Circles
Class 9th Mathematics Part 1 Kerala Board Solution
- Prove that the line joining the centres of two intersecting circles is the…
- The picture on the right shows two circles centred on the same point and a line…
- A chord and the diameter through one of its ends are drawn in a circle. A chord of the…
- Prove that the angle made by two equal chords drawn from a point on the circle is…
- Draw a square and a circle through all four vertices. Draw diameters parallel to the…
- Prove that chords of the same length in a circle are at the same distance from the…
- Two chords intersect at a point on a circle and the diameter through this point bisects…
- In the picture on the right, the angles between the radii and the chords are equal. e…
- In a circle, a chord 1 centimetre away from the centre is 6 centimetres long. What is…
- In a circle of radius 5 centimetres, two parallel chords of lengths 6 and 8 centimetres…
- The bottom side of the quadrilateral in the picture is a diameter of the circle and the…
- In a circle, two parallel chords of lengths 4 and 6 centimetres are 5 centimetres…
Questions Pg-82
Question 1.Prove that the line joining the centres of two intersecting circles is the perpendicular bisector of the line joining the points of intersection.
Answer:
O is the centre of the bigger circle and P is the centre of the smaller circle. DE is the line joining the points of intersection. We have to prove that, OP is the perpendicular bisector of DE.
In ΔODP and ΔOEP we have,
OD = OE [radius of same circle]
PD = PE [radius of same circle]
OP is the common side.
∴ΔODP≅ΔOEP [SSS congruency]
∴ ∠DOP = ∠EOP
∴ ∠DOG = ∠EOG ……. (i)
Now in ΔODG and ΔOEG we have,
OD = OE [radius of same circle]
∠DOG = ∠EOG from (i)
OG is the common side.
∴ΔODG≅ΔOEG [SAS congruency]
∴ DG = EG
∴ ∠OGD = ∠OGE
We know, ∴ ∠OGD + ∠OGE = 180°
∴ ∠OGD = ∠OGE = 90°
∴ DG = EG and ∠OGD = ∠OGE = 90°
∴ OP is perpendicular bisector of DE
Question 2.
The picture on the right shows two circles centred on the same point and a line intersecting them.
Prove that the parts of the line between the circles on either side are equal.
Answer:
D is the centre of both the circles.
DG is a perpendicular drawn on EF(HI).
We have to prove that, EH = IF.
In ΔDHG and ΔDIG we have,
DH = DI [radius of the same circle]
DG is the common side.
DG is perpendicular on HI.
∴ ∠DHG = ∠DIG = 90°
∴ ∆DHG≅∆DIG [SAS congruency]
∴ GH = GI …………… (1)
In ΔDEG and ΔDFG we have,
DE = DF [radius of the same circle]
DG is the common side.
DG is perpendicular on EF.
∴ ∠DEG = ∠DFG = 90°
∴ ∆DEG≅∆DFG [SAS congruency]
∴ GE = GF …………… (2)
From, (2) – (1) we get,
⇒ GE – GH = GF – GI
⇒ EH = IF
Question 3.
A chord and the diameter through one of its ends are drawn in a circle. A chord of the same inclination is drawn on the other side of the diameter.
Prove that the chords are of the same length.
Answer:
EF is the diameter of the circle.
Chord EG and EH both have same inclination. GF and HF are joined.
We have to prove that, EG = EH.
In ΔGEF and ΔHEF we have,
∠GEF = ∠HEF [∵ both have same inclination]
EF is the common side.
∠EGF = ∠EHF = 90° [∵ both are angle inscribed in a semi circle]
∴∆GEF≅∆HEF [AAS congruency]
∴ EG = EH [similar sides of congruent triangles]
Question 4.
Prove that the angle made by two equal chords drawn from a point on the circle is bisected by the diameter through that point.
Answer:
EG and EH are two equal chords drawn from point E.
EF is the diameter of the circle. GF and HF are joined.
Angle between two chords = ∠GEH.
We have to prove that, ∠GEF = ∠HEF.
In ΔGEF and ΔHEF we have,
EG = EH [given in the problem]
EF is the common side.
∠EGF = ∠EHF = 90° [∵ both are angle inscribed in a semi circle]
∴∆GEF≅∆HEF [SAS congruency]
∴ ∠GEF = ∠HEF [similar angles of congruent triangle]
Question 5.
Draw a square and a circle through all four vertices. Draw diameters parallel to the sides of the square and draw a polygon joining the end points of these diameters and the vertices of the square.
Prove that this polygon is a regular octagon.
Answer:
D is the centre of the circle. EFGH is the inscribed square.
Let, side of the square = a
∴ EF = FG = GH = HE = PM = NO = a
∴ Diagonal of the square = a√2
∴ EG = a√2
Now we have, EG = IJ = KL [diameter of the circle]
∴ EG = IJ = KL = a√2
IJ = a√2 and PM = a
[∵IM = PJ]
In ΔIME,
EM = a/2 [∵ EM = MF = EF/2]
∠IME = 90°
∴ tan ∠MIE = EM/IM
∴ tan ∠MIE = √2 + 1
∴ tan ∠MIE = tan 67.5°
∴ ∠MIE = 67.5°……… (1)
Similarly in ΔIMF,
∠MIF = 67.5°
∴ Inner angle of the polygon,
⇒ ∠EIF = ∠MIE + ∠MIF = 67.5°+67.5° = 135°
Similarly all the inner angles of the polygon = 135°
∴ All the outer angles of the Polygon = 180° - 135° = 45°
∴ It is a regular polygon.
Questions Pg-86
Question 1.Prove that chords of the same length in a circle are at the same distance from the centre.
Answer:
D is the centre of the circle. EF and GH are two chords of same length.
DI and DJ are two perpendiculars drawn on GH and EF respectively.
We have to prove that, DI = DJ.
In ΔDJF and ΔDIG we have,
DF = DG [radius of the same circle]
∠DJF = ∠DIG = 90° [∵DI and DJ are perpendiculars on GH and EF]
JF = IG [∵Perpendicular drawn from the centre bisects the chords]
∴∆DJF≅∆DIG [RHS congruency]
∴ DI = DJ [similar sides of congruent triangle]
∴ The chords are at same distance from the centre.
Question 2.
Two chords intersect at a point on a circle and the diameter through this point bisects the angle between the chords. Prove that the chords have the same length.
Answer:
EF and FG are two chord meet at point F on circle centred at D.
FH is the diameter through point F.
We have to prove that EF = FG.
We have joined EH and HG.
In ΔEHF and ΔGHF we have,
HF is the common side.
∠EFH = ∠ GFH [HF bisects the angle between two chords]
∠HEF = ∠HGF = 90° [both are angle inscribed on semicircle]
∴∆EHF≅∆GHF [RHS congruency]
∴ EF = FG [similar side of congruent triangle]
∴ The chords have same length.
Question 3.
In the picture on the right, the angles between the radii and the chords are equal.
Prove that the chords are of the same length.
Answer:
C is the radius of the circle. ED and FG are two chords.
In ΔCDE we have,
EC = CD [radius of the circle]
∴ ∠CDE = ∠CED ……… (1)
In ΔCFG we have,
FC = CG [radius of the circle]
∴ ∠CGF = ∠CFG ……… (2)
In ΔCDE and ΔCFG we have,
∠CED = ∠CFG [given in the problem] ……… (3)
∠CDE = ∠CGF [from (1), (2) and (3)]
CE = CF [radius of the same circle]
∴∆CDE≅∆CFG [AAS congruency]
∴ ED = FG [similar sides of the triangle.
∴ The chords are of same length.
Questions Pg-87
Question 1.In a circle, a chord 1 centimetre away from the centre is 6 centimetres long. What is the length of a chord 2 centimetres away from the centre?
Answer:
O is centre of the circle.
DE is the chord 1 cm away from centre. CH is perpendicular on DE.
∴ CH = 1 cm and DE = 6 cm
FG is the chord 2 cm away from centre. CI is perpendicular on FG.
∴ CI = 2 cm
From centre C, CE and CG are joined.
CE and CG both are radius of the circle.
In ΔCHE we have,
∠CHE = 90° [∵CH is perpendicular on DE]
CH = 1 cm
HE = DE/2 = 3 cm [∵perpendicular drawn from centre bisects chord]
In ΔCGI we have,
∠CIG = 90° [∵CI is perpendicular on FG]
CI = 2 cm
CG = CE = √10 cm
∴ FG = 2 × IG = 2√6 cm [∵perpendicular drawn from centre bisects chord]
∴ The length of the chord is = 2√6 cm
Question 2.
In a circle of radius 5 centimetres, two parallel chords of lengths 6 and 8 centimetres are drawn on either side of a diameter. What is the distance between them? If parallel chords of these lengths are drawn on the same side of a diameter, what would be the distance between them?
Answer:
When chords are drawn on either side of diameter.
O is the centre of the circle and IJ is the diameter.
CD and EF are two parallel chord on either side of the diameter.
CD = 6 cm and EF = 8 cm
OH is the perpendicular drawn on CD from centre.
OG is the perpendicular drawn on EF from centre.
In ΔOGE we have,
∠OGE = 90° [∵ OG is perpendicular on EF]
EG = EF/2 = 4 cm [∵perpendicular drawn from centre bisects chord]
OE = 5 cm [radius]
In ΔOHC we have,
∠OHC = 90° [∵ OH is perpendicular on CD]
HC = CD/2 = 3 cm [∵perpendicular drawn from centre bisects chord]
OC = 5 cm [radius]
∴ Distance between the chords = HG = OH + OG = 4 + 3 = 7 cm
When the chords are drawn on same side of diameter:
O is the centre of the circle and IJ is the diameter.
CD and EF are two parallel chords on same side of the diameter.
CD = 6 cm and EF = 8 cm
OH is the perpendicular drawn on CD from centre.
OG is the perpendicular drawn on EF from centre.
In ΔOGE we have,
∠OGE = 90° [∵ OG is perpendicular on EF]
EG = EF/2 = 4 cm [∵perpendicular drawn from centre bisects chord]
OE = 5 cm [radius]
In ΔOHC we have,
∠OHC = 90° [∵ OH is perpendicular on CD]
HC = CD/2 = 3 cm [∵perpendicular drawn from centre bisects chord]
OC = 5 cm [radius]
∴ Distance between the chords = HG = OH – OG = 4 – 3 = 1 cm
Question 3.
The bottom side of the quadrilateral in the picture is a diameter of the circle and the top side is a chord parallel to it. Calculate the area of the quadrilateral.
Answer:
O is the centre of the circle.
CD is the diameter, CD = 5 cm
EF is the chord parallel to CD, EF = 3 cm
OG is perpendicular drawn on EF from O.
In ΔOGE we have,
∠OGE = 90° [∵ OG is perpendicular on EF]
OE = 5/2 = 2.5 cm [∵ Diameter = 5 cm]
EG = 3/2 = 1.5 cm [∵perpendicular drawn from centre bisects chord]
∴ Area of the quadrilateral,
= 8 cm2
Question 4.
In a circle, two parallel chords of lengths 4 and 6 centimetres are 5 centimetres apart. What is the radius of the circle?
Answer:
O is the centre of the circle.
EF and GH are two parallel chords 5 cm apart.
EF = 6 cm and GH = 4 cm.
OJ is perpendicular on EF and OI is perpendicular drawn on GH.
IJ = 5 cm
Let, OJ = x cm and OI = (5 – x) cm
Let, radius of the circle = r cm
OE = OG = r cm
In ΔOIG we have,
∠OIG = 90° [∵OI is perpendicular drawn on GH]
IG = 4/2 = 2 cm [∵perpendicular drawn from centre bisects chord]
OI = (5 – x) cm
OG = r cm
…… (1)
In ΔOJE we have,
∠OJE = 90° [∵ OJ is perpendicular on EF]
JE = 6/2 = 3 cm [∵perpendicular drawn from centre bisects chord]
OJ = x cm
OE = r cm
…… (2)
From (1) and (2) we have,
⇒ 4 + 25 – 10x + x2 = 9 + x2
⇒ 29 – 10x = 9
⇒ 10x = 20
∴ x = 2
Putting the value x = 2 in (2) we get,
⇒ r = √13
∴ radius of the circle = √13 cm
Questions Pg-92
Question 1.Draw three triangles with lengths of two sides 4 and 5 centimetres and the angle between them 60°, 90°, 120°. Draw the circumcircle of each. (Note how the position of the circumcentre changes).
Answer:
In this figure AC = 5 cm and AB = 4 cm.
∠CAB = 60°
J is the circumcentre, which is inside the triangle.
AC = 5cm and AB = 4 cm.
∠CAB = 90°
H is the circumcentre, which lies on the hypotenuse of the triangle.
AB = 5 cm and BC = 4 cm.
∠ABC = 120°
H is the circumcentre, which is outside the triangle.
Question 2.
The equal sides of an isosceles triangle are 8 centimetres long and the radius of its circumcircle is 5 centimetres. Calculate the length of its third side.
Answer:
AB = AC = 8 cm
AJ = BJ = CJ = 5 cm
Let, JE = y cm and BE = x cm
In ΔJBE we have,
∠JEB = 90°
BE = x cm
JE = y cm
BJ = 5 cm
∴ x2 + y2 = 52
⇒ x2 = 25 – y2 ……… (1)
In ΔABE we have,
∠AEB = 90°
BE = x cm
AE =(AJ + JE)=(5 + y) cm
AB = 8 cm
∴ x2 + (5 – y)2 = 82
⇒ x2 = 64 – (5 – y)2……… (2)
From (1) and (2) we have,
⇒ 25 – y2 = 64 – (5 – y)2
⇒ 25 – y2 = 64 – 25 + 10y – y2
⇒ 10y = 14
⇒ y = 1.4
Putting the value y = 1.4 in (1) we get,
⇒ x2 = 25 – (1.4)2
⇒ x2 = 25 – 1.96
⇒ x2 = 23.04
⇒ x = 4.8
∴ Length of the third side, BC = 2x = 2 × 4.8 = 9.6 cm
Question 3.
Find the relation between a side and the circumradius of an equilateral triangle.
Answer:
AB = BC = CA
∠ABC = ∠ACB = ∠BAC = 60°
Let, AH = x cm
∴ AH = CH = BH = x cm
∴ DH = x/2 cm [∵H is circumcentre, ABC is equilateral]
In ΔAHD,
∠ADH = 90°
AH = x cm
HD = x/2 cm
∴ AB = 2 × AD = 2 × x√3/2 = x√3 cm
∴ side = √3 × circumradius