CBSE Class 10 Answered
Please solve this question
Asked by vikasg13.hardware | 28 Jun, 2018, 06:11: AM
Expert Answer
QR is the diameter of the semi-circular arc through R-O-Q
So if M is the midpoint of QR, OM = MQ = MR = r [Radii of the same circle] ---- (1)
But given PS is tangent to this circle at O;
Hence OM perpendicular to PS. [Radius and tangent perpendicular at point of contact]
==> OM perpendicular to QR [Since QR is parallel to PS] ---- (2)
So from (1) & (2), angle OMQ is a right triangle and applying Pythagoras theorem,
r² + r² = OQ² = 4 [Since OQ is the radius of the bigger circle of diameter 4 units]
==> r = √2 units. ---- (3)
Also QOR = 90 deg [Angle in the semi circle of R-O-Q] ----- (4)
iii) Area of the semi circle ROQ = (1/2)(r²) = (1/2)(2π) [Substituting r = √2 from (3)]
==> Area of the semi circle ROQ = π sq. units ----- (5)
iv) QXR is the minor segment of the circle whose radius is 2 units and the chord is QR.
Angle subtended by the chord QR at the center O = angle QOR = 90 deg
[Angle in the semi circle of the arc QOR]
Area of minor segment = (1/2)(r²){Ө - sin(Ө)}, where Ө is the angle at the center in radians and r is the radius of the circle]
Here Ө = 90 deg = π/2; so sin(Ө) = sin(π/2) = 1
r = radius of bigger circle = 2 units
Thus area of the segment QXR = (1/2)(4)(π/2 - 1) = π - 2 sq. units ----- (6)
Adding (5) & (6), Area of the region QXROQ = (2π - 2) sq. units = 2(π - 1) sq. units
So if M is the midpoint of QR, OM = MQ = MR = r [Radii of the same circle] ---- (1)
But given PS is tangent to this circle at O;
Hence OM perpendicular to PS. [Radius and tangent perpendicular at point of contact]
==> OM perpendicular to QR [Since QR is parallel to PS] ---- (2)
So from (1) & (2), angle OMQ is a right triangle and applying Pythagoras theorem,
r² + r² = OQ² = 4 [Since OQ is the radius of the bigger circle of diameter 4 units]
==> r = √2 units. ---- (3)
Also QOR = 90 deg [Angle in the semi circle of R-O-Q] ----- (4)
iii) Area of the semi circle ROQ = (1/2)(r²) = (1/2)(2π) [Substituting r = √2 from (3)]
==> Area of the semi circle ROQ = π sq. units ----- (5)
iv) QXR is the minor segment of the circle whose radius is 2 units and the chord is QR.
Angle subtended by the chord QR at the center O = angle QOR = 90 deg
[Angle in the semi circle of the arc QOR]
Area of minor segment = (1/2)(r²){Ө - sin(Ө)}, where Ө is the angle at the center in radians and r is the radius of the circle]
Here Ө = 90 deg = π/2; so sin(Ө) = sin(π/2) = 1
r = radius of bigger circle = 2 units
Thus area of the segment QXR = (1/2)(4)(π/2 - 1) = π - 2 sq. units ----- (6)
Adding (5) & (6), Area of the region QXROQ = (2π - 2) sq. units = 2(π - 1) sq. units
Answered by Arun | 09 Jul, 2018, 01:52: PM
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