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Sunday, August 7, 2022

Circles Solution of TS & AP Board Class 9 Mathematics

Circles Solution of TS & AP Board Class 9 Mathematics


Exercise 12.1

Question 1.

Name the following parts from the adjacent figure where ‘O’ is the centre of the circle.

i. 

ii. 

iii. BC

iv. 

v. DCB

vi. ACB

vii. 

viii. shaded region


Answer:

i. Radius

ii. Diametre

iii. Minor arc

iv. Chord

v. Major arc

vi. Semi-circle

vii. Chord

viii. Minor segment



Question 2.

State true or false.

i. A circle divides the plane on which it lies into three parts. ( )

ii. The area enclosed by a chord and the minor arc is minor segment. ( )

iii. The area enclosed by a chord and the major arc is major segment. ( )

iv. A diameter divides the circle into two unequal parts. ( )

v. A sector is the area enclosed by two radii and a chord ( )

vi. The longest of all chords of a circle is called a diameter. ( )

vii. The mid point of any diameter of a circle is the centre. ( )


Answer:

i. True

ii. True

iii. True

iv. False

v. False

vi. True

vii. True



Exercise 12.2

Question 1.

In the figure, if AB = CD and ∠AOB = 90° find ∠COD


Answer:

We know “Angles subtended by equal chords at the center of acircle are equal”.

 ∠COD = ∠AOB = 900



Question 2.

In the figure, PQ = RS and ∠ORS = 48°. Find ∠OPQ and ∠ROS.


Answer:

In ORS,

∠ORS = ∠OSR (radius of circle)

 ∠OSR = ∠ORS = 480

 ∠OSR + ∠ORS + ∠ROS = 180

⇒ 48 + 48 + ∠ROS = 180

⇒ 96 + ∠ROS = 180

⇒ ∠ROS = 180-96

⇒ ∠ROS = 180-96

⇒ ∠ROS = 84

We know “Angles subtended by equal chords at the center of a circle are equal”.

 ∠POQ = ∠ROS = 840

In POQ,

∠POQ + ∠OPQ + ∠OQP = 180

840 + x + x = 180

840 + 2x = 180

2x = 180-84

2x = 96

x = 

x = 48

 ∠OPQ = 48



Question 3.

In the figure PR and QS are two diameters. Is PQ = RS?


Answer:

Considering a circle, where PR and QS being a diameter.

Let center of circle be O.

In triangles POQ and SOR

PO = OR (Radius of circle)

OS = OQ (Radius of circle)

Angle POQ = Angle SOR (Vertically opposite angles are always

equal)

Hence triangle POQ is congruent to triangle SOR (By Side Angle Side

Axiom)

PQ = RS (By C.P.C.T.C)

Hence, proved.



Exercise 12.3

Question 1.

Draw the following triangles and construct circumcircles for them.

(i) In ∆ ABC, AB = 6cm, BC = 7cm and ∠A = 60°

(iii) In ∆ XYZ, XY = 4.8cm, ∠X = 60° and ∠Y = 70°

(iii) In ∆ XYZ, XY = 4.8cm, ∠X = 60° and ∠Y = 70°


Answer:

(i) 

(ii) 

(iii) 



Question 2.

Draw two circles passing through A, B where AB = 5.4cm


Answer:



Question 3.

If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.


Answer:

Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles.

OO' is the line segment joining the centers.
Let OO' intersect AB at M
Now Draw line segments OA, OB , O'A and O'B
In ΔOAO' and OBO' , we have
OA = OB (radii of same circle)
O'A = O'B (radii of same circle)
O'O = OO' (common side)
⇒ ΔOAO' ≅ ΔOBO' (SSS congruency)
⇒ ∠AOO' = ∠BOO'
⇒ ∠AOM = ∠BOM ......(i)
Now in ΔAOM and ΔBOM we have
OA = OB (radii of same circle)
∠AOM = ∠BOM (from (i))
OM = OM (common side)
⇒ ΔAOM ≅ ΔBOM (SAS congruency)
⇒ AM = BM and ∠AMO = ∠BMO
But
∠AMO + ∠BMO = 180°
⇒ 2∠AMO = 180°
⇒ ∠AMO = 90°
Thus, AM = BM and ∠AMO = ∠BMO = 90°
Hence OO' is the perpendicular bisector of AB.



Question 4.

If two intersecting chords of a circle make equal angles with diameter passing through their point of intersection, prove that the chords are equal.


Answer:

Given that AB and CD are two chords of a circle, with center O intersecting at a point E.
PQ is a diameter passing through E, such that ∠ AEQ = ∠ DEQ
Draw OL ⊥ AB and OM ⊥ CD.
In right angled OLE
∠LOE + 90° + ∠ LEO = 180° (Angle sum property of a triangle)
∴∠LOE = 90° – ∠LEO
= 90° – ∠AEQ = 90° – ∠DEQ
= 90° – ∠MEO = ∠MOE
In triangles OLE and OME,
∠LEO = ∠MEO
∠LOE = ∠MOE (Proved)
OE = OE (Common side)
∴ ΔOLE ≅ ΔOME
⇒ OL = OM (CPCT)
Thus, AB = CD



Question 5.

In the adjacent figure, AB is a chord of circle with centre O. CD is the diameter perpendicualr to AB. Show that AD = BD.


Answer:

Given, AB is a chord of circle with centre O. CD is the diameter

perpendicular to AB.

We know that line drawn from the center of a circle to the chord

Perpendicular to it bisects the chord.

AP = BP

In ADP and BDP,

AP = BP

APD = BPD = 90

PD = PD (Common)

ADP BDP

 AD = BD (CPCT)



Exercise 12.4

Question 1.

In the figure, ‘O’ is the centre of the circle. ∠AOB = 100° find ∠ADB.


Answer:

Given, ∠AOB = 100°

Join AB, which is a chord.

∠ ACB =  (Angle subtended at the center is twice the angle subtended at circumference by the same chord)

⇒∠ ACB = 50

Now, ACBD is a quadrilateral and angles on the opposite sides in the quadrilateral are supplementary.

∠ ADB = 180-∠ ACB

⇒∠ ADB = 180-50

⇒∠ ADB = 130o



Question 2.

In the figure, ∠BAD = 40° then find ∠BCD.


Answer:

Given, ∠BAD = 40°

Join BD, which is a chord.

We know “Two or more angles subtended by the chord at the circumference are same”.

 ∠BCD = ∠BAD = 40°



Question 3.

In the figure, O is the centre of the circle and ∠POR = 120°. Find ∠PQR and ∠PSR


Answer:

Given, ∠POR = 120°

We know that “Angle subtended at the center is twice the angle subtended at circumference by the same chord”.

∠PQR = 

⇒∠PQR = 

⇒∠PQR = 60

Now, PQRS is a quadrilateral and angles on the opposite sides in the quadrilateral are supplementary.

∠ PSR = 180-∠ PQR

⇒∠ PSR = 180-60

⇒∠ PSR = 120o



Question 4.

If a parallelogram is cyclic, then it is a rectangle. Justify.


Answer:

Let ABCD be a cyclic parallelogram.

A rectangle is a parallelogram with one angle 90. So, we have to prove angle 90.

Since ABCD is a parallelogram,

∠A = ∠C

In cyclic parallelogram ABCD,

∠A + ∠C = 180

∠A + ∠A = 180

2∠A = 180

∠A = 

∠A = 90

Hence, proved.



Question 5.

In the figure, ‘O’ is the centre of the circle. OM = 3cm and AB = 8cm. Find the radius of the circle


Answer:

Joining OA we see that OA is radius of the circle.

Let radius (OA) be r cm.

Given, AB = 8cm

OM = 3cm

We know that “perpendicular from center divides the chord into equal parts”.

 AM = 

⇒ AM = 4 cm

In right angled triangle OAM,

OA2 = OM2 + AM2 (Pythagoras Thm.)

⇒OA2 = 32 + 42

⇒OA2 = 9 + 16

⇒OA2 = 25

⇒OA = 

⇒OA = 5 cm



Question 6.

In the figure, ‘O’ is the centre of the circle and OM, ON are the perpendiculars from the centre to the chords PQ and RS. If OM = ON and PQ = 6cm. Find RS.


Answer:

Given, OM = ON

We know that “If two chords are equidistant from the center then the two chords are equal in length”.

PQ = RS = 6 cm



Question 7.

A is the centre of the circle and ABCD is a square. If BD = 4cm then find the radius of the circle.


Answer:

Given, BD = 4cm

We know that “a square has equal diagonals” and BD is a diagonal.

AC, which is the radius of the circle, is also a diagonal of the square.

 AC = BD = 4 cm

Thus, Radius of the circle = 4 cm



Question 8.

Draw a circle with any radius and then draw two chords equidistant from the centre.


Answer:

Circle with center O and two chords CD and EF equidistant from center .



Question 9.

In the given figure ‘O’ is the centre of the circle and AB, CD are equal chords. If ∠AOB = 70°. Find the angles of the ∆OCD.


Answer:

We know “Angles subtended by equal chords at the center of a circle are equal”.

 ∠DOC = ∠AOB = 70

In OCD,

∠OCD + ∠ODC + ∠DOC = 180

x + x + 70 = 180

700 + 2x = 180

2x = 180-70

2x = 110

x = 

x = 55

 ∠OCD = ∠ODC = 55



Exercise 12.5

Question 1.

Find the values of x and y in the figures given below.

i. 

ii. 

iii. 


Answer:

(i) We know “Sum of all angles of a triangle is 180”.

 x + y + 30 = 180

Since it is an isosceles triangle, x = y.

 x + x + 30 = 180

⇒ 2x + 30 = 180

⇒ 2x = 180-30

⇒ 2x = 150

⇒ x = 

⇒ x = 75

 x = y = 75

(ii) We know that “Angles on the opposite sides in the cyclic quadrilateral are supplementary”.

 x + 110 = 180

⇒ x = 180-110

⇒ x = 70

y + 85 = 180

⇒ y = 180-85

⇒ y = 95

(iii) Given, x = 90°

 x + y + 50 = 180

⇒ 90 + y + 50 = 180

⇒ y + 140 = 180

⇒ y = 180-140

⇒ y = 40



Question 2.

Given that the vertices A, B, C of a quadrilateral ABCD lie on a circle. Also ∠A + ∠C = 180°, then prove that the vertex D also lie on the same circle.


Answer:

In a quadrilateral, if the sum of opposite angles is 180°, then it is a

cyclic quadrilateral.

In quad. ABCD, A + C = 180°,

Therefore, ABCD is a cyclic quad.

In a cyclic quad., the vertices lie on the same circle. Thus, D also

lies on the same circle.

Hence, proved.



Question 3.

Prove that a cyclic rhombus is a square.


Answer:

To prove rhombus inscribed in a circle is a square, we need to prove that either any one of its interior angles is equal to 90° or its diagonals are equal.

∠ABD = ∠DBC = b

∠ADB = ∠BDC = a

In the figure, diagonal BD is angular bisector of angle B and angle D.

In triangle ABD and BCD,

AD = BC (sides of rhombus are equal)

AB = CD (sides of rhombus are equal)

BD = BD (common side)

△ABD ≅ △BCD. (SSS congruency)

In the figure,

2a + 2b = 180° (as, opposite angles of a cyclic quadrilateral are always supplementary)

2(a + b) = 180°

a + b = 90°

In △ABD,

Angle A = 180°-(a + b)

= 180°-90°

= 90°

Therefore, proved that one of its interior angle is 90°

Hence, rhombus inscribed in a circle is a square.



Question 4.

For each of the following, draw a circle and inscribe the figure given. If a polygon of the given type can’t be inscribed, write not possible.

(a) Rectangle

(b) Trapezium

(c) Obtuse triangle

(d) Non-rectangular parallelogram

(e) Accute issosceles triangle

(f) A quadrilateral PQRS with  as diameter.


Answer:

(a) Rectangle = Possible

(b) Trapezium = Possible

(c) Obtuse triangle = Not Possible

(d) Non-rectangular parallelogram = Not Possible

(e) Acute isosceles triangle = Possible

(f) A quadrilateral PQRS with  as diameter = Possible


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