Maharashtra Board Class 10 Math Part – 2 Solution Chapter 3 Practice Set 3.3 – 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 |
3.3 |
Practice Set – 3.3
(1) In figure 3.37, points G, D, E, F are concyclic points of a circle with centre C.
∠ECF = 70°, m (arc DGF) = 200° find m (arc DE) and m (arc DEF)
Solution:
In the adjacent fig it is given that’s DE F, G are concylic points C is centre.
∠ECF = 70°, m (arc DGF) = 200°
∴ m (arc EF) = 70° [∵ ∠ECF = 70°]
∴ m (arc DE) = 360° – m (arc ED) + m (arc DGF)
= 360° – 70° + 200
= 360° – 270° = 90°
∴ m (arc DEF) = m (arc DE) + m (arc EF)
= 90° + 70°
= 160°
(2) In fig 3.38 ∆QRS is an equilateral triangle. Prove that,
(1) arc RS ≅ are QS ≅ arc QR
(2) m (arc QRS) = 240°
Solution:
In the adjacent fig it is given that, ∆RQS is an equilateral triangle.
∴ RQ = QS = RS [Sides of equilateral triangle are equal]
(1) ∴ m (arc RQ) = m (arc RS) = m (arc QS)
Or, m (arc RQ) ≅ m (arc RS) ≅ m (arc QS) [∵ Corresponding areas of equal chords are equal] (Proved)
(2) Draw RA, QC & SB Medians of ∆RQS that intersect at centroid o.
Now, We know medians of equilateral triangle bisects the vertex angles.
∴ In ∆ROQ,
∠R = 60° = ∠Q = 60° [Angles of equilateral triangle]
∠DRQ = ∠OQR = ∠R/2 = 60/2 = 30°
∴ ∠ROQ = 180° – (∠ORQ + ∠OQR) [Sum of interior angles of triangle is 180°]
= 180° (30+30°)
= 180° – 60 = 120°
Similarly
In ∆QOS
∠QOS = 180° – 60°
= 120°
Similarly,
In ∆ROS
∠ROS = 180° – 60°
= 120°
∴ m (arc QR) = 120° [∵ ∠ROQ = 120°]
m (arc RS) = 120° [∵ ∠ROS = 120°]
∴m (arc QRS) = m (arc QR) + m (arc RS)
= 120° + 120°
Or, m (arc QRS) = 240° (Proved)
(3) In fig 3.39 chord AB ≅ Chord CD,
Prove that,
arc AC ≅ arc BD
Solution:
Given chord AB ≅ chord CD.
∴ m (arc AB) ≅ m (arc CD)
m (arc AC) = m (arc AB) – m (arc BC) —– (i)
Also, m = m (arc BD) = m (arc CD) – m (arc BC)
= m (arc AB) – m (arc ac) [∵ —- (ii) m (arc bc) = m (arc BC)]
Comparing equation (i) & (ii) we get.
m (arc AC) = m (arc BD)
Or, arc AC ≅ arc BD (Proved)
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 3 Circle Practice Set 3.3
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