Please select a subject first
Advertisements
Advertisements
In an equilateral triangle PQR, prove that PS2 = 3(QS)2.
Concept: Right-angled Triangles and Pythagoras Property
A person starts his trip from home. He moves 24 km in south direction and then starts moving towards east. He travels 7 km in that direction and finally reaches his destination. How far is the destination from his home?
Concept: Pythagoras Theorem
In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm, then find RS.
Concept: Property of 30°- 60°- 90° Triangle Theorem
In the following figure, m(arc PMQ) = 130o, find ∠PQS.
Concept: Angle Subtended by the Arc to the Point on the Circle
In the following figure, secants containing chords RS and PQ of a circle intersects each other in point A in the exterior of a circle if m(arc PCR) = 26°, m(arc QDS) = 48°, then find:
(i) m∠PQR
(ii) m∠SPQ
(iii) m∠RAQ
Concept: Angle Subtended by the Arc to the Point on the Circle
In the given figure, altitudes YZ and XT of ∆WXY intersect at P. Prove that,
- `square`WZPT is cyclic.
- Points X, Z, T, Y are concyclic.
Concept: Angle Subtended by the Arc to the Centre
In the given figure, chord MN and chord RS intersect at point D.
(1) If RD = 15, DS = 4, MD = 8 find DN
(2) If RS = 18, MD = 9, DN = 8 find DS
Concept: Intersecting Chords and Tangents
In the figure Q is the contact point. If
PQ = 12, PR = 8, then PS = ?
Concept: Tangent and Secant Properties
In the adjoining figure, point O is the centre of the cirlcle, seg OM ⊥ chord AB. If OM = 8cm, AB = 12 cm, then find OB.
Concept: Angle Subtended by the Arc to the Point on the Circle
In the adjoining figure chord EF || chord GH.
Prove that chord EG ≅ chord FH.
Fill in the boxes and write the complete proof.
Concept: Inscribed Angle Theorem
In the given figure, O is the centre of the circle, ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following m(arc QXR).
Concept: Inscribed Angle Theorem
In the given figure, O is centre of circle. ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following ∠QOR.
Concept: Inscribed Angle Theorem
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
