = 92 : Pythagora's theorem
    3. (y + x)2 = 122 + 92 : Pythagora's theorem
    4. x + y = 15 : Solve equation C by extracting the square root
    5. y2 - x2 = 63 : subtracting equations A and B
    6. (y - x)(y + x) = 63 : factoring the left term of equation E.
    7. y - x = 21/5
    8. x = 27/5 , y = 48/5 and z = 36/5 : solve the system made up of equations D and G.
  1.  
    1. Split the given rectangle into 4 other rectangles as shown.
    2. a2 + c2 = 42 : Pythagora's theorem applied to top left right triangle.
    3. b2 + c2 = x2 : Pythagora's theorem applied to bottom left right triangle.
    4. b2 + d2 = 52 : Pythagora's theorem applied to bottom right right triangle.
    5. a2 + d2 = 62 : Pythagora's theorem applied to bottom right right triangle.
    6. a2 - b2 = 42 - x2 : subtract equations B and C.
    7. a2 - b2 = 62 - 52 : subtract equations D and E.
    8. 42 - x2 = 62 - 52 : combine equations F and G.
    9. x = sqrt(5) : solve above equation for x.

solution to problem 9.

  2.  
    1. Because of symmetry, the shaded region may be considered as made up of two equal (in area) regions. The area of the left half of the shaded region is given by the area of the sector BOC minus the area of the triangle BOC.
    2. length of OM = 3 (by symmetry) since distance between centers is 6, and radius r = 4.
    3. Let t be the measure of angle BOM.
    4. cos(t) = OM/OB = 3/4 , t = arccos(3/4) : using right triangle BOM.
    5. area of sector BOC = (1/2)(2t)r2
    6. area of triangle BOC = (1/2)sin(2t)r2
    7. area of shaded region = 2 [ (1/2)(2t)r2 - (1/2)sin(2t)r2 ]
    8. = [2t - sin(2t)] r2 = 7.25 square units (rounded to 3 decimal places)

solution to problem 10.





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