Telangana SCERT Class 9 Math Solution Chapter 12 Circles Exercise 12.3

Telangana SCERT Solution Class IX (9) Math Chapter 12 Circles Exercise 12.3

(1.) Draw the following triangles and construct circumcircles for them.

(i) In Δ ABC, AB = 6cm, BC = 7 cm and A = 60o

(ii) In Δ PQR, PQ = 5cm, QR = 6 cm and RP = 8.2cm

(iii) In Δ XYZ, XY = 4.8cm, X = 60o and Y = 70o

Solution:

(i)

In ABC, AB = 6cm, BC = 7cm, ∠A = 60o

Step 1: – Draw line AB = 6cm

Step 2: – Draw a line 60o from point A on line AB

Step 3: – Cut the line at form B taking radius 7cm on compass. Hence, BC = 7cm

Step 4: – Join AC and BC at point C to from △ABC

Step 5: – Draw perpendicular bisectors of AB and AC from base AB and AC

Step 6: – Point where they intersects is the centre of circle

Step 7: – Draw a circle taking radius OA which interact at point B and C

(ii)

In △PQR, PQ= 5cm, QR = 6Cm and RP = 8.2cm

Step 1:- Draw a line PQ = 5cm

Step 2: – Take radius 8.2 cm on compass and draw an arc from point P

Step 3:- Take radius 6cm on compass draw an arc that cuts the previous arc from point Q

Step 4: – Joint the intersection point R at the arc with point P & Q to get △PQR of side 5cm, 8.2cm, & 6cm

Step 5: – Draw, perpendicular bisects for PQ & QR using compass of any radius from each ends of line segments PQ and QR

Step 6: – The intersection of as bisects is the centre of circle O

Step 7: – Draw a circle of radius OP from centre O connecting points R and Q

(iii)

In △XYZ, XY = 4.8Cm, ∠X = 60o & ∠Y = 70o

Step 1: – Draw One segment XY = 4.8cm

Step 2: – Draw a line at 60o from point X at Line segment XY

Step 3: – Draw a line at 70o from point Y at Line segment XY

Step 4: – Draw perpendicular bisectors of line XY and YZ by drawing arc from each end of of XY and YZ taking any radius

Step 5: – The point where the perpendicular bisects cuts each other is the centre of circle O

Step 6: – Draw a circle of radius OX connecting point Z & Y

(2.) Draw two circles passing through A, B where AB = 5.4 cm

Solution:

Step 1: – Draw a line AB = 5.4 cm

Step 2: – Draw perpendicular bisects of line AB by drawing arcs of any radius from each ends of AB on opposite sides of AB

Step 3: – The point where the arcs into cuts each other arc the centre of circles P & Q

Step 4: – Draw a circle taking radius PA with P as centre

Step 5: – Draw a circle taking radius QA at centre Q

∴ AB being the chord of two circle.

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

Solution:

From the above demonstration we can see that the centres of circle lie on the same perpendicular bisects of chord AB.

(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.

Solution:

Given,

PQ diameter

AB, CD are two chords intersecting each other at E on diameter PQ.

OL & OM are perpendiculars from centre of circle O to chords AB & CD respectively

∠AEQ = ∠DEQ [given]

∴ In △EOL & △EON

(i) ∠LED = ∠NEO [∠AEQ = ∠DEQ, similar angle]

(ii) ∠ELO = ∠ENO = 90o [⊥ was at chord AB & CD]

(iii) EO is common side

∴ △EOL ≅ △EOM by AAS

∠O = MO [∵ corresponding sides of congruent triangles are equal]

∴ AB = CD [∵∠ O = MO which are perpendicular of AB & CD & chords equidistance from the centre of circle are equal in length]

(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.

Solution:

Solution:

Given, CD diameter perpendicular to chord AB. O being the centre of circle

Let, the point where the diameter & Chord intersect be P.

∴In △APD & △BPD

(i) DP is the common side

(ii) ∠BPD = ∠APD = 90o [diameter DC ⊥ to chord AB]

(iii) AP = BP [∵perpendicular drawn from the centre of circle O bisect chord AB at point P]

△APD ≅ △BPD by SAS

AD = BD [∵corresponding sides of congruent triangle are equal in length]


Updated: October 2, 2021 — 4:36 pm

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