1)In a figure,triangle PQR has a circle circumscribed to it.PQ=5.9cm,QR=4.5cm, and PR=7.0cm.
a)Calculate the size of angle OPR,correct to the nearest degree
b)Calculate the area of the triangle correct to 2 decimal places
c)Calculate the radius of the circle,correct to 1 decimal place
d)Find the area of the circle outside the triangle to 4 significant figures
2)Triangle PQR with coordinates P(0,1),Q(8,1) and R(2,7).M is the midpoint of R,Q and N divides PQ such that PN:PQ=3:5.PM and RN intersect at X.
Find;a)Vector PM
b)Vector RN
c)Given that PX=sPM and RX=tRN,express QX in two different ways?Hence find the values of s and t?
d)Determine the ratio in which N divides RX
3)A hockey coach has 5 white corks,4 orange corks and some luminous green corks in the bag for drills during practice.The balls are similar in shape and size;
a)If the probability of picking a luminous green ball is 0.25,find the total number of balls in the bag
b)Three corks picked at random,one after the other without replacement and their colour noted.Find the probability that the three balls are of:(i)The same colour
(ii)Different colour
c)The probability of Jack winning a game of pool is 2/3.If he plays the game until he wins,find his probability of winning the game in either the third or fourth attempt?
1)
a) In a circle, the angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any point on the circumference of the circle. Therefore, angle OPR can be found by finding the angle subtended by arc PR at any point on the circumference, and then dividing by 2.
Using the Law of Cosines, we can find the angle subtended by arc PR at point Q:
PR^2 = PQ^2 + QR^2 - 2*PQ*QR*cos(angle PQR)
7.0^2 = 5.9^2 + 4.5^2 - 2*5.9*4.5*cos(angle PQR)
49 = 34.81 + 20.25 - 53.1*cos(angle PQR)
49 = 55.06 - 53.1*cos(angle PQR)
-6.06 = -53.1*cos(angle PQR)
cos(angle PQR) = -6.06/(-53.1) = 0.1146
Now, we can find the angle subtended by arc PR at point Q:
angle PQR = acos(0.1146) ≈ 83.75 degrees
Therefore, angle OPR ≈ 83.75/2 = 41.88 degrees (to the nearest degree).
b) The area of a triangle can be found using Heron's formula:
s = (PQ + QR + PR)/2 = (5.9 + 4.5 + 7.0)/2 = 8.7
Area = sqrt(s(s - PQ)(s - QR)(s - PR)) = sqrt(8.7(8.7 - 5.9)(8.7 - 4.5)(8.7 - 7.0)) ≈ 13.99 square cm (to 2 decimal places).
c) The radius of a circle circumscribed to a triangle can be found using the following formula:
R = (PQ*QR*PR)/(4*Area)
R = (5.9*4.5*7.0)/(4*13.99) ≈ 4.75 cm (to 1 decimal place).
d) The area of the circle outside the triangle can be found by subtracting the area of the triangle from the area of the circle:
Area of circle outside triangle = πR^2 - Area of triangle
Area of circle outside triangle = π(4.75)^2 - 13.99 ≈ 55.91 square cm (to 4 significant figures).
2)
a) Vector PM can be found by finding the midpoint of the coordinates of points R and Q:
PM = (1/2)(RQ)
PM = (1/2)(8, 7 - 1) = (1/2)(8, 6) = (4, 3).
b) Vector RN can be found by subtracting the coordinates of point N from the coordinates of point R:
RN = R - N
RN = (2, 7) - (5/8)(8, 1 - 7)
RN = (2, 7) - (5/8)(8, -6)
RN = (2, 7) - (5, -3/2)
RN = (-3, 17/2).
c) Given that PX = sPM and RX = tRN, we can express QX in two different ways:
QX = PX + RX
Substituting the values:
QX = sPM + tRN
We can find the values of s and t by solving the equations:
8 - 0 = s(4) + t(-3)
1 - 0 = s(3) + t(17/2)
Simplifying the equations:
8 = 4s - 3t
1 = 3s + (17/2)t
We can solve these equations simultaneously to find the values of s and t.
d) To determine the ratio in which N divides RX, we can use the proportionality of the vectors:
RX = tRN
PM = s(2, 7 - 1)
We can set the corresponding components equal to each other:
-3 = ts
17/2 = t(7 - 1)
Simplifying these equations:
-3 = ts
17/2 = 6t
Dividing the second equation by 6:
17/12 = t
Substituting the value of t into the first equation:
-3 = s(17/12)
Solving for s:
s = -3(12/17) = -36/17
Therefore, the ratio in which N divides RX is -36/17.