Given : In circle C (O,r), OA = AB. To find: ∠ADB and ∠ACB. Solution: In ΔOAB OA = AB [∵ Given] OA = OB [∵ Radii of circle] Hence, OA = OB = AB ⇒ ABC is an equilateral triangle. Therefore, ∠AOB = 60° [∵ Each angle of an equilateral triangle is 60°] ∠AOB = 2 ∠ADB [∵ The angle subtended by an arc at tRead more
Given : In circle C (O,r), OA = AB.
To find: ∠ADB and ∠ACB.
Solution: In ΔOAB
OA = AB [∵ Given]
OA = OB [∵ Radii of circle]
Hence, OA = OB = AB
⇒ ABC is an equilateral triangle.
Therefore, ∠AOB = 60° [∵ Each angle of an equilateral triangle is 60°]
∠AOB = 2 ∠ADB
[∵ The angle subtended by an arc at the centre is double the angle subtended by it at
any]
⇒ ∠ADB = 1/2 ∠AOB ⇒ ∠ADB = (1/2) × 60° = 30
ACBD is a cyclic quadrilateral.
Therefore, ∠ACB + ∠ADB = 180°
[∵ the sum of either pair of opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠ACB + 30° = 180° [∵ ∠ADB = 30°]
⇒ ∠ACB = 180° – 30° = 150°
Given : Circle C (P, 3) and circle C (Q, 5) are intersecting at points A and B. Construction: Join PA and QA. Draw PM as bisector of chord AB. Proof: AB is chord of circle C (P, 3) and PM is bisector of chord AB. Therefore, PM ⊥AB [∵ The line drawn through the center of a circle to bisect a chord isRead more
Given : Circle C (P, 3) and circle C (Q, 5) are intersecting at points A and B.
Construction: Join PA and QA. Draw PM as bisector of chord AB.
Proof: AB is chord of circle C (P, 3) and PM is bisector of chord AB.
Therefore, PM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the
chord.]
Hence, ∠PMA = 90°
Let, PM = X, therefore, QM = 4 – x
In ΔAPM, using Pythagoras theorem
AM² = AP² – PM² …(1)
And in ΔAPM, using Pythagoras theorem
AM² = AQ² – QM² …(2)
From the equation (1) and (2), we get
AP² – PM² = AQ² – QM²
⇒ 3² – x² = 5² -(4 – x)² ⇒ 9 – x² = 25 – (16 + x² – 8x)
⇒ 9 – 9 = 8x ⇒ x = 0/8 = 0
from the equation (1), AM² = 3² – 0² = 9 ⇒ AM = 3
⇒ AB = 2AM = 6
Given : In circle C (O, r) equal chords AB and CD intersects at P. To Prove: AP = CP and BP = DP. Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD. Proof: In ΔOPM and ΔONP, ∠OMP = ∠ONP [∵ Each 90] AP = AP [∵ Common] OM = ON [∵ Equal chords of a circle are equidistant from the centre] Hence, ΔOMP ≅ ΔONRead more
Given : In circle C (O, r) equal chords AB and CD intersects at P.
To Prove: AP = CP and BP = DP.
Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD.
Proof: In ΔOPM and ΔONP,
∠OMP = ∠ONP [∵ Each 90]
AP = AP [∵ Common]
OM = ON [∵ Equal chords of a circle are equidistant from the centre]
Hence, ΔOMP ≅ ΔONP [∵ RHS Congruency rule]
PM = PN …(1) [∵ CPCT]
And AB = CD …(2) [∵ Given]
⇒ (1/2)AB = (1/2)CD
⇒ AM = CN …(3)
Adding the equations (1) and (3), we have
AM + PM = CN + PM
⇒ AP = CP …(4)
Subtracting equation (4) from (2), we have
AB – AP = CD – CP
⇒ PB = PD
Given : In circle C (O, r) equal chords AB and CD intersects at point P. To Prove: ∠OPM = ∠OPN Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD. Proof: In ΔOMP and ΔONP, ∠OMP = ∠ONP [∵ Each 90°] AP = AP [∵ Common] OM = ON [∵ Equal chords of a circle are equidistant from the centre] Hence, ΔOMP ≅ ΔONPRead more
Given : In circle C (O, r) equal chords AB and CD intersects at point P.
To Prove: ∠OPM = ∠OPN
Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD.
Proof: In ΔOMP and ΔONP,
∠OMP = ∠ONP [∵ Each 90°]
AP = AP [∵ Common]
OM = ON [∵ Equal chords of a circle are equidistant from the centre]
Hence, ΔOMP ≅ ΔONP [∵ RHS Congruency rule]
∠OPM = ∠OPM [∵ CPCT]
Given : In figure, points R, S and M are showing the position of reshma, salma and mandeep respectively. Therefore RS = SM = 6 cm Construction: Join OR, OS, RS, RM, and OM. Draw OL ⊥RS. Proof: In ΔORS, OS = OR and OL ⊥RS [∵ By construction] Therefore, RL = LS = 3 cm [∵ RS = 6 cm] In ΔOLS, using PythRead more
Given : In figure, points R, S and M are showing the position of reshma, salma and mandeep respectively.
Therefore RS = SM = 6 cm
Construction: Join OR, OS, RS, RM, and OM. Draw OL ⊥RS.
Proof: In ΔORS,
OS = OR and OL ⊥RS [∵ By construction]
Therefore, RL = LS = 3 cm [∵ RS = 6 cm]
In ΔOLS, using Pythagoras theorem, OL² = OS² – SL²
⇒ OL² = 5² – 3² = 25 – 9 = 16
⇒ OL = 4
In ΔORK and ΔOMK,
OR = OM [∵ Radii of circle]
∠ROK = ∠MOK [∵ Equal chords subtend equal angle at the centre]
Ok = OK [∵ Common]
Hence, ∠ORK ∠OMK [∵ SAS congruency rule]
RK = MK [∵ CPCT]
Hence, OK ⊥RM
[∵ The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord]
Now, the area of ΔORS = (1/2)× RS × OL …(1)
And the area of ΔORS = (1/2) × OS × KR …(2)
From the equation (1) and (2),
1/2 × RS × OL = 1/2 × OS × KR
⇒ RS × OL = OS × KR ⇒ 6 × 4 = 5 × KR ⇒ KR = (6×4/5) = 4.8
Hence, RM = 2 × KR = 2 × 4.8 = 9.6 cm
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Given : In circle C (O,r), OA = AB. To find: ∠ADB and ∠ACB. Solution: In ΔOAB OA = AB [∵ Given] OA = OB [∵ Radii of circle] Hence, OA = OB = AB ⇒ ABC is an equilateral triangle. Therefore, ∠AOB = 60° [∵ Each angle of an equilateral triangle is 60°] ∠AOB = 2 ∠ADB [∵ The angle subtended by an arc at tRead more
Given : In circle C (O,r), OA = AB.
To find: ∠ADB and ∠ACB.
Solution: In ΔOAB
OA = AB [∵ Given]
OA = OB [∵ Radii of circle]
Hence, OA = OB = AB
⇒ ABC is an equilateral triangle.
Therefore, ∠AOB = 60° [∵ Each angle of an equilateral triangle is 60°]
∠AOB = 2 ∠ADB
[∵ The angle subtended by an arc at the centre is double the angle subtended by it at
any]
See less⇒ ∠ADB = 1/2 ∠AOB ⇒ ∠ADB = (1/2) × 60° = 30
ACBD is a cyclic quadrilateral.
Therefore, ∠ACB + ∠ADB = 180°
[∵ the sum of either pair of opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠ACB + 30° = 180° [∵ ∠ADB = 30°]
⇒ ∠ACB = 180° – 30° = 150°
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Given : Circle C (P, 3) and circle C (Q, 5) are intersecting at points A and B. Construction: Join PA and QA. Draw PM as bisector of chord AB. Proof: AB is chord of circle C (P, 3) and PM is bisector of chord AB. Therefore, PM ⊥AB [∵ The line drawn through the center of a circle to bisect a chord isRead more
Given : Circle C (P, 3) and circle C (Q, 5) are intersecting at points A and B.
Construction: Join PA and QA. Draw PM as bisector of chord AB.
Proof: AB is chord of circle C (P, 3) and PM is bisector of chord AB.
Therefore, PM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the
chord.]
See lessHence, ∠PMA = 90°
Let, PM = X, therefore, QM = 4 – x
In ΔAPM, using Pythagoras theorem
AM² = AP² – PM² …(1)
And in ΔAPM, using Pythagoras theorem
AM² = AQ² – QM² …(2)
From the equation (1) and (2), we get
AP² – PM² = AQ² – QM²
⇒ 3² – x² = 5² -(4 – x)² ⇒ 9 – x² = 25 – (16 + x² – 8x)
⇒ 9 – 9 = 8x ⇒ x = 0/8 = 0
from the equation (1), AM² = 3² – 0² = 9 ⇒ AM = 3
⇒ AB = 2AM = 6
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Given : In circle C (O, r) equal chords AB and CD intersects at P. To Prove: AP = CP and BP = DP. Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD. Proof: In ΔOPM and ΔONP, ∠OMP = ∠ONP [∵ Each 90] AP = AP [∵ Common] OM = ON [∵ Equal chords of a circle are equidistant from the centre] Hence, ΔOMP ≅ ΔONRead more
Given : In circle C (O, r) equal chords AB and CD intersects at P.
See lessTo Prove: AP = CP and BP = DP.
Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD.
Proof: In ΔOPM and ΔONP,
∠OMP = ∠ONP [∵ Each 90]
AP = AP [∵ Common]
OM = ON [∵ Equal chords of a circle are equidistant from the centre]
Hence, ΔOMP ≅ ΔONP [∵ RHS Congruency rule]
PM = PN …(1) [∵ CPCT]
And AB = CD …(2) [∵ Given]
⇒ (1/2)AB = (1/2)CD
⇒ AM = CN …(3)
Adding the equations (1) and (3), we have
AM + PM = CN + PM
⇒ AP = CP …(4)
Subtracting equation (4) from (2), we have
AB – AP = CD – CP
⇒ PB = PD
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Given : In circle C (O, r) equal chords AB and CD intersects at point P. To Prove: ∠OPM = ∠OPN Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD. Proof: In ΔOMP and ΔONP, ∠OMP = ∠ONP [∵ Each 90°] AP = AP [∵ Common] OM = ON [∵ Equal chords of a circle are equidistant from the centre] Hence, ΔOMP ≅ ΔONPRead more
Given : In circle C (O, r) equal chords AB and CD intersects at point P.
See lessTo Prove: ∠OPM = ∠OPN
Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD.
Proof: In ΔOMP and ΔONP,
∠OMP = ∠ONP [∵ Each 90°]
AP = AP [∵ Common]
OM = ON [∵ Equal chords of a circle are equidistant from the centre]
Hence, ΔOMP ≅ ΔONP [∵ RHS Congruency rule]
∠OPM = ∠OPM [∵ CPCT]
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Given : In figure, points R, S and M are showing the position of reshma, salma and mandeep respectively. Therefore RS = SM = 6 cm Construction: Join OR, OS, RS, RM, and OM. Draw OL ⊥RS. Proof: In ΔORS, OS = OR and OL ⊥RS [∵ By construction] Therefore, RL = LS = 3 cm [∵ RS = 6 cm] In ΔOLS, using PythRead more
Given : In figure, points R, S and M are showing the position of reshma, salma and mandeep respectively.
See lessTherefore RS = SM = 6 cm
Construction: Join OR, OS, RS, RM, and OM. Draw OL ⊥RS.
Proof: In ΔORS,
OS = OR and OL ⊥RS [∵ By construction]
Therefore, RL = LS = 3 cm [∵ RS = 6 cm]
In ΔOLS, using Pythagoras theorem, OL² = OS² – SL²
⇒ OL² = 5² – 3² = 25 – 9 = 16
⇒ OL = 4
In ΔORK and ΔOMK,
OR = OM [∵ Radii of circle]
∠ROK = ∠MOK [∵ Equal chords subtend equal angle at the centre]
Ok = OK [∵ Common]
Hence, ∠ORK ∠OMK [∵ SAS congruency rule]
RK = MK [∵ CPCT]
Hence, OK ⊥RM
[∵ The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord]
Now, the area of ΔORS = (1/2)× RS × OL …(1)
And the area of ΔORS = (1/2) × OS × KR …(2)
From the equation (1) and (2),
1/2 × RS × OL = 1/2 × OS × KR
⇒ RS × OL = OS × KR ⇒ 6 × 4 = 5 × KR ⇒ KR = (6×4/5) = 4.8
Hence, RM = 2 × KR = 2 × 4.8 = 9.6 cm