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In the figure above, A is the center of the circle,

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GreenlightTestPrep
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In the figure above, A is the center of the circle, [#permalink] New post   27 Jan 2019, 10:15
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In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

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In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 05:23
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GreenlightTestPrep wrote:
Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

Show: ::
7


Let x = the length of AF
Image
This means AD = x + 5 = radius of the circle.

This is convenient, because AE is also a radius of the circle.
So, AE must have length x + 5
Image

At this point, we can focus on the RIGHT TRIANGLE below:
Image
When we apply the Pythagorean Theorem we get: (x + 5 )² + x² = 25²
Expand to get: x² + 10x + 25 + x² = 625
Simplify: 2x² + 10x + 25 = 625
Set equal to zero: 2x² + 10x - 600 = 0
Divide both sides by 2 to get: x² + 5x - 300 = 0
Factor: (x + 20)(x - 15) = 0
So, EITHER x = -20 or x - 15
Since x cannot be negative, we know that x = 15

So, let's add this to our diagram.
Image
Also, notice that I added some symbols to represent the 3 angles in ∆EAF

At this point we might recognize that ∆EAF and that ∆ECB are similar triangles
Here's why:
Since EB is the diameter of the circle, we know that ∠C is 90°
Also, both triangles have ∠E in common.
If ∆EAF and ∆ECB share two angles, then the 3rd angles must also be equal.
So, the two triangles must be similar.
Image

Let y = the length of CF
So, side EC has length 25 + y

If two triangles are similar, the ratio of their corresponding sides must be equal.
We can write: 25/40 = 20/(25 + y)
Cross multiply to get: 25(25 + y) = (40)(20)
Expand: 625 + 25y = 800
Then: 25y = 175
So y = 7

In other words, CF = 7

Answer: 7

Cheers,
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Re: In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 03:33
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GreenlightTestPrep wrote:
Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

Show: ::
7



Explanation::

Let's take the triangle EDA, since it is a right angled triangle and the sides are arranged in triplets

as one side is 25 so the other two sides are arranged in 15 : 20 : 25

AE and AD is the radius of the circle and

therefore AE = 20 and AF = 15

Now for the triangle EBD , this is also a right angled triangle with ∠C=90

Since AE = 20 therefore EB = 40

Again as this is also a right angled triangle so the sides are

24 : 32 : 40

EC cannot be 24, since EF = 25

Therefore EC = 32 and BC = 24

Hence FC = EC - 25 = 32 -25 = 7
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Re: In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 04:57
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GreenlightTestPrep wrote:
Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

Show: ::
7



Take triangle EFA, it is a right angled triangle.
But can we take the sides as 3:4:5, just because the hypotenuse 25 fits in .. NO
so if EA is the radius AF = r-5, so r2+(r−5)2=252.....r2+r2−10r+25=625......2r2−10r−600=0.....r2−5r−300=0.....(r−20)(r+15)=0
So, r is 20 or -15, but r cannot be negative, hence r=20..

Now, if you know a shortcut, answer is straight..
the product of two intersecting chords is same..
For example, If Ab and CD are two cords intersecting at O, AO*OB=CO*OD
here extend DA further to meet the circumference at G, then DG and CE are two chords intersecting at F, so DF*FG=CF*FE......(r+r-5)*5=CF*25
35*5=CF*25....CF=5∗3525=7
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Re: In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 04:58
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pranab01 wrote:
GreenlightTestPrep wrote:
Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

Show: ::
7



Explanation::

Let's take the triangle EDA, since it is a right angled triangle and the sides are arranged in triplets

as one side is 25 so the other two sides are arranged in 15 : 20 : 25

AE and AD is the radius of the circle and

therefore AE = 20 and AF = 15

Now for the triangle EBD , this is also a right angled triangle with ∠C=90

Since AE = 20 therefore EB = 40

Again as this is also a right angled triangle so the sides are

24 : 32 : 40

EC cannot be 24, since EF = 25

Therefore EC = 32 and BC = 24

Hence FC = EC - 25 = 32 -25 = 7



You are wrong in the highlighted portion.
It is not necessary that a hypotenuse of 25 would mean it is a triplet 15:20:25
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Re: In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 05:25
chetan2u wrote:
pranab01 wrote:
GreenlightTestPrep wrote:
Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

Show: ::
7



Explanation::

Let's take the triangle EDA, since it is a right angled triangle and the sides are arranged in triplets

as one side is 25 so the other two sides are arranged in 15 : 20 : 25

AE and AD is the radius of the circle and

therefore AE = 20 and AF = 15

Now for the triangle EBD , this is also a right angled triangle with ∠C=90

Since AE = 20 therefore EB = 40

Again as this is also a right angled triangle so the sides are

24 : 32 : 40

EC cannot be 24, since EF = 25

Therefore EC = 32 and BC = 24

Hence FC = EC - 25 = 32 -25 = 7



You are wrong in the highlighted portion.
It is not necessary that a hypotenuse of 25 would mean it is a triplet 15:20:25



Plz elaborate, in this case it is justified as the other 2 legs are radius of the circle and AD = AF + 5 = AE = Radius of the circle
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Re: In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 05:41
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pranab01 wrote:
chetan2u wrote:
pranab01 wrote:
Explanation::

Let's take the triangle EDA, since it is a right angled triangle and the sides are arranged in triplets

as one side is 25 so the other two sides are arranged in 15 : 20 : 25

AE and AD is the radius of the circle and

therefore AE = 20 and AF = 15

Now for the triangle EBD , this is also a right angled triangle with ∠C=90

Since AE = 20 therefore EB = 40

Again as this is also a right angled triangle so the sides are

24 : 32 : 40

EC cannot be 24, since EF = 25

Therefore EC = 32 and BC = 24

Hence FC = EC - 25 = 32 -25 = 7



You are wrong in the highlighted portion.
It is not necessary that a hypotenuse of 25 would mean it is a triplet 15:20:25



Plz elaborate, in this case it is justified as the other 2 legs are radius of the circle and AD = AF + 5 = AE = Radius of the circle


Hi..

Quote:
Let's take the triangle EDA, since it is a right angled triangle and the sides are arranged in triplets

as one side is 25 so the other two sides are arranged in 15 : 20 : 25


You have to derive it through pythagorean theorem, and cannot by merely looking at a 25 in hypotenuse claim it to be a triplet.
Anyway it was for learning, and if you already know it, we can drop the point here.
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Re: In the figure above, A is the center of the circle, [#permalink] New post   29 Jan 2019, 17:40
chetan2u wrote:

You have to derive it through pythagorean theorem, and cannot by merely looking at a 25 in hypotenuse claim it to be a triplet.
Anyway it was for learning, and if you already know it, we can drop the point here.


I am still not convinced, as my approach is valid for this ques only. Will get some more feedback from others.
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Re: In the figure above, A is the center of the circle, [#permalink] New post   03 Sep 2019, 09:15
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Regarding the pythagorean triple with only a right angle and hypotenuse. It seems to me like you need at least two sides.

As you can see in the attached file it is possible to make a right angled triangle with a hypotenuse of 25 and legs that are not 15 and 20.
Attachments

Screen Shot 2019-09-03 at 1.10.09 PM.jpg
Screen Shot 2019-09-03 at 1.10.09 PM.jpg [ 99.03 KiB | Viewed 7020 times ]

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Re: In the figure above, A is the center of the circle, [#permalink] New post   04 Sep 2019, 00:48
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arc601 wrote:
Regarding the pythagorean triple with only a right angle and hypotenuse. It seems to me like you need at least two sides.

As you can see in the attached file it is possible to make a right angled triangle with a hypotenuse of 25 and legs that are not 15 and 20.


Appreciate, if you can read the reasoning explained above.

Moreover,

In my explanation, I tried in a different way.

Yes, you are correct you cannot assume a right angled to be triplet with only one side. But, the reasoning I mentioned is only for this problem.

I guess, you will agree AE = AD = Radius of the circle. and FD =5

For the right angled triangle AEF,

let use Pythagorean triplet in the ratio :- 15:20:25

you can see, the hypotenuse EF = 25.

Again, AE = AD radius of the circle

if AE = 20 , then AD = 20 or AD = FD + AF = 5 + 15 = 20

This fits the case .

***Remember, we won't be able to solve if the information FD =5 is not mentioned.

Another way to solve:

Let x = the length of side AE

So x - 5 = the length of side AF

Applying the Pythagorean Theorem,

we get: x2+(x−5)2=252


or 2x2−10x−600=0
Divide both sides by 2 to get: x2−5x−300=0

Factor: (x−20)(x+15)=0

Since x must be POSITIVE, so x=20

Therefore AE = 15

Cheers!!!
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Re: In the figure above, A is the center of the circle, [#permalink] New post   27 Jun 2023, 07:11
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