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Chapter: Mensuration· Mathematics Part 2 (Geometry)

In figure 7.49, two circles with centres O and P touch internally at A. If BQ = 9 and DE = 5, complete the activity to find the radii of the circles.

In the figure 7.49, two circles with centres O and P are touching internally at point A. If BQ = 9, DE = 5, complete the following activity to find the radii of the circles.

[Figure: Bigger circle with centre O and smaller circle with centre P touch internally at A. Diameter BA of the bigger circle passes through Q, O and P in the order B – Q – O – P – A, where QA is a diameter of the smaller circle. Chord DC of the bigger circle is perpendicular to BA at O and cuts the smaller circle at E (above) and F (below). BQ = 9, DE = 5.]

Solution: Let the radius of the bigger circle be R and that of smaller circle be r.
OA, OB, OC and OD are the radii of the bigger circle \(\therefore\) OA = OB = OC = OD = R
PQ = PA = r
OQ = OB − BQ = [ ]
OE = OD − DE = [ ]
As the chords QA and EF of the circle with centre P intersect in the interior of the circle, so by the property of internal division of two chords of a circle,
OQ × OA = OE × OF
[ ] × R = [ ] × [ ] ... (\(\because\) OE = OF)
R2 − 9R = R2 − 10R + 25
R = [ ]
AQ = 2r = AB − BQ
2r = 50 − 9 = 41
r = [ ] = [ ]

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Let the radius of the bigger circle be R and that of the smaller circle be r.

OA = OB = OC = OD = R and PQ = PA = r.

OQ = OB − BQ = R − 9

OE = OD − DE = R − 5

Chords QA and EF of the circle with centre P intersect at O, so by the property of internal division of two chords,

OQ × OA = OE × OF

\((R - 9) \times R = (R - 5) \times (R - 5)\) ... (\(\because\) OE = OF, as the perpendicular OP from the centre P bisects chord EF)

\[R^2 - 9R = R^2 - 10R + 25\]\[-9R + 10R = 25 \quad \therefore R = 25\]

AB = 2R = 50

AQ = 2r = AB − BQ = 50 − 9 = 41

\[r = \frac{41}{2} = 20.5\]

Radius of the bigger circle = 25 units, radius of the smaller circle = 20.5 units.

Boxes in order: R − 9; R − 5; (R − 9), (R − 5), (R − 5); 25; \(\dfrac{41}{2}\), 20.5.

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