AB=6cm
CD=8 cm
Radius=5cm
Answers
Here, Given that =>
AB= 6cm
CD=8 cm and
Radius(R) =5cm
Construction =>
Join O to C, O to R, O to D and O to B.
Since the perpendicular from the centre to the chord bisect the cord.
Hence, we get=>
A) AP=PB=1/2×(AB)= 1/2×(6cm)
=>AP= PB = 3cm.
B) CQ= QD=1/2×(CD)=1/2×(8cm)
=>CQ=QD=4cm.
Now, In triangle OAP and triangle OQC we have=>
OA^2 = OP^2 + AP^2
=>OP^2 = OA^2 - AP^2
--------(i)
OC^2 = OQ^2 + CQ^2
=>OQ^2 = OC^2 - CQ^2
- - - - - - - -(ii)
So, on putting values on (i) and (ii) we get=>
OP^2 = (5)^2 - (3)^2
=> OP^2 = 16
=>OP =4cm.
And
OQ^2 = (5)^2 - (4)^2
=>OQ^2 = 9
=>OQ=3cm.
Hence ,
PQ = OP- OQ
=>PQ = 4cm-2cm
=>PQ = 1cm.
Remember
#Theorem
A perpendicular from the centre to the chord bisects the cord.
Here, O is center with OP ⊥ Ab and OQ ⊥ CD. Thus, by theorem OQ and OP are perpendicular bisectors.
Join OC and OC to form triangles OCQ and OAP respectively.
Since OQ is perpendicular bisector, CQ = 1/2 CD
CQ = 1/2 × 8 cm
CQ = 4 cm
Since OP is perpendicular bisector, AP = 1/2 × AB
AP = 1/2 × 6 cm
AP = 3 cm
Since OCQ and OAP are right triangles.
By Pythagorus theorem,
H² = B² + P²
5² = 4² + OQ²
25 = 16 + OQ²
9 = OQ²
3 cm = OQ___(1)
Now, OA² = OP² + AP²
r² = ( OQ + QP )² + 3²
5² = OQ² + QP² + 2(OQ)(QP) + 9
25 = 3² + QP²+ 2(3)(QP) + 9
25 - 18 = QP² + 6QP
7 = PQ² + 6QP
0 = PQ² + 6PQ - 7
0 = PQ² + 7PQ - 1PQ - 7
0 = PQ( PQ + 7 ) - 1( PQ + 7 )
0 = ( PQ - 1 )( PQ + 7 )
Each getting zero,
PQ + 7 = 0 and PQ - 1 = 0
PQ = - 7 and PQ = 1 cm
Since length can't be negative
So, Length of PQ is 1 cm.