Maharashtra Board Class 10 Math Part – 2 Solution Chapter 3 Practice Set 3.1 – 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 |
3.1 |
Practice Set – 3.1
(1) In the adjoining figure the radius of a circle with centre C is 6 cm, line AB is a tangent at A. Answer the following questions.
(1) What is the measure of ∠CAB? Why ?
(2) What is the distance of point C from line AB ? Why?
(3) d (A, B) = 6 cm, find d (B, C).
(4) What is the measure of ∠ABC? Why?
Solution:
In the adjacent fig it is given,
AB is tangent meeting at point A with C is the center of the circle.
Radius circle = 6 cm.
AB = 6m
(1) ∠CAB = 90° [∵ AC is a radius and angle made by a tangent to is center is 90° by tangent theorem]
(2) Distance of point C that is center AB tangent is 6cm. Given in question because CA is a radius meeting at point A which is a tangent of the circle.
(3) In ∆CAB
By Pythagoras theorem,
CA2 + AB2 = BC2
Or, 62 + 62 = BC2 [Given]
Or, BC = √36+36
= √72 = 6√2 cm
(4) ∵ AB = CA = 6 cm
∴ ∠ACB = ∠ABC [Opposite angles of equal sides of ∆are equal]
∴ ∠ABC = 180° – ∠CAB/2
= 180 – 40/2 = 45°
(2) In the adjoining figure, O is the centre of the circle. From point R, se RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10cm and radius of the circle = 5cm, them
(1) What is the length of each tangent segment?
(2) What is the measure of ∠MRO?
(3) What is the measure of ∠MRN?
Solution:
In the adjacent fig it is given that, O is the centre of the circle. RM & RN are two tangent of circle.
OR = 10cm, radius of circle = 5cm
Connect OM and ON,
Now, OM and ON are radius of the circle.
Since M and N are tangents of the circle.
∴ OM = ON = 5cm [Given], ∠OMR = ∠ONR = 90° [By tangent theorem].
(1) ∴ In∆ RON, By Pythagoras theorem,
OR2 + ON2 = RN2
Or, 102 + 52 = RN2
Or, RN = √100 + 25
= √125
= 5√5 cm
Also, ∆ROM
∠OMR = 90°
By Pythagoras theorem,
OM2 + OR2 = RM2
Or, RM2 = 52 + 102
Or, RM = √25+100
= √125
= 5√5 cm.
Now, In∆ROM
OR is the hypotenuse,
OM = 5 = 10/2 = OR/2
(2)∵ OM is half of hypotenuse or, ∴ Opposite angle to OM is 30 [30° – 60° – 90° theorem]
∴∠MRO = 30°
Similarity, ∠NRO = 30°
(3)∴ ∠MRN = ∠MRO + ∠NRO
= 30° + 30°
= 60°
(3) Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠MRN as well as ∠MON.
Solution:
In the adjacent fig. it is given that, O is the center of the circle. RM & RN are two tangents meeting the circle at point M & N respectively.
OM & ON are radius of circle
Since, they are tangents of circle.
∴ OM = ON = r let.
∠OMR = ∠ONR = 90° [tangent theorem]
In ∆OMR,
∠OMR = 90°
By Pythagoras theorem,
OM2 + RN2 = OR2
Or, r2 + RM2 = OR2
Or, RM = √OR2 – r2 —– (i)
In ∆ONR,
∠ONR = 90°
By Pythagoras theorem,
ON2 + RN2 = OR2
Or, r2 + RN2 = OR2
Or, RN = √OR2 – r2 —– (ii)
Comparing equation (i) & (ii) we get,
RM = RN
∴ In ∆OMR & ∆ONR
OM = ON = r
RM = RN
∠OMR ≅ ∠ONR = 90°
∴ ∆OMR ~ ∆ONR by SAS test.
∴ ∠MRO = ∠NRO [Corresponding angles of similar triangle are congruent]
∴ We can say that,
∠MRN is bisects by OR.
Also, ∠MOR = ∠NOR [Corresponding angles of similar triangle are congruent]
∴ We can say that,
∠NON is bisected by seg OR.
(4) What is the distance between two parallel tangents of a circle having radius 4.5 cm? Justify your answer.
Solution:
In the adjacent fig line a and b are two tangents of circle meeting at point A & B respectively.
O is the center of the circle.
C is a line drawn parallel to a going through center O.
Also line a || line b
line a||line c
∴ Line b||line c
∠OAP = ∠OBQ = 90° [tangent theorem]
∠ROB = ∠ROA = 90° [Interior angle property]
∴ ∠ROB + ∠ROA = 90° + 90°= 180°
∴ Points A – O – B are co-linear points.
∴ AB is a straight line connecting two tangent and the center O.
∴ AB is the diameter of circle.
Also OA = OB = radius [∵ A & B are point of tangent where it meets the circle]
∴ OA = OB = 4.5 cm [Given]
∴ AB = OA + OD = [∵ A – O – B]
Or, AB = 4.5 + 4.5
= 9 cm
∴ Distance between two parallel tangents line a and line b is 9cm.
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 3 Circle Practice Set 3.1
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