NCERT solution class 9 chapter 10 Circles exercise 10.6 mathematics ~ evidya

EXERCISE 10.6

Question 1:

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

Answer:

Let two circles having their centres as O and O' intersect each other at point A and B respectively. Let us join O O'.

In ΔAOO' and BOO' ,
OA = OB (Radius of circle 1)
O'A = O' B (Radius of circle 2)
OO' = OO' (Common)
ΔAOO' ≅ ΔBOO' (By SSS congruence rule)
∠OAO' = ∠OBO' (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.

Answer:

Draw OM ⊥ AB and ON ⊥ CD. Join OB and OD.

$\mathrm{BM}=\frac{\mathrm{AB}}{2}=\frac{5}{2}$ (Perpendicular from the centre bisects the chord)
$\mathrm{ND}=\frac{\mathrm{CD}}{2}=\frac{11}{2}$

Let ON be x. Therefore, OM will be 6− x.
In ΔMOB,
$\mathrm{OM}^{2}+\mathrm{MB}^{2}=\mathrm{OB}^{2}$

$(6-x)^{2}+\left(\frac{5}{2}\right)^{2}=\mathrm{OB}^{2}$

$36+x^{2}-12 x+\frac{25}{4}=\mathrm{OB}^{2}$..(i)

In ΔNOD,
$\mathrm{ON}^{2}+\mathrm{ND}^{2}=\mathrm{OD}^{2}$

$x^{2}+\left(\frac{11}{2}\right)^{2}=\mathrm{OD}^{2}$

$x^{2}+\frac{121}{4}=\mathrm{OD}^{2}$..(2)

We have OB = OD (Radii of the same circle)
Therefore, from equation (1) and (2),

$36+x^{2}-12 x+\frac{25}{4}=x^{2}+\frac{121}{4}$

$12 x=36+\frac{25}{4}-\frac{121}{4}$

$=\frac{144+25-121}{4}=\frac{48}{4}=12$

x=1

From equation (2),

$(1)^{2}+\left(\frac{121}{4}\right)=\mathrm{OD}^{2}$

$\mathrm{OD}^{2}=1+\frac{121}{4}=\frac{125}{4}$

$\mathrm{OD}=\frac{5}{2} \sqrt{5}$
Therefore, the radius of the circle is $\frac{5}{2} \sqrt{5}$cm.

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 centre?

Answer:

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 = $\frac{A B}{2}=\frac{6}{2}=3 \mathrm{cm}$
In ΔOMB,
$\mathrm{OM}^{2}+\mathrm{MB}^{2}=\mathrm{OB}^{2}$

$(4)^{2}+(3)^{2}=\mathrm{OB}^{2}$

16+9=$\mathrm{OB}^{2}$

$\mathrm{OB}=\sqrt{25}$

OB=5cm

In ΔOND,
OD=OB=5cm (Radii of the same circle)

$\mathrm{ND}=\frac{\mathrm{CD}}{2}=\frac{8}{2}=4 \mathrm{cm}$

$\mathrm{ON}^{2}+\mathrm{ND}^{2}=\mathrm{OD}^{2}$

$\mathrm{ON}^{2}+(4)^{2}=(5)^{2}$

$\mathrm{ON}^{2}=25-16=9$

ON=3

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 centre.

Answer:

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 − 2y = 2a − 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 + 2x − 180º
= $\frac{1}{2}$[4a + 4x − 360°]
∠ABC = $\frac{1}{2}$[∠DOE − ∠ AOC] [Using equation (5)]

Question 4:

Answer:

Question 6:

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

Answer:

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.

Answer:

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° $-\frac{1}{2} A$ ,$90^{\circ}-\frac{1}{2} \mathrm{B}$ and $90^{\circ}-\frac{1}{2} \mathrm{C}$.

Answer:

It is given that BE is the bisector of ∠B.
∴ ∠ABE = $\frac{\angle \mathrm{B}}{2}$
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
⇒ ∠ADE = $\frac{\angle \mathrm{B}}{2}$
Similarly, ∠ACF = ∠ADF = $\frac{\angle \mathrm{C}}{2}$(Angle in the same segment for chord AF)
∠D = ∠ADE + ∠ADF
$=\frac{\angle \mathrm{B}}{2}+\frac{\angle \mathrm{C}}{2}$

$=\frac{1}{2}(\angle \mathrm{B}+\angle \mathrm{C})$

$=\frac{1}{2}\left(180^{\circ}-\angle \mathrm{A}\right)$

$=90^{\circ}-\frac{1}{2} \angle \mathrm{A}$

Similarly, it can be proved that

$\angle \mathrm{E}=90^{\circ}-\frac{1}{2} \angle \mathrm{B}$

$\angle \mathrm{F}=90^{\circ}-\frac{1}{2} \angle \mathrm{C}$

Page No 187:

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.

Answer:

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.

Answer:

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 = $\frac{\angle \mathrm{A}}{2}$
⇒ 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.

Sunday, May 31, 2020