O is the center of the circle of radius 5 cm. AB and CD are two par

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O is the center of the circle of radius 5 cm. AB and CD are two parallel chords of the circle on the same side of O. OP is perpendicular to AB, 
OQ is perpendicular to CD, AB = 6 cm, CD = 8 cm. Find PQ.

A. 1
B. 1.2
C. 2
D. 2.2
E. 2.7

Raxit85 · Answer

This seems more tricky one. Still, let me try.

AB & CD are parallel and line OP is bisecting AB & CD. 
So, CQ=QD=4 & AP=PB=3
If we join line from O to B and O to C and radius = 5, so, OB and OD can be 5.
As OPB is right angle triangle, OP^2 + PB^2 = OB^2 
So, OP=4.
Similarly, OQD is right angle triangle, OQ^2 + QD^2 = OD^2 
So, OQ=3.
PQ=OP-OQ=1.

Imo. A

ashwini2k6jha · Answer

ANSWER:A
PQ= OP- OQ
OP^2=OB^2-PB^2
OP^2=25-9
OP=4
OQ^2=OD^2-QD^2
OQ=25-16
OQ=3
PQ=4-3=1

uc26 · Answer

A perpendicular from the center of the circle to the chords bisects the chords.
OQD forms a 3-4-5 right triangle where OQ=3
Similarly OPB forms a 3-4-5 right triangle where OP=4
Thus PQ= OP-OQ= 1

Answer: A

Archit3110 · Answer

given that Chords CD and AB are on same side of O 
and OP is perpendicular to AB, OQ is perpendicular to CD, AB = 6 cm, CD = 8 cm
we know OA = OC = 5cm target is to find distance b/w two chords PQ

AP= 3 cm and CQ = 4cm
so ∆OQC ; OC^2 = OQ^2+CQ^2 ; we can find OQ ; 25-16 ; 9cm ; 3 cm
similarly for ∆ AOP ; AO^2 = OP^2 + AP^2 ; 25-9 ; 16 ; 4cm

PQ distance ; OP-OQ ; 4-3 ; 1 cm
OPTION A

O is the center of the circle of radius 5 cm. AB and CD are two parallel chords of the circle on the same side of O. OP is perpendicular to AB,
OQ is perpendicular to CD, AB = 6 cm, CD = 8 cm. Find PQ.

A. 1
B. 1.2
C. 2
D. 2.2
E. 2.7

Gamerboyttt · Answer

A. 
Op=4
OQ=3 
By help of Pythagoras theorem
Difference pq is equal to 1

Posted from my mobile device

SohGMAT2020 · Answer

O is the center of the circle of radius 5 cm. AB and CD are two parallel chords of the circle on the same side of O. OP is perpendicular to AB,
OQ is perpendicular to CD, AB = 6 cm, CD = 8 cm. Find PQ.

A. 1
B. 1.2
C. 2
D. 2.2
E. 2.7

Solution:

Attachment:

Capturegmat.JPG [ 21.4 KiB | Viewed 3128 times ]

Here AB and CD are parallel, and OP and OQ are perpendicular to AB and CD.

Hence, OQP is the same line perpendicular to AB and CB.

So OPA is a right-angled triangle,

In which AO = 5, AB = 6/2 = 3, therefore, OP =\(\sqrt{5^2-3^2} = 4 \)

OQC is also a right-angled triangle,

In which CO = 5, CQ = 8/2 = 4, therefore, OQ =\(\sqrt{5^2-4^2} = 3 \)

PQ = OP - OQ = 4- 3 = 1.

Hence, answer is A

yashikaaggarwal · Answer

Draw line OC and OA. Such that two right angle triangle forms OCQ right angle at Q
OAP right angle at P

In triangle OCQ,
OC^2=OQ^2+CQ^2
5^2 = OQ^2 + 4^2 (since OQ is perpendicular on CD, bisecting CD in two equal parts)
25-16 = OQ^2 
9 = OQ^2 => OQ = 3 --------(1)

In triangle OAP,
OA^2=OP^2+AP^2
5^2 = OP^2 + 3^2 (since OP is perpendicular on AB, bisecting AB in two equal parts)
25-9 = OP^2 
16 = OP^2 => OP = 4 --------(2)

OP = OQ+PQ
4 = 3+PQ
1 = PQ

Answer is A

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ArunSharma12 · Answer

radius = 5cm
AB and CD are on the same side and AB is further away from diameter with CD in between AB and diameter.

as OP is perpendicular to AB, it divides AB in two equal parts of 3cm each. 
OP^2 + PB^2 = OB^2
OP^2 = 25 - 9 = 16; OP =4

similarly OQ divides CD in two equal parts of 4 cm each
OQ^2 + QD^2 = OD^2
OQ^2 = 25 - 16 = 9; OQ = 3

PQ = OP - OQ = 4-3 = 1
Ans: A

gmatway · Answer

IMO A =1 
we can make 2 right angles and find diffrence .
as ab and cD are ll and op and oq are perpendicular to both cords they will divide it in equal parts respectively .
hence in triangle OQD we get to know OD=5 ,QD=4 ,hence using right ans triangle OQ =3 
in another right triangle OPB , OB=5 ,PB=3 hence OP =4 .
we need diffrence OP-OQ = 4-3 =1

tejas29d · Answer

OA is 1 
pythagoras theorem is to be used for this case 
OP =4 OQ =3 PQ =1

Kinshook · Answer

Given:
1. O is the center of the circle of radius 5 cm.
2. AB and CD are two parallel chords of the circle on the same side of O.
3. OP is perpendicular to AB,
4. OQ is perpendicular to CD,
5. AB = 6 cm,
6. CD = 8 cm.

Asked: Find PQ.

Attachment:

Screenshot 2020-06-07 at 12.47.22 PM.png [ 26.93 KiB | Viewed 3033 times ]

In \(\triangle OAP \):
\(\angle OPA = 90^0\)
\(PA = PB = \frac{AB}{2} = \frac{6}{2 }= 3\)
\(PO = \sqrt{AO^2 - AP^2 } = \sqrt{5^2 - 3^4} = 4\)

In \(\triangle OCQ \):
\(\angle OQC = 90^0\)
\(QC = QD = \frac{CD}{2} = \frac{8}{2} = 4\)
\(QO = \sqrt{CO^2 - CQ^2 } = \sqrt{5^2 - 4^4} = 3\)

PQ = PO - QO = 4 - 3 = 1

IMO A

HoneyLemon · Answer

A,  IMO

AB = 6 , so AP = 3 . Hence OP =4 
CD=4 , so CQ=4 Hence OQ = 3 
NOW PQ = OP-OQ = 4-3 =1

coreyander · Answer

Since, AB is shorter than CD, CD will be closer to center O than AB is. 
 A perpendicular from radius to a chord bisects the chord into two equal parts.

Thus, with the above information, CQ = 4 & AP= 3.

From  	riangle AOP , we know AO=5 (radius), AP = 3.
By Pythagorus theorem, OP = 4.

From  	riangle COQ , we know CO=5 (radius), CQ = 4.
By Pythagorus theorem, OQ = 3.

Thus, PQ=PO-QO
 Thus, PQ = 4 -3 = 1.
Answer is (A) 1

bM22 · Answer

W.r.t given figure: AP = 3cm, PQ = 4cm, OC = 5cm, CQ = 4cm.
By Pythagoras theorem, in triangle CQO, we get: OQ = 3cm.

Let PQ = x,
In AOP, 3^2 + (3+x) ^2 = 5^2.
=> 9 + 9 + x^2 + 6x = 25
=> x^2 + 6x - 7 = 0.
we get x as 1,-7.

O is the center of the circle of radius 5 cm. AB and CD are two par

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O is the center of the circle of radius 5 cm. AB and CD are two par

Prep Toolkit

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