# Telangana SCERT Class 9 Math Solution Chapter 12 Circles Exercise 12.4

## Telangana SCERT Solution Class IX (9) Math Chapter 12 Circles Exercise 12.4

(1) In the figure, ‘O’ is the centre of the circle. AOB = 100° find ADB Solution:

Given,

∠AOB = 100o

O is the centre of the circle

∴∠ACB = (∠AOB)/2 [∵angle subtended by an X arc is twice of angle subtended by it on the remaining circle]

= 100/2

= 50o

∴ ABCD is a cyclic quadrillater

∴ Sum of opposite angles is 180o

∴ ∠ACB + ∠ADB = 180o

Or, 50o + ∠ADB = 180o

Or, ∠ADB = 180o – 50o

= 130o

Alternative method

∠AOB on arc ∠ACB

= 360o – ∠AOB on arc ∠ADB

= 360o – 100o

= 260o

∠ADB = ½ ∠AOB on arc ∠ACB

= ½  x 260o

= 130o

[∵angle subtended by an arc is twice the angle subtended by it on remaining circle]

(2) In the figure, BAD = 40° then find BCD In △AOB ⊥ △COD given,

(i) AO = OD [radius a circle]

(ii) BO = OC [radius a circle]

(iii) ∠COD = ∠AOB [vertically opposite angles, since AD & BC arc diameters of circle]

∴ △AOB ≅ △COD by SAS

∴ ∠BAD = ∠BAO = ∠OCD = ∠BCD = 40o [∵corresponding angles of congruent triangles are equal]

(3) In the figure, O is the centre of the circle and POR = 120° . Find PQR and PSR Solution:

Given, O is the centre of the circle

∠POR = 120o

∴∠PQR = ½ ∠POR = ½ x 120o = 60o [∵angle substended by an arc at the centre of circle is twice he angles substanded by it on the remaining of the circle]

Now, PQRS is a cyclic quadrilateral since all verticals lie on the same circle.

∴ Opposite angles are supplementary

∴ ∠PQR + ∠PSR = 180o

Or, 60 + ∠PSR = 180o

Or, ∠PSR = 180o – 60o

= 120o

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

Given,

OM = 3cm

AB = 8cm

∴AM = MB = 8/2 = 4cm

Since perpendicular form the centre of the circle to any chord bisects the chord

Now, join OA ∴OA is one radius of the circle

∴In △AOM, By Pythagoras theorem

OM2 + AM2 = DA2

Or, OA =  √32 + 42

= √9 + 16

= √25

= 5 cm

∴ The radius of circle is 5cm

(5) 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 Solution:

Given, O is the centre of the circle OM and ON are perpendiculars to chord PQ and RS respectively

OM = ON; PQ = 6cm

∴ RS = PQ = 6cm [∵OM = ON, distance between chords PQ & RS is from the centre of circle is equal therefore, the chords are equal in length]

(6) A is the centre of the circle and ABCD is a square. If BD = 4cm then find the radius of the circle. Solution:

Given, A is the centre of circle

ABCD is a square

BD = 4cm

∴ AC = BD = 4cm [∵AC & BD two diagonals of square ABCD & diagonals of a square are equal in length]

Also AC is the radius of circle line C is a point M the circle & A is the centre of circle.

∴ Radius of circle = AC = 4cm

(7) Draw a circle with any radius and then draw two chords equidistant from the centre Solution:

Chord AB & CD are equidistant from centre O of the circle

Step 1: – Draw a circle of any radius

Step 2: – Draw two radius OA & OB and join them to get AB as a chord

Step 3: – Draw perpendicular bisect of AB as OM passing through the centre O

Step 4: – Take double the length of OM as radius on the compass and cut the circle from each end on AB on the opposite side of centre o

Step 5: – Join the two point CD

AB & CD are the two chords equidistance from the centre.

(8) 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 ΔOCD. Updated: October 2, 2021 — 4:36 pm