Circles HELP! (was: Polygons Help)

• January 12th 2007, 05:22 AM
Ruler of Hell
Circles HELP! (was: Polygons Help)
Ok, I'm back again...
Here are a few polygons sums.

1. An isosceles triangle whose sides are 13 cm, 13cm and 10cm is inscribed in a circle. Find the radius of the circle

2. Two parallel chords of a circle are of lengths 6cm and 8cm respectively and lie on the same side of the center. If the radius of the circle is 5cm, find the distance between the chords.

3. AB and CD are two parallel cords. AB = 10cm, CD = 24 cm. Distance between AB and CD is 17 cm. Find the radius of the circle

See, in the last one, my teacher first tried it out assuming both were on the same side of the center just to explain to us in detail. But I didn't understand one part in it.
If suppose both chords lie on the same side of the center... Then,
Draw a perpendicular. So O (The center of the circle), to P (The bisecting point on the smaller chord AB of the perpendicular ) and O to Q (The bisecting point of the larger chord CD of the perpendicular ).

Now according to Pythagoras' Theorem...
S²+S² = Hypotenuse²

Draw a radius (which is also the hypotenuse) to both the chords...
Therefore,

OQ²+QD² = OD²

Now if you take the radius as x, then you need to take OP as y right?
Since OQ = y + OP

Now if somebody can explain further. I know you can do substitution, but I got stuck in some place. See, I know that this is just gonna be a waste of time, since I KNOW that the chords need to be on different sides of the center, but still, if somebody can explain

Also please help me with the remaining sums... THANKS A LOT GUYS! YOU ROCK! (Please Hurry :p Sorry)
• January 12th 2007, 07:08 AM
ThePerfectHacker
Quote:

Originally Posted by Ruler of Hell
Ok, I'm back again...
Here are a few polygons sums.

1. An isosceles triangle whose sides are 13 cm, 13cm and 10cm is inscribed in a circle. Find the radius of the circle

Do you know the "Extended Law of Sines"?

To find the radius you need to use,
$2R=\frac{a}{\sin \alpha}$.

1)Draw an Isoseles triangles with 13 13 and 10.
2)From the vertex drop a perpendicular to base side.
3)The base will be bisected and now length 5.
4)The height will be, by Pythagorean theorem 12.

Now the cosine of half the vertex angle is,
$\cos \frac{\alpha}{2} = \frac{12}{13}$

Thus, by half-angle identity,
$\sin^2 \alpha = \frac{1- \cos \frac{\alpha}{2} }{2}$
$\sin^2 \alpha = \frac{1-\frac{12}{13}}{2}$
$\sin^2 \alpha =\frac{1/13}{2}=\frac{1}{26}$
Thus,
$\sin \alpha = \frac{\sqrt{26}}{26}$

Using the extended law of sines,
$2R=\frac{a}{\sin \alpha}=\frac{10}{\frac{\sqrt{26}}{26}}=10\sqrt{26 }$
Thus,
$R=5\sqrt{26}$
• January 12th 2007, 07:21 AM
ThePerfectHacker
The upper red line is divided into three parts.

Call the first section $x$.
Then the remaining (diameter) is $10-x$.

By the chords theorem in a circle we have,
$x(10-x)=3\cdot 3=9$
$-x^2+10x=9$
$x^2-10x+9=0$
$(x-1)(x-9)=10$
$x=1,9$.
But we are looking at the smaller peice.
Thus, $x=1$.

Now let us call the middle red line (the distance between the red lines because it is perpendicular) to be $x$.

Thus, by chords theorem we have,
$(x+1)(5+5-(x+1))=(4)(4)$
Solve.
• January 12th 2007, 08:11 AM
CaptainBlack
Quote:

Originally Posted by Ruler of Hell

2. Two parallel chords of a circle are of lengths 6cm and 8cm respectively and lie on the same side of the center. If the radius of the circle is 5cm, find the distance between the chords.

3. AB and CD are two parallel cords. AB = 10cm, CD = 24 cm. Distance between AB and CD is 17 cm. Find the radius of the circle

See, in the last one, my teacher first tried it out assuming both were on the same side of the center just to explain to us in detail. But I didn't understand one part in it.
If suppose both chords lie on the same side of the center... Then,
Draw a perpendicular. So O (The center of the circle), to P (The bisecting point on the smaller chord AB of the perpendicular ) and O to Q (The bisecting point of the larger chord CD of the perpendicular ).

Now according to Pythagoras' Theorem...
S&#178;+S&#178; = Hypotenuse&#178;

Draw a radius (which is also the hypotenuse) to both the chords...
Therefore,

OQ&#178;+QD&#178; = OD&#178;

Now if you take the radius as x, then you need to take OP as y right?
Since OQ = y + OP

Now if somebody can explain further. I know you can do substitution, but I got stuck in some place. See, I know that this is just gonna be a waste of time, since I KNOW that the chords need to be on different sides of the center, but still, if somebody can explain

Also please help me with the remaining sums... THANKS A LOT GUYS! YOU ROCK! (Please Hurry :p Sorry)

You need the intersecting chord theorem. In this case the diameter that
bisects the chord is divided into two pieces one of length x and the other
of length 2r-x. Then if the length of the chord is c

x(2r-x)=(c/2)^2

You solve for x for both the given chords (choose x so its < 2r-x, that is the
less of the roots of the quadratic) then the distance you require is the difference
between the x's for the two chords.

RonL
• January 12th 2007, 08:32 AM
Ruler of Hell
I'm 13 and I'm in the 8th grade. I haven't learned anything about the Extended Law of Sines...:confused: :confused: Btw, what sum was the next one you showed?

I've done not much complicated stuff in this topic so I'm sorry..
• January 12th 2007, 02:09 PM
ticbol
Quote:

Originally Posted by Ruler of Hell
Ok, I'm back again...
Here are a few polygons sums.

1. An isosceles triangle whose sides are 13 cm, 13cm and 10cm is inscribed in a circle. Find the radius of the circle

2. Two parallel chords of a circle are of lengths 6cm and 8cm respectively and lie on the same side of the center. If the radius of the circle is 5cm, find the distance between the chords.

3. AB and CD are two parallel cords. AB = 10cm, CD = 24 cm. Distance between AB and CD is 17 cm. Find the radius of the circle

See, in the last one, my teacher first tried it out assuming both were on the same side of the center just to explain to us in detail. But I didn't understand one part in it.
If suppose both chords lie on the same side of the center... Then,
Draw a perpendicular. So O (The center of the circle), to P (The bisecting point on the smaller chord AB of the perpendicular ) and O to Q (The bisecting point of the larger chord CD of the perpendicular ).

Now according to Pythagoras' Theorem...
S²+S² = Hypotenuse²

Draw a radius (which is also the hypotenuse) to both the chords...
Therefore,

OQ²+QD² = OD²

Now if you take the radius as x, then you need to take OP as y right?
Since OQ = y + OP

Now if somebody can explain further. I know you can do substitution, but I got stuck in some place. See, I know that this is just gonna be a waste of time, since I KNOW that the chords need to be on different sides of the center, but still, if somebody can explain

Also please help me with the remaining sums... THANKS A LOT GUYS! YOU ROCK! (Please Hurry :p Sorry)

It's Saturday morning, no work, plenty of spare time. So let me show you one way to do your 3 problems here. By the way, one problem per posting is the best. Three is terrible. But what the heck. It's Saturday here!

----------------------------
[Problem #1.]
An isosceles triangle whose sides are 13 cm, 13cm and 10cm is inscribed in a circle. Find the radius of the circle

Draw the figure on paper. Draw the 10 side as horizontal, as the base of the isosceles triangle. Draw the altitude from the apex to the midpoint of the base--this altitude will pass through the center of the circle. From center of the circle, draw a radius to any of the base corners. So now you have two right triangles: a small one inside a large one.
Let r = radius of the circle
And x = altitude of the small right triangle.
So,

Small right triangle has these:
---vertical leg = x cm
---horizontal leg = 10/2 = 5 cm
---hypotenuse = r cm

Large right triangle has these:
---vertical leg = (r+x) cm
---horizontal leg = 10/2 = 5 cm
---hypotenuse = 13 cm

Using the Pythagorean Theorem on the large right triangle, [although by inspection, if you have the experience, (r+x) = 12 because of 5-12-13 triangle.]
13^2 = (r+x)^2 +5^2
169 = (r+x)^2 +25
(r+x)^2 = 169 -25 = 144
(r+x) = 12
x = 12 -r ------------(1)

Using the Pythagorean Theorem on the small right triangle,
r^2 = x^2 +5^2
Plug in the x from (1),
r^2 = (12-r)^2 +25
r^2 = 144 -24r +r^2 +25
24r = 169
r = 169/24 cm, or 7.04 cm -----------answer.

-----------------------------------------------------------
[Problem #2.]
Two parallel chords of a circle are of lengths 6cm and 8cm respectively and lie on the same side of the center. If the radius of the circle is 5cm, find the distance between the chords.

Imagine, or draw the figure on paper. For simplicity, draw the two parallel chords as vertical. Then draw a horizontal diameter. (As you know, a diameter is a chord that passes through the center of the circle.) Then, say above that diameter, draw radii to the ends of the two parallel chords.
So you have two right triangles.
Let x = horizontal leg of right triangle from the 8-cm chord, say, of triangle 1.
And y = horizontal leg of right triangle from the 6-cm chord, say, of triangle 2.

We are asked to find (y-x).

In triangle 1,
---vertical leg = 8/2 = 4 cm.
---horizontal leg = x cm
---hypotenuse = 5 cm
5^2 = 4^2 +x^2
25 = 16 +x^2
x^2 = 25 -16 = 9
x = 3 cm ----------------***

In triangle 2,
---vertical leg = 6/2 = 3 cm.
---horizontal leg = y cm
---hypotenuse = 5 cm
5^2 = 3^2 +y^2
25 = 9 +y^2
y^2 = 25 -9 = 16
y = 4 cm ----------------***

So, y-x = 4-3 = 1 cm.
Therefore, the distance between the two parallel chords is 1 cm. ------answer.

--------------------------------------------------------
[Problem #3.]
AB and CD are two parallel cords. AB = 10cm, CD = 24 cm. Distance between AB and CD is 17 cm. Find the radius of the circle.

The two parallel chords have to be on the same side of the center. If they are on opposite sides of the center, it is very difficult to solve the problem---if it can be solved at all.

Draw the figure on paper---almost the same figure as in Problem #2 above.
Let x = horizontal leg of right triangle from the 24-cm chord, or of triangle 3.
So, (x+17) = horizontal leg of right triangle from the 10-cm chord, or of triangle 4.

In triangle 3,
---vertical leg = 24/2 = 12 cm.
---horizontal leg = x cm
---hypotenuse = r cm
r^2 = 12^2 +x^2
r^2 = 144 +x^2 -------------------(i)

In triangle 4,
---vertical leg = 10/2 = 5 cm.
---horizontal leg = (x+17) cm
---hypotenuse = r cm
r^2 = 5^2 +(x+17)^2
r^2 = 25 +x^2 +34x +289
r^2 = 314 +34x +x^2 -------------(ii)

(ii) minus (i),
0 = 170 +34x
x = 170/(-34) = -5 cm -----uh-oh, cannot be. There are no negative distances if we are not talking about directions or positions. Something wrong.

Could it be that the distance between the two parallel chords is 12 cm---not 17 cm as given?
If so,then, in triangle 4,
---vertical leg = 10/2 = 5 cm.
---horizontal leg = (x+12) cm
---hypotenuse = r cm
r^2 = 5^2 +(x+12)^2
r^2 = 25 +x^2 +24x +144
r^2 = 169 +244x +x^2 -------------(iii)

(iii) minus (i),
0 = 25 +24x
x = 25/(-24) cm ---------negative, cannot be again.

Could it be that the distance between the two parallel chords is 7 cm---not 17 cm as given?
If so,then, in triangle 4,
---vertical leg = 10/2 = 5 cm.
---horizontal leg = (x+7) cm
---hypotenuse = r cm
r^2 = 5^2 +(x+7)^2
r^2 = 25 +x^2 +14x +49
r^2 = 74 +14x +x^2 -------------(iv)

(iv) minus (i),
0 = -70 +14x
x = -70/(-14) = 5 cm ---------OK.

Then, substitute that in (i),
r^2 = 144 +x^2 -------------------(i)
r^2 = 144 +5^2 = 169
r = sqrt(169) = 13

Uh-oh again.
r = 13 cm is okay with Triangle 3, but not with Triangle 4. The horizontal leg og Triangle 4 would be 5+17 = 22 cm, which is longer than the "hypotenuse" 13 cm.

There is really something wrong with Problem #3.
• January 12th 2007, 06:42 PM
Ruler of Hell
I've found the answer to Problem #3. It has to be on different sides of the center. Because when I tried keeping both the chords on the same side of the center, the answer came in negative..
Draw a circle... Draw one chord on one side of the center, and another on the other side. Now, Draw a perpendicular to both the chords.
So the figure should look like an "I".. Or something like that atleast.
Now, the chord on top is CD and the one on the bottom is AB.
From the midpoint of the circle (which I'll be naming as 'O') draw a radius to point D and to point B. These radii are also the hypotenuse of the two right angled triangles.
Now, the point where the perpendicular lines touch the chord... Name the point on CD as 'M' and the point where the perpendicular touches AB as 'N'.

Line OM = x
Therefore, Line ON = 17-x (Since the distance between the two chords is 17cm)
Now, using Pythagoras' Theorem, find the required equation for Triangle MDO and Triangle NBO.
You've found the equational value for both the hypotenuse.
But,
Hypotenuse 1 (Which Radius 1) = Hypotenuse 2
Solve the equation

Voila!

Btw, THANKS A LOT ticbol for helping me out in easier language..
*Click Thanks Button*
• January 12th 2007, 06:51 PM
Soroban
Hello, Ruler of Hell!

I have a different version of #3 . . .

Quote:

3. AB and CD are two parallel cords: $AB = 10$cm, $CD = 24$ cm.
Distance between $AB$ and $CD$ is 17 cm.
Find the radius of the circle,

The chords must be on opposite sides of the center.
Code:

              * * *         A*-----+-----*B         *  *    |  5 *  *       *    *  y|  *    *             *  |  * R       *      * | *      *       *        *O        *       *      *  |  * R    *           *    |    *       C*--------+--------*D         *      |  12  *           *    |    *               * * *

We have two isosceles triangles: $\Delta AOB$ and $\Delta COD$.
. . $OA = OB = OC = OD = R.$

Let $y$ = altitude of $\Delta AOB.$
Then $17 - y$ = altitude of $\Delta COD.$

In the upper-right right triangle: .[1] . $R^2\:=\:y^2 + 5^2$
In the lower-right right triangle: .[2] . $R^2\:=\:(17-y)^2 + 12^2$

Equate [1] and [2]: . $y^2 + 25 \:=\:(17-y)^2 + 144$

. . which simplifies to: . $34y \:=\:408\quad\Rightarrow\quad y \,=\,12$

Substitute into [1]: . $R^2\:=\:12^2 + 5^2 \:=\:169\quad\Rightarrow\quad\boxed{R \,=\,13}$

• January 12th 2007, 07:51 PM
Ruler of Hell
Soroban, thanks a lot!! I did the exact same thing except for I interchanged the position of the chords.
Btw, I think I must have accidentally clicked the Thanks button twice, so it no longer shows that I've thanked you....:( :( :confused: :confused: ... Thanks a lot all of you people!! Hurray, I've completed my sums! Thanks once again!:D