Practice Set 3.5 Geometry 10th Std Maths Part 2 Answers Chapter 3 Circle
Maharashtra Board Class 10 Maths Chapter 3 Circle Intext Questions and Activities
Question 1
Maharashtra Board Solution
Theorem: If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more point, then the measure of the angle is half the measure of its intercepted arc. (Textbook pg.no. 75 and 76) Given: ∠ABC is any angle, whose vertex B lies on the circle with centre M. Line BC is tangent at B and line BA is secant intersecting the circle at point A. Arc ADB is intercepted by ∠ABC. To prove: ∠ABC = m(arc ADB) Proof: Case I: Centre M lies on arm BA of ∠ABC. ∠MBC = 90° [Trangnet theorem] i.e. ∠ABC 90° (i) [A – M – B] arc ADB is a semicircular arc. ∴ m(arc ADB) = 180° (ii) [Measure ofa semicircle is 180°] ∴ ∠ABC = m(arc ADB) [(From (i) and (ii)] |
Case II: Centre M lies in the exterior of ∠ABC.
Draw radii MA and MB.
∴ ∠MBA = ∠MAB [Isosceles triangle theorem]
Let, ∠MHA = ∠MAB =x, ∠ABC = y In ∆ABM,
∠AMB + ∠MBA + ∠MAB = 180° [Sum of the measures of all the angles of a triangle is 1800]
∴ ∠AMB + x + x = 180°
∴ ∠AMB = 180° – 2x …… (i)
Now, ∠MBC = ∠MBA + ∠ABC [Angle addition property]
∴ 90° = x + y [Tangent theorem]
∴ x = 90° – y ……(ii)
∠AMB = 180° – 2 (90° – y) [From (i) and (ii)]
∴ ∠AMB = 180° – 180° + 2y
∴ 2y = ∠AMB
∴ y = ∠AMB
∴ ∠ABC = ∠AMB
∴ ∠ABC = m(arc ADB) [Definition of measure of minor arc]
Case III: Centre M lies in the interior of ∠ABC.
Ray BE is the opposite ray of ray BC.
Now, ∠ABE = m (arc AFB) (i) [Proved in case II]
∠ABC + ∠ABE = 180° [Angles in a linear pair]
∴ 180 – ∠ABC = ∠ABE
∴ 180 – ∠ABC = m(arc AFB) [From (i)]
= [360 – m (arc ADB)]
∴ 180 – ∠ABC = 180 – m(arc ADB)
∴ -∠ABC = – m(arc ADB)
∴ ∠ABC = m(arc ADB)
Question 2. We have proved the above theorem by drawing seg AC and seg DB. Can the theorem be proved by drawing seg AD and seg CB, instead of seg AC and seg DB? (Textbook pg. no. 77)
Solution & Step-by-Step Answer:
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Solution & Step-by-Step Answer:
Yes, the theorem can be proved by drawing seg AD and seg CB. Given: P is the centre of circle, chords AB and CD intersect internally at point E. To prove: AE × EB = CE × ED Construction: Draw seg AD and seg CB. Proof: In ∆CEB and ∆AED, ∠CEB = ∠DEA [Vertically opposite angles] ∠CBE = ∠ADE [Angles inscribed in the same arc] ∴ ∆CEB ~ ∆AED [by AA test of similarity] ∴ = [Corresponding sides of similar triangles] ∴ AE × EB = CE × ED
Question 3
Maharashtra Board Solution
In figure, seg PQ is a diameter of a circle with centre O. R is any point on the circle, seg RS ⊥ seg PQ. Prove that, SR is the geometric mean of PS and SQ. [That is, SR2 = PS × SQ] (Textbook pg. no. 81) Given: seg PQ is the diameter. seg RS ⊥ seg PQ To prove: SR2 = PS × SQ Construction: Extend ray RS, let it intersect the circle at point T. Proof: seg PQ ⊥ seg RS [Given] ∴ seg OS ⊥ chord RT [R – S – T, P – S – O] ∴ segSR = segTS (i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord] Chords PQ and RT intersect internally at point S. ∴ SR × TS = PS × SQ [Theorem of internal division of chords] ∴ SR × SR = PS × SQ [From (i)] ∴ SR2 = PS × SQ
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Question 4. Theorem: If secants containing chords AB and CD of a circle intersect outside the circle in point E, then AE × EB = CE × ED. (Textbook pg. no. 78) Given: Chords AB and CD of a circle intersect outside the circle in point E. To prove: AE × EB = CE × ED Construction: Draw seg AD and seg BC. Proof: In ∆ADE and ∆CBE, ∠AED = ∠CEB [Common angle] ∠DAE ≅ ∠BCE [Angles inscribed in the same arc] ∴ ∆ADE ~ ∆CBE [AA testof similaritv] ∴ = [Corresponding sides of similar triangles] ∴ AE × EB = CE × ED |
Question 5. Theorem: Point E is in the exterior of a circle. A secant through E intersects the circle at points A and B, and a tangent through E touches the circle at point T, then EA × EB = ET2. Given: Secant through point E intersects the circle in points A and B. Tangent drawn through point E touches the circle in point T. To prove: EA × EB = ET2 Construction: Draw seg TA and seg TB. Proof: In ∆EAT and ∆ETB, ∠AET ≅ ∠TEB [Common angle] ∠ETA ≅ ∠EBT [Theorem of angle between tangent and secant, E – A – B] ∴ ∆EAT ~ ∆ETB [AA test of similarity] ∴ = [Corresponding sides of similar triangles] ∴ EA × EB = ET2 |
Question 6. In the figure in the above example, if seg PR and seg RQ are drawn, what is the nature of ∆PRQ. (Textbook pg. no, 81)
Solution & Step-by-Step Answer:
seg PQ is the diameter of the circle. ∴ ∠PRQ = 90° ∴ ∆PRQ is a right angled triangle. [Angle inscribed in a semicircle]
Question 7
Maharashtra Board Solution
Have you previously proved the property proved in the above example? (Textbook pg. no. 81)
Solution & Step-by-Step Answer:
Yes. It is the theorem of geometric mean. ∆PSR ~ ∆RSQ [Similarity of right angled triangles] ∴ = [Corresponding sides of similar triangles] ∴ SR2 = PS × SQ
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