Here we are going to see some examples problems on finding radius when length of two parallel chords are given.
Example 1 :
AB and CD are two parallel chords of a circle which are on either sides of the centre. Such that AB = 10 cm and CD = 24 cm. Find the radius if the distance between AB and CD is 17 cm.
Solution :
Consider the right triangles OEB and OFD,
In triangle OEB, OB2 = OE2 + EB2 OB2 = x2 + 52 ---(1) |
In triangle OFD, OD2 = OF2 + FD2 OD2 = (17-x)2 + 122 ---(2) |
OB = OD (radius of the given circle)
(1) = (2)
x2 + 52 = (17-x)2 + 122
x2 + 52 = 172 + x2 - 2(17) x + 122
x2 + 25 = 289 + x2 - 34x + 144
x2 - x2 + 34x + 25 - 144 - 289 = 0
34x - 408 = 0
34(x - 12) = 0
x = 12 cm
By applying the value of x in the 1st equation, we get
OB2 = 122 + 52
OB2 = 144 + 25 = 169
OB = √169 = 13 cm
Example 2 :
In the figure given below, AB and CD are two parallel chords of a circle with centre O and radius 5 cm such that AB = 6 cm and CD = 8 cm. If OP ⊥ AB and CD = OQ determine the length of PQ.
Solution :
Here we have two right triangles,
Triangle OPB and triangle OQD.
OB = OD = radius of the circle = 5 cm
In Δ OPB,
OB2 = OP2 + PB2 OB2 = OP2 + PB2 52 = OP2 + 32 OP2 = 25 - 9 OP2 = 16 OP = √16 OP = 4 cm |
OD2 = OQ2 + QD2 52 = OQ2 + 42 25 = OQ2 + 16 OQ2 = 25 - 16 OQ2 = 9 OQ = √9 OQ = 3 cm |
PQ = OP - OQ
= 4 - 3
= 1 cm
Hence the length of PQ is 1 cm
Example 3 :
In the figure given below, AB and CD are two parallel chords of a circle with centre O and radius 5 cm. Such that AB = 8 cm and CD = 6 cm. If OP = AB and OQ ⊥ CD.determine the length PQ.
Solution :
Consider the triangles APO and COQ
OA = OC = radius of the circle = 5 cm
AP = PB = 4 cm
CQ = QD = 3 cm
In triangle APO, OA2 = AP2 + PO2 52 = 42 + PO2 PO = √(25 - 16) PO = √9 PO = 3 cm |
In triangle COQ, OC2 = OQ2 + CQ2 52 = OQ2 + 32 OQ = √(25 - 9) OQ = √16 OQ = 4 cm |
PQ = PO + OQ
= 3 + 4
= 7 cm
Hence the length of PQ is 7 cm.
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