Maharashtra Board Class 10 Math Part – 2 Solution Chapter 3 Practice Set 3.2 – Circle
Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 3: Circle. Marathi or English Medium Students of Class 10 get here Circle full Exercise Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 2 Solution |
Chapter |
Circle |
Practice Set |
3.2 |
Practice set 3.2
(1) Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.
Solution:
Adjacent circles touch each other internally at point c.
∴ By theorem of touching circles,
A – B – C
∴ Radius of small circle = BC
Radius of bigger circle = AC.
[∵ C is point tangent l which touches two circle c]
∴ Distance between their centers.
= AC – BC
= 4.8 – 3.5 = 1.3 cm
(2) Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.
Solution:
The adjacent circles touch each other at point c on tangent l.
A, B are centers.
∴ by theorem of touching circles.
A – B – C
∴ Radius of circle A = AC
Radius of circle B = BC [∵ C is a point on tangent l where the tangent touches the circles]
∴ Distance between their center
= AC + BC
= 5.5 + 4.2
= 9.7 cm
(3) If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other –
(i) Externally
(ii) Internally.
Solution:
Given, radii of bigger circle = 4cm
radii smaller circle = 2.8 cm
(i) Touching each other externally.
(ii) Touching each other internally
The above figures shows two circle with centers A and B touching each other externally and in internally in tangents n and in respectively at point c.
(4) In fig 3.27, the circles with centres P and Q touch each other at R. A line passing through R meets the circles at A and B respectively. Prove that –
(1) seg AP||seg BQ,
(2) ∆APR ~∆RQB, and
(3) Find ∠RQB if ∠PAR = 35°
Solution:
In the adjacent fig,
P, Q are centers of two circle.
A, B are any point of the perimeter of the circle.
∴ AP and BQ are radius of circles.
The two circles to each other at point R.
∴ PR and QR are radius of two circles. [∵ R is the point of contact of two circles]
(1) ∴ In ∆BQR & ∆APR
∠QBR = ∠QRB [Opposite angles of equal sides BQ and QR which is the radius of bigger circle]
Also, ∠PAR = ∠ARP = [—–]
Also, ∠ARP = ∠QRB [Vertically opposite angles]
∴ ∠QBR = ∠PAR = ∠APR = ∠QRB
Now, since ∠QBR = ∠PAR
∴ They are alternate angles of line AB and radius PA & QR.
∴ We can say that seg AP || seg BQ.
(2) In ∆APR & ∆RQB,
∠ARP = ∠QRB [Vertically opposite angles]
∠QBR = ∠PAR [Alternate angles since AP||BQ]
∴ ∆APR ~ ∆RQB by AA test.
(3) ∠PAR = 35° given,
∴ ∠QBR = ∠PAR = 35°
∠QRB = ∠ARP = ∠PAR = 35°
∴ In ∆BQR,
∠BQR + ∠QRB + ∠QBR = 180° [Sum of angles of triangle is 180°]
Or, ∠BQR + 35° + 35° = 180°
Or, ∠BQR = 180° – 70°
Or, ∠BQR = 110°
(5) In fig 3.28 the circles with centres A and B touch each other at E. Line l is a common tangent which touches the circles at C and D respectively. Find the length of seg CD if the radii Fig. 3.28 of the circles are 4 cm, 6 cm.
Solution:
In the adjacent fig. A, B are centre of circles with radius of bigger circle. 6cm and smaller circle 4cm.
C, D are points on tangent of both circle l where l meets both circle.
∴ ∠ACD = ∠BDC = 90° [Tangent theorem]
AC = AE = 4cm [Radius]
BE = BD = 6cm [Radius]
∴ AB = AE + BE = 4 + 6 = 10a
Now, draw a perpendicular line from A to radius BD. Which meets BD at F.
∴ ∠AFD = 90°
∴ In quadrilateral AFCD
∠AFD = 90, ∠FDC = 90, ∠ACD = 90°
∴ ∠CAD = 360° – (90 + 90 + 90) [Sum of interior angles of a quadrilateral is 360°]
= 90°
∴ Quad AFCD is a rectangle since all integer angles are 90°.
∴ AC = FD = 4cm
∴ BF = BD – FD
= 6 – 4 = 2cm
∴ In ∆AFB,
By Pythagoras theorem,
AB2 = AF2 + BF2
Or, 102 = AF2 + 22
Or, AF = √100-4
Or, AF = 4√6
∴ AF = CD = 4√6 cm
[∵ AF and CD are opposite sides of rectangle AFCD]
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 3 Circle Practice Set 3.2
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