9th Maths 10.4
NCERT Class 9th solution of Exercise 10.1
NCERT Class 9th solution of Exercise 10.2
NCERT Class 9th solution of Exercise 10.3
Exercise 10.4
Q1. Two circles of radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm. Find the length of the common chord.
Sol. :
Given:
Let radius OP=5cm, O′P=3cm. The distance of OO′ is 4cm.
To Find:
PQ
Solve:
In `triangleOPO'
(OP)2=(OO′)2+(O′P)2 [By Pythagoras Theorem]
(5)2=(4)2+(3)2
25=16+9
25=25
`triangleOPO' is a right-angled triangle.
this gives OO′⊥OP
PQ is the diameter of the circle of radius 3cm as it passes through centre O′.
PQ=2×3=6cm.
Answer:
The length of the common chord is 6cm.
Q2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Sol. :
Given:
Chords AB=CD of a circle with centre O which intersect at P
To Prove:
AP=PD and PB=PC
Construction:
Draw ON⊥CD and OM⊥AB. Join OP.
Proof:
In △OPM and △OPN
∠OMP=∠ONP=90∘ [By construction]
OP=OP [Common]
OM=ON [Equal chords are equidistance from the centre of circle]
△OPM≅△OPN [By RHS rule]
PM=PN_________(1) [By CPCT]
AB=CD [Given]
12AB=12CD
MB=CN________(2)
AM=ND
Add equation (1) and (2)
PM+AM=PN+ND
AP=PD
Subtract equation (1) and (2)
MB-PM=CN-PN
PB=PC
Proved.
Q3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Sol. :
Given:
Chords AB=CD of a circle with centre O which intersect at P
To Prove:
∠OPM=∠OPN
Construction:
Draw ON⊥CD and OM⊥AB. Join OP.
Proof:
In △OPM and △OPN
∠OMP=∠ONP=90∘ [By construction]
OP=OP [Common]
OM=ON [Equal chords are equidistance from the centre of circle]
△OPM≅△OPN [By RHS rule]
∠OPM=∠OPN [By CPCT]
Proved.
Q4. If a line intersects two concentric circles (circles with the same centre) with centre O at A,B,C and D, prove that AB=CD[See figure]
Sol. :
Given:
Line intersect two concentric circles with centre O at A,B,C,D.
To Prove:
AB=CD
Construction:
Draw OM⊥AD
Proof:
Perpendicular OM from the centre of a circle to a chord AD bisects the chord.
AM=DM___________(1)
Perpendicular OM from the centre of a circle to a chord BC bisects the chord.
BM=CM___________(2)
Subtract equation (1) - (2)
AM-BM=DM-CM
AB=CD
Proved.
Q5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and distance between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Sol. :
Given:
OR=OM=5m and SR=SM=6m
To Find:
MR
Construction:
Join OR,OM and OS=5cm. OL⊥SR
Solve:
In △ORL,
(OR)2=(OL)2+(LR)2 [By Pythagorus Theorem]
(5)2=(OL)2+(3)2
(OL)2=25-9
(OL)2=16
OL=4
Area of △ROS=Area of △ROS
12×OS×RN=12×OR×OL
12×5×RN=12×6×4
RN=245
RN=4.8m
RM=2RN
RM=2×4.8=9.6m
Answer:
The distance between Reshma and Mandip is 9.6m.
Q6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other . Find the length of the string of each phone.
Sol. :
radius OA=OB=20m,
AS=SD=DA
Construction:
Draw AM⊥SD, join OS.
Solve:
Let AS=SD=DA=2x, and SM=MD=x
In Right △ASM
(AS)2=(SM)2+(AM)2[By Pythagorus Theroem]
(2x)2=(x)2+(AM)2
4x2=x2+(AM)2
(AM)2=4x2-x2
(AM)2=3x2
AM=√3x
OM=AM-OA
OM=(√3x-20)m__________(1)
In Right △OSM
(OS)2=(SM)2+(OM)2
(20)2=(x)2+(√3x-20)2 [from equation (1)]
400=x2+3x2-2×20√3x+400 [(a-b)2=a2-2ab+b2]
400-400=4x(x-10√3x)
0=x-10√3x
x=10√3x
SD=2SM
SD=2x=2×10√3m
SD=20√3m
SD=20×1.732m
SD=34.640m
Answer :
The length of the string of each phone is 34.640m.
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