А,
C
12. Two chords AB and CD of lengths 5 cm and 11 cm
respectively of a circle are parallel to each other and are on
the same side of its centre. If the distance between AB and
CD is 3 cm, find the radius of the circle.
Answers
The perpendicular from the centre of a circle to a chord bisects the chord.
Radius is the line segment joining the centre and any point on the circle is called the radius of the circle. All radius have a same length in a circle.
=========================================================
Let there is a circle having center O and let radius is b .
Draw ON perpendicular to AB and OM perpendicular to CD.
Now since ON perpendicular to AB and OM perpendicular to CD and AB || CD
So N, O,M are collinear.
Given distance between AB and CD is 6.
So MN = 6
Let ON = a, then OM= (6-a)
Join OA and OC.
Then OA = OC = b
Since we know that perpendicular from the centre to a chord of the circle bisects the chord.
and CM = MD = 11/2 = 5.5
AN= NB=5/2= 2.5
From ΔONA and ΔOMC
OA² =ON² +AN²
b²=a² + (2.5)².........(i)
and OC² = OM²+CM²
b²= (6-x)² + (5.5)²......(ii)
from eq i and ii we get
a²+ (2.5)²= (6-a)² + (5.5)²
a²+ 6.25 = 36 +a² - 12a + 30.25
6.25 = -12a+ 66.25
12a = 66.25 - 6.25
12a = 60
a= 60/12
a= 5
Put a = 5 in eq i,
b²= 5²+ (2.5)²
b²= 25 + 6.25
b² = 31.25
b= √31.25
b= 5.6 (approx)
RADIUS=b= 5.6cm (approx)
Hence, radius of the circle is 5.6 cm (approx).
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Answer
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Answer:
Given: AB = 11cm
CD = 5cm
AB ║ CD
Draw OP ⊥ AB and OQ ⊥ CD
PQ = 3cm
To Find: Radius (r) of the circle.
Sol: In triangle AOP
AP = 1/2 AB (The perpendicular from the centre of the circle bisects the chord.)
AP = 1/2 x11 = 5.5cm
Let OP = x cm
OA² = AP² + x²
r² = (5.5)² + x²
r² - (5.5)² = x²
In ΔCOQ,
CQ = 1/2 CD (theorem)
CQ = 1/2 x 5 = 2.5
OC² = CQ² + OQ²
r² = (2.5)² + (x+3)²
Now,
(5.5)² + x² = (2.5)² + (x+3)²
(5.5)² - (2.5)² = (x+3)² (x)²
8x3 = (2x+3) (3)
∴2x + 3 = 8
x = 5/2 = 2.5 cm
Substituting x = 2.5
r² = (5.5)² + (2.5)²
r² = 36.5
r = √36.5 = 6.04 cm
*Hope it helps*
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