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In the given figure, O is centre of circle, ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following ∠PQR.
Concept: Inscribed Angle Theorem
Choose the correct alternative:
If the points, A, B, C are non-collinear points, then how many circles can be drawn which passes through points A, B, and C?
Concept: Circles Passing Through One, Two, Three Points
If the length of an arc of the sector of a circle is 20 cm and if the radius is 7 cm, find the area of the sector.
Concept: Angle Subtended by the Arc to the Point on the Circle
In the following figure, O is the centre of the circle. ∠ABC is inscribed in arc ABC and ∠ ABC = 65°. Complete the following activity to find the measure of ∠AOC.
∠ABC = `1/2`m ______ (Inscribed angle theorem)
______ × 2 = m(arc AXC)
m(arc AXC) = _______
∠AOC = m(arc AXC) (Definition of measure of an arc)
∠AOC = ______
Concept: Angle Subtended by the Arc to the Centre
In the above figure, the circles with P, Q, and R intersect at points B, C, D, and E as shown. Lines CB and ED intersect in point M. Lines are drawn from point M to touch the circles at points A and F. Prove that MA = MF.
Concept: Circles Passing Through One, Two, Three Points
In figure, chord EF || chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof.
Proof: Draw seg GF.
∠EFG = ∠FGH ......`square` .....(I)
∠EFG = `square` ......[inscribed angle theorem] (II)
∠FGH = `square` ......[inscribed angle theorem] (III)
∴ m(arc EG) = `square` ......[By (I), (II), and (III)]
chord EG ≅ chord FH ........[corresponding chords of congruent arcs]
Concept: Inscribed Angle Theorem
Prove the following theorem:
Angles inscribed in the same arc are congruent.
Concept: Corollaries of Inscribed Angle Theorem
In the above figure, chord PQ and chord RS intersect each other at point T. If ∠STQ = 58° and ∠PSR = 24°, then complete the following activity to verify:
∠STQ = `1/2` [m(arc PR) + m(arc SQ)]
Activity: In ΔPTS,
∠SPQ = ∠STQ – `square` ......[∵ Exterior angle theorem]
∴ ∠SPQ = 34°
∴ m(arc QS) = 2 × `square`° = 68° ....... ∵ `square`
Similarly, m(arc PR) = 2∠PSR = `square`°
∴ `1/2` [m(arc QS) + m(arc PR)] = `1/2` × `square`° = 58° ......(I)
But ∠STQ = 58° .....(II) (given)
∴ `1/2` [m(arc PR) + m(arc QS)] = ∠______ ......[From (I) and (II)]
Concept: Inscribed Angle Theorem
Given: In the figure, point A is in the exterior of the circle with centre P. AB is the tangent segment and secant through A intersects the circle in C and D.
To prove: AB2 = AC × AD
Construction: Draw segments BC and BD.
Write the proof by completing the activity.
Proof: In ΔABC and ΔADB,
∠BAC ≅ ∠DAB .....becuase ______
∠______ ≅ ∠______ ......[Theorem of tangent secant]
∴ ΔABC ∼ ΔADB .......By ______ test
∴ `square/square = square/square` .....[C.S.S.T.]
∴ AB2 = AC × AD
Proved.
Concept: Tangent and Secant Properties
In the figure, the centre of the circle is O and ∠STP = 40°.
- m (arc SP) = ? By which theorem?
- m ∠SOP = ? Give reason.
Concept: Inscribed Angle Theorem
Find the value of y, if the points A(3, 4), B(6, y) and C(7, 8) are collinear.
Concept: Circles Passing Through One, Two, Three Points
In the following figure, a quadrilateral LMNO circumscribes a circle with centre C. ∠O = 90°, LM = 25 cm, LO = 27 cm and MJ = 6 cm. Calculate the radius of the circle.
Concept: Tangent and Secant Properties
A pizza has 8 slices all equally spaced. Suppose pizza is a flat circle of radius 28 cm, find the area covered between 3 slices of pizza.
Concept: Secant and Tangent
In the above figure, ∠L = 35°, find :
- m(arc MN)
- m(arc MLN)
Solution :
- ∠L = `1/2` m(arc MN) ............(By inscribed angle theorem)
∴ `square = 1/2` m(arc MN)
∴ 2 × 35 = m(arc MN)
∴ m(arc MN) = `square` - m(arc MLN) = `square` – m(arc MN) ...........[Definition of measure of arc]
= 360° – 70°
∴ m(arc MLN) = `square`
Concept: Inscribed Angle Theorem
In the above figure, ∠ABC is inscribed in arc ABC.
If ∠ABC = 60°. find m ∠AOC.
Solution:
∠ABC = `1/2` m(arc AXC) ......`square`
60° = `1/2` m(arc AXC)
`square` = m(arc AXC)
But m ∠AOC = \[\boxed{m(arc ....)}\] ......(Property of central angle)
∴ m ∠AOC = `square`
Concept: Angle Subtended by the Arc to the Centre
Construct the circumcircle and incircle of an equilateral triangle ABC with side 6 cm and centre O. Find the ratio of radii of circumcircle and incircle.
Concept: Division of a Line Segment
Draw `angle ABC` of measure 80° and bisect it
Concept: Geometric Constructions
Draw ∠ABC of measures 135°and bisect it.
Concept: Geometric Constructions
∆AMT ~ ∆AHE. In ∆AMT, AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm. `"AM"/"AH" = 7/5`. Construct ∆AHE.
Concept: Division of a Line Segment
∆ABC ~ ∆LBN. In ∆ABC, AB = 5.1 cm, ∠B = 40°, BC = 4.8 cm, \[\frac{AC}{LN} = \frac{4}{7}\]. Construct ∆ABC and ∆LBN.
Concept: Division of a Line Segment
