Thursday, July 29, 2010

Exercise 10.6


Question 1:
Prove that line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Ans.
Let two circles having their centres as O and  intersect each other at point A and B respectively. Let us join O.
In ΔAO and BO,
OA = OB (Radius of circle 1)
A =  B (Radius of circle 2)
O = O (Common)
ΔAO ≅ ΔBO (By SSS congruence rule)
∠OA = ∠OB (By CPCT)
Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Question 2:
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Ans.
Draw OM ⊥ AB and ON ⊥ CD. Join OB and OD.
(Perpendicular from the centre bisects the chord)
Let ON be x. Therefore, OM will be 6− x.
In ΔMOB,
In ΔNOD,
We have OB = OD (Radii of the same circle)
Therefore, from equation (1) and (2),
From equation (2),
Therefore, the radius of the circle iscm.


Question 3:
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the center?

Ans.
Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.
Distance of smaller chord AB from the centre of the circle = 4 cm
OM = 4 cm
MB = 
In ΔOMB,
In ΔOND,
Therefore, the distance of the bigger chord from the centre is 3 cm.


Question 4:
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the center.

Ans.
In ΔAOD and ΔCOE,
OA = OC (Radii of the same circle)
OD = OE (Radii of the same circle)
AD = CE (Given)
∴ ΔAOD ≅ ΔCOE (SSS congruence rule)
∠OAD = ∠OCE (By CPCT) ... (1)
∠ODA = ∠OEC (By CPCT) ... (2)
Also,
∠OAD = ∠ODA (As OA = OD) ... (3)
From equations (1), (2), and (3), we obtain
∠OAD = ∠OCE = ∠ODA = ∠OEC
Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = x
In Δ OAC,
OA = OC
∴ ∠OCA = ∠OAC (Let a)
In Δ ODE,
OD = OE
∠OED = ∠ODE (Let y)
ADEC is a cyclic quadrilateral.
∴ ∠CAD + ∠DEC = 180° (Opposite angles are supplementary)
x + a + x + y = 180°
2x + a + y = 180°
y = 180º − 2x − a ... (4)
However, ∠DOE = 180º − 2y
And, ∠AOC = 180º − 2a
∠DOE − ∠AOC = 2a − 2= 2− 2 (180º − 2x − a)
= 4a + 4x − 360° ... (5)
∠BAC + ∠CAD = 180º (Linear pair)
⇒ ∠BAC = 180º − ∠CAD = 180º − (a + x)
Similarly, ∠ACB = 180º − (a + x)
In ΔABC,
∠ABC + ∠BAC + ∠ACB = 180º (Angle sum property of a triangle)
∠ABC = 180º − ∠BAC − ∠ACB
= 180º − (180º − a − x) − (180º − a −x)
= 2a + 2− 180º
[4a + 4x − 360°]
∠ABC = [∠DOE − ∠ AOC] [Using equation (5)]


Question 5:
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Ans.
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.
∴ ∠COD = 90°
Also, in rhombus, the diagonals intersect each other at 90°.
∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Clearly, point O has to lie on the circle.

Question 6:
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Ans.
It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°.
∠AEC + ∠CBA = 180°
∠AEC + ∠AED = 180° (Linear pair)
∠AED = ∠CBA ... (1)
For a parallelogram, opposite angles are equal.
∠ADE = ∠CBA ... (2)
From (1) and (2),
∠AED = ∠ADE
AD = AE (Angles opposite to equal sides of a triangle)


Question 7:
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.

Ans.
Let two chords AB and CD are intersecting each other at point O.
In ΔAOB and ΔCOD,
OA = OC (Given)
OB = OD (Given)
∠AOB = ∠COD (Vertically opposite angles)
ΔAOB ≅ ΔCOD (SAS congruence rule)
AB = CD (By CPCT)
Similarly, it can be proved that ΔAOD ≅ ΔCOB
∴ AD = CB (By CPCT)
Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C
However, ∠A + ∠C = 180° (ABCD is a cyclic quadrilateral)
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
As ACBD is a parallelogram and one of its interior angles is 90°, therefore, it is a rectangle.
∠A is the angle subtended by chord BD. And as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.


Question 8:
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° .

Ans.
It is given that BE is the bisector of ∠B.
∴ ∠ABE = 
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
⇒ ∠ADE = 
Similarly, ∠ACF = ∠ADF = (Angle in the same segment for chord AF)
∠D = ∠ADE + ∠ADF
Similarly, it can be proved that


Question 9:
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Ans.
AB is the common chord in both the congruent circles.
∴ ∠APB = ∠AQB
In ΔBPQ,
∠APB = ∠AQB
∴ BQ = BP (Angles opposite to equal sides of a triangle)


Question 10:
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.

Ans.
Let perpendicular bisector of side BC and angle bisector of ∠A meet at point D. Let the perpendicular bisector of side BC intersect it at E.
Perpendicular bisector of side BC will pass through circumcentre O of the circle. ∠BOC and ∠BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively. We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠BOC = 2 ∠BAC = 2 ∠A ... (1)
In ΔBOE and ΔCOE,
OE = OE (Common)
OB = OC (Radii of same circle)
∠OEB = ∠OEC (Each 90° as OD ⊥ BC)
∴ ΔBOE ≅ ∠COE (RHS congruence rule)
∠BOE = ∠COE (By CPCT) ... (2)
However, ∠BOE + ∠COE = ∠BOC
⇒ ∠BOE +∠BOE = 2 ∠A [Using equations (1) and (2)]
⇒ 2 ∠BOE = 2 ∠A
⇒ ∠BOE = ∠A
∴ ∠BOE = ∠COE = ∠A
The perpendicular bisector of side BC and angle bisector of ∠A meet at point D.
∴ ∠BOD = ∠BOE = ∠A ... (3)
Since AD is the bisector of angle ∠A,
∠BAD = 
⇒ 2 ∠BAD = ∠A ... (4)
From equations (3) and (4), we obtain
∠BOD = 2 ∠BAD
This can be possible only when point BD will be a chord of the circle. For this, the point D lies on the circum circle.
Therefore, the perpendicular bisector of side BC and the angle bisector of ∠A meet on the circum circle of triangle ABC.










Exercise 10.5


Question 1:
In the given figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Ans.
It can be observed that
∠AOC = ∠AOB + ∠BOC
= 60° + 30°
= 90°
We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.

Question 2:
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.

Ans.
In ΔOAB,
AB = OA = OB = radius
∴ ΔOAB is an equilateral triangle.
Therefore, each interior angle of this triangle will be of 60°.
∴ ∠AOB = 60°
In cyclic quadrilateral ACBD,
∠ACB + ∠ADB = 180° (Opposite angle in cyclic quadrilateral)
⇒ ∠ADB = 180° − 30° = 150°
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.

Question 3:
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Ans.
Consider PR as a chord of the circle.
Take any point S on the major arc of the circle.
PQRS is a cyclic quadrilateral.
∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠PSR = 180° − 100° = 80°
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠POR = 2∠PSR = 2 (80°) = 160°
In ΔPOR,
OP = OR (Radii of the same circle)
∴ ∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)
∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of a triangle)
2 ∠OPR + 160° = 180°
2 ∠OPR = 180° − 160° = 20º
∠OPR = 10°


Question 4:
In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Ans.
In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180° (Angle sum property of a triangle)
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠BAC = 180° − 100º
⇒ ∠BAC = 80°
∠BDC = ∠BAC = 80° (Angles in the same segment of a circle are equal)


Question 5:
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Ans.
In ΔCDE,
∠CDE + ∠DCE = ∠CEB (Exterior angle)
⇒ ∠CDE + 20° = 130°
⇒ ∠CDE = 110°
However, ∠BAC = ∠CDE (Angles in the same segment of a circle)
⇒ ∠BAC = 110°

Question 6:
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

Ans.
For chord CD,
∠CBD = ∠CAD (Angles in the same segment)
∠CAD = 70°
∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral)
∠BCD + 100° = 180°
∠BCD = 80°
In ΔABC,
AB = BC (Given)
∴ ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)
⇒ ∠BCA = 30°
We have, ∠BCD = 80°
⇒ ∠BCA + ∠ACD = 80°
30° + ∠ACD = 80°
⇒ ∠ACD = 50°
⇒ ∠ECD = 50°


Question 7:
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Ans.
Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.
(Consider BD as a chord)
∠BCD + ∠BAD = 180° (Cyclic quadrilateral)
∠BCD = 180° − 90° = 90°
(Considering AC as a chord)
∠ADC + ∠ABC = 180° (Cyclic quadrilateral)
90° + ∠ABC = 180°
∠ABC = 90°
Each interior angle of a cyclic quadrilateral is of 90°. Hence, it is a rectangle.


Question 8:
If the non-parallel sides of a trapezium are equal, prove that it is cyclic

Ans.
.
Consider a trapezium ABCD with AB | |CD and BC = AD.
Draw AM ⊥ CD and BN ⊥ CD.
In ΔAMD and ΔBNC,
AD = BC (Given)
∠AMD = ∠BNC (By construction, each is 90°)
AM = BM (Perpendicular distance between two parallel lines is same)
∴ ΔAMD ≅ ΔBNC (RHS congruence rule)
∴ ∠ADC = ∠BCD (CPCT) ... (1)
∠BAD and ∠ADC are on the same side of transversal AD.
∠BAD + ∠ADC = 180° ... (2)
∠BAD + ∠BCD = 180° [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.


Question 9:
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
Ans.
Join chords AP and DQ.
For chord AP,
∠PBA = ∠ACP (Angles in the same segment) ... (1)
For chord DQ,
∠DBQ = ∠QCD (Angles in the same segment) ... (2)
ABD and PBQ are line segments intersecting at B.
∴ ∠PBA = ∠DBQ (Vertically opposite angles) ... (3)
From equations (1), (2), and (3), we obtain
∠ACP = ∠QCD

Question 10:
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Ans.
Consider a ΔABC.
Two circles are drawn while taking AB and AC as the diameter.
Let they intersect each other at D and let D not lie on BC.
Join AD.
∠ADB = 90° (Angle subtended by semi-circle)
∠ADC = 90° (Angle subtended by semi-circle)
∠BDC = ∠ADB + ∠ADC = 90° + 90° = 180°
Therefore, BDC is a straight line and hence, our assumption was wrong.
Thus, Point D lies on third side BC of ΔABC.


Question 11:
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Ans.
In ΔABC,
∠ABC + ∠BCA + ∠CAB = 180° (Angle sum property of a triangle)
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90° ... (1)
In ΔADC,
∠CDA + ∠ACD + ∠DAC = 180° (Angle sum property of a triangle)
⇒ 90° + ∠ACD + ∠DAC = 180°
⇒ ∠ACD + ∠DAC = 90° ... (2)
Adding equations (1) and (2), we obtain
∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°
⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°
∠BCD + ∠DAB = 180° ... (3)
However, it is given that
∠B + ∠D = 90° + 90° = 180° ... (4)
From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.
Consider chord CD.
∠CAD = ∠CBD (Angles in the same segment)

Question 12:
Prove that a cyclic parallelogram is a rectangle.

Ans.
Let ABCD be a cyclic parallelogram.
∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral) ... (1)
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C and ∠B = ∠D
From equation (1),
∠A + ∠C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
Parallelogram ABCD has one of its interior angles as 90°. Therefore, it is a rectangle.