Problems Based on Circles Examples :
Here we are going to see some example problems based on circles.
Question 1 :
Th e diameter of the circle is 52 cm and the length of one of its chord is 20 cm. Find the distance of the chord from the centre.
Solution :
Diameter = 52 cm
Radius of circle = 26 cm
In triangle BOC,
OC2 = OB2 + BC2
262 = OB2 + 102
676 - 100 = OB2
OB = √576
OB = 24
Question 2 :
The chord of length 30 cm is drawn at the distance of 8 cm from the centre of the circle. Find the radius of the circle
Solution :
In triangle BOC,
OC2 = OB2 + BC2
OC2 = 82 + 152
OC2 = 64 + 225
OC = √289
OC = 17
Question 3 :
Find the length of the chord AC where AB and CD are the two diameters perpendicular to each other of a circle with radius 4√2 cm and also find <OAC and <OCA.
Solution :
AC2 = AO2 + CO2
AC2 = (4√2)2 + (4√2)2
AC2 = 16(2) + 16(2)
AC2 = 64
AC = √64
AC = 8 cm
In triangle AOC,
<COA = 90
OC = OA
Equal sides will form a equal angles.
<OCA = <OBC
Question 4 :
A chord is 12 cm away from the centre of the circle of radius 15 cm. Find the length of the chord.
Solution :
Let BC = x
OC2 = OB2 + BC2
152 = 122 + x2
225 - 144 = x2
x = √81
x = 9 cm
AC = 2BC = 2(9) = 18 cm
Question 5 :
In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords?
Solution :
In triangle OPB,
OB2 = OP2 + PB2
OA = OB = OC = OD = 10 (Radius)
102 = OP2 + 82
OP2 = 100 - 64
OP2 = 36
OP = √36
OP = 6 cm
In triangle OQD,
OD2 = OQ2 + QD2
102 = OQ2 + 62
OQ2 = 100 - 36
OQ2 = 64
OQ = √64
OQ = 8 cm
Distance between two chords PQ = OQ - OP
= 8 - 6
= 2 cm
Question 6 :
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution :
OP = 3 cm, PS = 5 cm and OS = 4 cm
Let RS be x
OR = 4 - x
In triangle POR,
OP2 = OR2 + PR2
32 = x2 + PR2
PR2 = 9 - x2 ------(1)
In triangle PRS,
PS2 = PR2 + RS2
52 = PR2 + (4 - x)2
PR2 = 25 - (4 - x)2
PR2 = 25 - (16 - 8x + x2)
PR2 = 25 - 16 + 8x - x2
PR2 = 9 + 8x - x2 ----(2)
(1) = (2)
9 - x2 = 9 + 8x - x2
8x = 0
x = 0
By applying the value of x in (1), we get
PR2 = 9 - 0
PR2 = 9 ==> PR = 3
PQ = 2PR
PQ = 2(3)
PQ = 6 cm
After having gone through the stuff given above, we hope that the students would have understood, "Problems Based on Circles Examples"
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