Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6, find the radius of the circle.
Answers
Step-by-step explanation:
Join OA and OC.
Let the radius of the circle be r cm and O be the centre
Draw OP⊥AB and OQ⊥CD.
We know, OQ⊥CD, OP⊥AB and AB∥CD.
Therefore, points P,O and Q are collinear. So, PQ=6 cm.
Let OP=x.
Then, OQ=(6–x) cm.
And OA=OC=r.
Also, AP=PB=2.5 cm and CQ=QD=5.5 cm.
(Perpendicular from the centre to a chord of the circle bisects the chord.)
In right triangles QAP and OCQ, we have
OA
2
=OP
2
+AP
2
and OC
2
=OQ
2
+CQ
2
∴r
2
=x
2
+(2.5)
2
..... (1)
and r
2
=(6−x)
2
+(5.5)
2
..... (2)
⇒x
2
+(2.5)
2
=(6−x)
2
+(5.5)
2
⇒x
2
+6.25=36−12x+x
2
+30.25
12x=60
∴x=5
Putting x=5 in (1), we get
r
2
=52+(2.5)
2
=25+6.25=31.25
⇒r
2
=31.25⇒r=5.6
Hence, the radius of the circle is 5.6 cm
solution
Answer:
Here, OM ⊥ AB and ON ⊥ CD. is drawn and OB and OD are joined.
As we know, AB bisects BM as the perpendicular from the centre bisects the chord.
Since AB = 5 so,
BM = AB/2
Similarly, ND = CD/2 = 11/2
Now, let ON be x.
So, OM = 6− x.
Consider ΔMOB,
OB2 = OM2 + MB2
Or,
OB2 = 36 + x2 – 12x + 25/4 ……(1)
Consider ΔNOD,
OD2 = ON2 + ND2
Or,
OD2 = x2+121/4 ……….(2)
We know, OB = OD (radii)
From eq. (1) and eq. (2) we have;
36 + x2 -12x + 25/4 = x2 + 121/4
12x = 36 + 25/4 – 121/4
12x = (144 + 25 -121)/4
12x = 48/4 = 12
x = 1
Now, from eq. (2) we have,
OD2 = 11 + (121/4)
Or OD = (5/2) × √5