Question

Hewwo!

Answer 1

Answer:

I) 1 cm

II) 7 cm

Step-by-step explanation:

We are given a circle with a radius of 5 cm. AB and CD are two parallel chords of length 8cm and 6cm respectively. And we want to calculate the length of the chords if they are:

I) On the same side of the center.

Please refer to the first figure.

In it, O is the center, AB measures 8 and CD measures 6.

To find the distance between the two chords, we essentially need to find KJ, which is OJ - OK.

We will construct a segment from O to K and from K to J. This is perpendicular to the chords. Therefore, by the Perpendicular Bisector Theorem, it also bisects the chords.

ΔAOK is a right triangle. Therefore:

$AO^2=OK^2+KA^2$

AO is simply the radius.

KA is half the length of AB since OK perpendicularly bisects the chord.

Therefore:

$(5)^2=OK^2+(4)^2$

So:

$25=OK^2+16\Rightarrow OK^2=9\Rightarrow OK=3$

Likewise, ΔCOJ is also a right triangle. Therefore:

$CO^2=OJ^2+JC^2$

CO is the radius, and JC will be half the length of CD. Thus:

$(5)^2=OJ^2+(3)^2\Rightarrow OJ^2=16\Rightarrow OJ=4$

Therefore, the distance KJ between the two chords are:

$KJ= OJ-OK=4-3=1\text{ cm}$

II) On opposite sides of the center.

Please refer to the second figure.

Again, O is the center, AB measures 8, and CD measures 6.

Likewise, segments OK and OJ are perpendicular bisectors of AB and CD, respectively.

And to find the distance, we essentially need to find JK or OJ + OK.

ΔKOB is a right triangle. Thus:

$OB^2=OK^2+KB^2$

OB is the radius, and KB is half of AB:

$(5)^2=OK^2+(4)^2\Rightarrow OK=3$

ΔJOD is also a right triangle. Thus:

$OD^2=OJ^2+JD^2$

OD is the radius and JD is half of CD:

$(5)^2=OJ^2+(3)^2\Rightarrow OJ=4$

Therefore, the distance between the two chords are:

$JK=OJ+OK=4+3=7\text{ cm}$

Answer 2

Answer:

AB=8cm⇒AE=4cmCD=6m⇒CF=3cmIn△AEDAE=4cm,AO=5cmOE=(AO)2−(AE)2=52−42=3cmIn△OFCCF=3cm,CO=5cmOF=OC2−CF2=52−32=4cm

∴ Distance between the chords=4cm−3cm=1cm