Last visit was: 21 Nov 2024, 16:12 It is currently 21 Nov 2024, 16:12
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

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

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [23]
Given Kudos: 136
Send PM

Top Member of the Month
In the figure above, A is the center of the circle, [#permalink] New post   27 Jan 2019, 10:15
4
19
Bookmarks
00:00

Question Stats:

44% (05:16) correct 55% (03:22) wrong based on 43 sessions

Hide Show timer Statistics

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
Most Helpful Community Reply
User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [6]
Given Kudos: 136
Send PM

Top Member of the Month
In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 05:23
6
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,
Brent
General Discussion
avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2273 [3]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
Re: In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 03:33
2
1
Bookmarks
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 \(\angle C = 90\)

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
avatar
chetan2u
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 470 [3]
Given Kudos: 0
Send PM
Re: In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 04:57
2
Expert Reply
1
Bookmarks
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 \(r^2+(r-5)^2=25^2.....r^2+r^2-10r+25=625......2r^2-10r-600=0.....r^2-5r-300=0.....(r-20)(r+15)=0\)
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....\(CF=\frac{5*35}{25}=7\)
avatar
chetan2u
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 470 [2]
Given Kudos: 0
Send PM
Re: In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 04:58
1
Expert Reply
1
Bookmarks
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 \(\angle C = 90\)

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
avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2273 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
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 \(\angle C = 90\)

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
avatar
chetan2u
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 470 [0]
Given Kudos: 0
Send PM
Re: In the figure above, A is the center of the circle, [#permalink] New post   28 Jan 2019, 05:41
Expert Reply
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 \(\angle C = 90\)

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.
avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2273 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
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.
User avatar
arc601
Manager
Manager
Joined: 19 Nov 2018
Posts: 102
Own Kudos [?]: 158 [0]
Given Kudos: 0
Send PM

Top Member of the Month
Re: In the figure above, A is the center of the circle, [#permalink] New post   03 Sep 2019, 09:15
1
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 6722 times ]

avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2273 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
Re: In the figure above, A is the center of the circle, [#permalink] New post   04 Sep 2019, 00:48
1
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: \(x^2 + (x - 5)^2 = 25^2\)


or \(2x^2 - 10x - 600 = 0\)
Divide both sides by 2 to get: \(x^2 - 5x - 300 = 0\)

Factor: \((x - 20)(x + 15) = 0\)

Since x must be POSITIVE, so \(x = 20\)

Therefore AE = 15

Cheers!!!
User avatar
bumpbot
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5030
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: In the figure above, A is the center of the circle, [#permalink] New post   27 Jun 2023, 07:11
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
gmatclubot
Re: In the figure above, A is the center of the circle, [#permalink]
27 Jun 2023, 07:11
Moderators:
adewale223
GRE Instructor
84 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1111 posts
VibhuAnurag
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact