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]