Problem 1 :
In the figure above, BD is a diameter, and PA and PD are tangents to circle O. m∠CDE = 52 and m∠APD = 45 and AP = 9.
1) What is the measure of ∠ODC ?
2) What is the measure of ∠OCD ?
3) What is the measure of ∠AOD ?
4) What is the length of PD ?
Solution :
(1) A line drawn from center of circle to the tangent line will be a perpendicular.
So,
∠ODE = 90
∠ODC + ∠CDE = 90
∠ODC + 52 = 90
∠ODC = 90 - 52
∠ODC = 38
(2) Since OC and OD are radius, they are having same measures and they will create equal angles.
∠OCD = 38
(3) Considering the quadrilateral AODP,
Sum of interior angles of a quadrilateral = 360
OD and OA are perpendiculars to PD and PA.
So, 90 + 90 + ∠AOD + ∠DPA = 360
180 + ∠AOD + 45 = 360
∠AOD = 360 - 225
∠AOD = 135
(4) Length of PD is also 9.
Because the tangents drawn from an external point to a circle are equal.
Problem 2 :
In the figure given below O is inscribed in triangle PQR. If PA = 12, QA = 6 and RB = 9.5. What is the perimeter of triangle PQR ?
Solution :
RB = RC = 9.5
PB = PA = 12
QC = QA = 6
Perimeter of triangle PQR :
= PQ + QR + PR ----(1)
PQ = PA + AQ ==> 12 + 6 ==> 18
PR = PB + BR ==> 12 + 9.5 ==> 21.5
RQ = RC + QC ==> 9.5 + 6==> 15.5
Applying the values in (1), we get
= 18 + 21.5 + 15.5
= 55
Problem 3 :
A certain pizza restaurant cuts slices out at every 4 inches along the edge of a pizza, as shown in the figure below. What is maximum number of of full pizza slices that can be cut out from a circular pizza with a radius of 10 inches ?
Solution :
Length of arc = 4 inches = piece of pizza having arc length
Number of pizzas = Perimeter of the circular pizza / 4
= 2πr/4
= [2(3.14) 10]/4
= 15.7
So, 15 pizzas can be cut out.
Problem 4 :
In the figure given below, circle A has a radius of 2, circle B has a radius of 4, the circle C has a radius of 6. what is the area of the shaded region ?
Solution :
Area of small circle having center at A.
Radius of circle having center A is 2 units.
Area of small circle = πr^{2}
= π(2)^{2}
= 4π
Area of remaining part (shaded)
= Area of circle having center C - Area of circle having center B
= π6^{2 }- π4^{2}
= π(36-16)
= 20π
Sum of shaded portions = 4π + 20π
= 24π
Problem 5 :
In the figure given above, A and B are points on the circle C. If the area of the circle is 54π, what is the area of the sector formed by the central angle ACB ?
Solution :
Area of the circle = 54π
πr^{2} = 54π
r^{2} = 54
r = √54
r = 3√6
The given triangle ACB is a isosceles triangle, because CA and CB are equal(radii)
∠ACB = 80
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM