Points on Parallel Chords

Note that because $\angle OEA$ and $\angle OFC = 90 ^ \circ$ and that $O$ is the center of the circle, $E$ and $F$ are the midpoints of $AB$ and $CD$ , respectively. Thus, $AE = 91$ and $CF = 60$ . From the Pythagorean theorem, we then have that $OE = \sqrt{109^2-91^2} = 60$ and $OF = \sqrt{109^2 - 60^2} = 91$ . Since $E$ lies between $O$ and $F$ , $EF = OF - OE = 91 - 60 = 31$ .

The first step is to draw the scenario described. Then, it becomes clear that we have four right triangles, two to the left of $\overline{OEF}$ and two to its right. Looking at two of the triangles on either side, like, for example, DFO and BEO, we can see that they are both right and that they share $\overline{OE}$ as part or all of one of their legs, but only DFO has the whole $\overline{OEF}$ as its leg. Since we have the hypotenuse given as 109 (the radius of the circle), and the other legs given (half of the lengths of the respective chords), then we can use the pythagorean theorem to solve for $\overline{OEF}$ and $\overline{OE}$ . Then, we simply subtract $\overline{OE}$ from $\overline{OEF}$ to get the length of $\overline{EF}$ .

We are given that the radius is 109. Angle OEA = 90 and angle OFC = 90, this suggests that E and F form an right triangle on AB and CD respectively. Thus, we have triangle AOB and triangle COD. OE and OF are perpendicular bisectors of chords AB and CD respectively. We can focus on the right triangles AOE and triangle COF. To get EF:

EF = OF - OE

CF^2 + OF^2 = radius^2

AE^2 + OE^2 = radius^2

CF = 120/2 = 60

AE = 182/2 = 91

radius = 109

Solving for OF and OE, we substitute the values on the equation above.

OF = 91

OE = 60

So, EF = OF - OE = 91 - 60 = 31.

since OE is perpendicular to chord AB ,it follows that EB=EA also,OF is perpendicular to chord CD,it follows that FD=FC

in right triangle EOB, EB=182/2=91 OB(radius)=109 (91)^2 + (OE)^2 = (109)^2 OE=60

in right triangle OFD FD=120/2=60 OD(radius)=109 (60)^2 +(OF)^2 = 109^2 OF=91

now, EF=OF-OE EF=91-60 EF=31

Since OE and OF are perpendicular , hence they will bisect the chord So CF=1/2 of CD =60 cm And AE = 1/2 of AB =91 cm AO = OC = 109 cm (radius) OE square= OA square -AE square ( Pythagoras theorem) = (109+91)(109-91) = 3600 OE= 60cm OF square=OC square-CE square = (109+60)(109-60) =7943 OF=91 cm EF=OF-OE = 91-60 = 31

Since angles OEA and OFC are both equal to 90 degrees, then the radius of the circle must bisect the two chords. Then, we can form two right triangles (OEA and OFC). Since the measurements of the radius and the two chords are given, we can find the distance of each chord from the center of the circle. OE^2=OA^2-EA^2=(109)^2-(91)^2=3600 OE=60 OF^2=OC^2-CF^2=(109)^2-(60)^2=8281 OF=91 EF=OF-OE=91-60=31

An angle drawn from the center of a circle to a point on the chord to the endpoint of a chord will be a right angle if and only if the point is at the center of the chord. Therefore, $E$ and $F$ are the midpoints of the chords they are on. This means that $AE=91$ and $CF=60$ . By Pythagorean Theorem, we get $OE=60$ and $OF=91$ . $EF=OF-OE=91-60=31$ .

According to the problem,we get two triangles-OAE and OCF .Using Pythogoras theorem,for triangle OCF,CO^2=CF^2+OF^2 or 109^2=60^2+OF^2 or OF^2=109^2-60^2=8281 or OF=91.Again,using Pythogras theorem for triangle OAE, AE^2+OE^2=OA^2 or OE^2=OA^2-AE^2=109^2-91^2=3600 or OE=60.Now,OF=EF+OE OR EF=OF-OE=91-60=31.So,the the length of EF is 31.

The radius of the circle is 109.

So, AO = CO = 109

Since OEA is a right angle, AE = 1/2 AB

AE = 182/2 = 91

Following this, CF = 1/2 CD

CF = 120/2 = 60

Using Pythagoras Theorem,

OE = Square root of ( 109^2 - 91^2 ) = 60

OF = Square root of ( 109^2 - 60^2 ) = 91

Hence, EF = 91-60 = 31

Since OA=OB for they are radii, triangle OAB should be an isosceles triangle, so then the height/projection of point O to AB/line OE should make point E divides line AB evenly into two parts with the same length. And it's the same with triangle OCD, for making F divides CD evenly.

And so we get AE=EB= $\frac{1}{2} . 182 = 91$

And CF=FD= $\frac{1}{2} . 120 = 60$

Then see triangle OCF. It's a right-angled triangle because $\angle$ OFC is known to have $90^\circ$ . So since we know the length of the radius OC (=109) and one of the side : CF (=60), we can simply use the Phytagoras theorem to know the length of OF. $OF = \sqrt{OC^2 - CF^2} = \sqrt{109^2 - 60^2} = 91$

Then see triangle OAE. It's a right-angled triangle because $\angle$ OEA is known to have $90^\circ$ . So since we know the length of the radius OA (=109) and one of the side : AE (=91), we can simply use the Phytagoras theorem to know the length of OE. $OE = \sqrt{OA^2 - AE^2} = \sqrt{109^2 - 91^2} = 60$

Since E lies between O and F, we should know that OF = OE + EF. So then EF=OF-OE=91-60=31, hence the answer to the question.

Line OE bisects AB and CD at a right angle. Therefore CF = 60 and AE = 91

So OEA is a right angle triangle with hypotenuse 109 and one side 91. To work out OE, you use the Pythagoras theorem, i.e. 109^2 - 91^2 = OE^2= 3600. This means OE = 60 And OFC is a right angle triangle with hypotenuse 109 and one side 60. That means OF = 91 because of Pythagoras i.e.109^2 - 60^2= 91^2

So EF = 31.

We draw the diagram with the given data.. From the diagram we observe that triangle OAE is a right angled triangle. Applying Pythagora's theorem: OE^2 + EA^2 = OA^2 Putting the given data we get OE = 60 Again triangle OCF is a right angled triangle. Applying Pythagora's theorem: OF^2 + FC^2 = OC^2 Putting the given data we get OF = 91 Now EF = OF - OE = 91 - 60 = 31

$EF=OF-OE=\sqrt{OC^2-CF^2}-\sqrt{OA^2-AE^2}\\=\sqrt{109^2-(\frac{120}{2})^2}-\sqrt{109^2-(\frac{182}{2})^2}=31$

The two chords lie on only one half of the circle. Since $AB$ and $CD$ are parallel and $\angle OEA=\angle OFC=90^\circ$ , it follows that $OE$ and $OF$ are collinear. By forming right triangles $OEA$ and $OFC$ and applying the Pythagorean Theorem, we get $OE=60$ and $OF=91$ . Thus, $EF=31$ .

Refer to diagram here: https://cacoo.com/diagrams/nOKwbEq1iAQcUAI6 There are two right triangles: AEO and OFD. For triangle AEO, we know that AO=109 because it is a radius of the circle. AE = AB/2 = 182/2 = 91 (a radius perpendicular to a chord bisects the chord). By the Pythagorean theorem, (AE)^2 + (OE)^2 = (AO)^2, and OE = sqrt((AO)^2 - (AE)^2) = sqrt(109^2 - 91^2) = 60. Similarly for triangle OFD, OD=109 because it is a radius of the circle. FD = CD/2 = 120/2 = 60. By the Pythagorean theorem, (FD)^2 + (OF)^2 = (OD)^2, and OF = sqrt((OD)^2 - (FD)^2) = sqrt(109^2 - 60^2) = 91. Since OE + EF = OF, EF = OF - OE = 91-60 = 31.

OA = OC =109 Also, EA = EB = 91 FC = FD = 60

In OAE, /(OE^2/) = /(OA^2/) - /(AE^2/) OE = 60 In OCF, /(OF^2/) = /(OC^2/) - /(CF^2/) OF = 91

EF = OF - OE EF = 31

Since $\angle OEA = \angle OFC = 90^\circ$ thus $OE$ bisects $AB$ and $OF$ bisects $CD$ . Since $AB$ and $CD$ are parallel chords, we have that $OEF$ is a straight line.

Applying the Pythagorean theorem on triangle $OEA$ , we have $109 ^2 = OA^2 = OE^2 + AE^2 = OE^2 + 91 ^2 \Rightarrow OE = 60$ .

Applying the Pythagorean theorem on triangle $OCF$ , we have $109 ^2 = OC^2 = OF^2 + CF^2 = OF^2 + 60 ^2 \Rightarrow OF = 91$ .

Thus $EF = OF - OE = 91 - 60 = 31$ .

given: AB=182 CD=120 NOW;

        SQRT( R^2-(CD/2)^2)=91

SINCE

      LEANGTH OF CHORD IS INVERSLY PROPOTIONAL OF DIST^N
       FROM CENTER

SO

   AB/CD = 91/H

H=60

. . . EF= 90 - 60 = 31 ANS.