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 GMAT 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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Quadrilateral ABCD is inscribed in circle K. The diameter of K is 20. AC is perpendicular to BD. What is the area of ABCD?

First note that as AC is perpendicular to BD and CA is a diameter then triangles CDA and CBA are congruent (they are mirror images of each other). Thus the area of quadrilateral ABCD is twice the area of triangle CDA, which equals to 1/2*DE*CA=1/2*DE*20=10*DE (AC is perpendicular to BD means DE is the height of CDA), so area of ABCD=2*CDA=20*DE. All we need to find is the lengths of the line segment DE.

Alternately as ABCD is a kite (triangles CDA and CBA are congruent then CB=CD and AB = AD --> ABCD is a kite) then its area equals to \(\frac{d_1*d_2}{2}\) and as one diagonal is a diameter CA, which equals to 20, then we need the second diagonal BD which equals to 2*DE.

Next, you should know the following properties to solve this question: • A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So, as CA is a diameter then angles CDA and CBA are right angles.

• Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular DE divides right triangle CDA into two similar triangles CDE and DAE. Now, as in similar triangles, corresponding sides are all in the same proportion then CE/DE=DE/EA --> DE^2=CE*EA.

(1) AB = AD --> we knew this ourselves (from the fact that triangles CDA and CBA are congruent). Not sufficient.

(2) The length of CE is 8 --> EA=CA-CE=20-8=12 --> DE^2=CE*EA=8*12=96. Sufficient.

" Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle."
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Well, this is a nice one. Howevr, the given qtn can be easily solved with out using this.

the area of the Quadrilateral ABCD = 2 * area of trianlge ACD.

area of triangle ACD = 1/2*AC * DE = 1/2 * 20 * DE we knw that AC = 20 (i.e. diameter)

==>the area of the Quadrilateral ABCD = 2 * area of trianlge ACD = 2 ( 1/2*AC * DE) = 20*DE

Stmnt1: AB = AD..

We can move the BD line up/down on the diameter (AC) keeping that line (BD) perpendicular to AC and also keeping AB=AD. Well but the area of the Quad will change as the line (BD) moves Up/Down...JUST IMAGINE.... Hence this stmnt is of no use.

Stmnt2: CE=8

connect k and D making a line KD that is radius = 20/2 = 10

now we also know that CK (radius) = 10. hence KE = CK-CE = 10-8 = 2

Now we have EKD, the right angle triangle, with EK = 2 and KD = 10

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

05 Oct 2012, 00:35

Hi Bunuel,

As always a great explanation. However i could not understand this part..Could you please explain it. Thanks.

"First note that as AC is perpendicular to BD and CA is a diameter then triangles CDA and CBA are congruent (they are mirror images of each other). Thus the area of quadrilateral ABCD is twice the area of triangle CDA, "

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

05 Oct 2012, 01:19

1

This post was BOOKMARKED

There are two important properties that can help in answering this question:

1) In any quadrilateral with perpendicular diagonals, the area is given by half the product of the diagonals. 2) A diameter perpendicular to a chord bisects that chord. In addition, any point on the diameter, is equally distant from the endpoints of the chord. In the attached drawing, it means that necessarily AB = AD and BC = DC. AC being the perpendicular bisector of BD means that BE = ED.

Now to our question:

(1) AB = AD doesn't help. It follows from the property of the diameter perpendicular to a chord. In addition, think of moving the chord BD parallel to itself, it will remain perpendicular to the diameter AC, AB will remain equal to AD, but the area of the quadrilateral will change. Not sufficient.

(2) Since CE = 8 and CK is radius, then EK = 10 - 8 = 2. From the right triangle KED we can compute ED (using Pythagoras's) and having ED in fact we have BD = 2ED. Then the area of the quadrilateral can be computed as we have both diagonals. Sufficient.

Answer B.

Attachments

KiteInCircle.jpg [ 25.07 KiB | Viewed 15283 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

05 Oct 2012, 16:18

Bunnel, I am not sure about the solutions being proposed here. I think it should be E. All the solutions i see here are based on the assumptions that the quadrilateral is a kite and nowhere this is mentioned. I have my doubts about this assumption as nothing is given about other properties of kite. Ex :

The two line segments connecting opposite points of tangency have equal length.

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

06 Oct 2012, 00:44

vdadwal wrote:

Bunnel, I am not sure about the solutions being proposed here. I think it should be E. All the solutions i see here are based on the assumptions that the quadrilateral is a kite and nowhere this is mentioned. I have my doubts about this assumption as nothing is given about other properties of kite. Ex :

The two line segments connecting opposite points of tangency have equal length.

THanks,

In the attached drawing, in the triangle BKD, BK = DK = radius. It follows that triangle BKD is isosceles, so KE being perpendicular to BD (it is given that the diameter AC is perpendicular to BD) it bisects the base BD of the triangle. Or, you can compare the two triangles BKE and DKE, they are congruent. Therefore, AC is the perpendicular bisector of BD, and every point on it, is equally distant from B and D. Or, you can continue with comparing pairs of triangles - ABE congruent to ADE, therefore ABD isosceles (AB = AD). Similarly, BCD is isosceles, so BC = DC. It follows that ABCD is a kite.

Attachments

KiteInCircle-more.jpg [ 26.73 KiB | Viewed 15166 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Quadrilateral ABCD is inscribed in circle K. The diameter of K is 20. AC is perpendicular to BD. What is the area of ABCD?

First note that as AC is perpendicular to BD and CA is a diameter then triangles CDA and CBA are congruent (they are mirror images of each other). Thus the area of quadrilateral ABCD is twice the area of triangle CDA, which equals to 1/2*DE*CA=1/2*DE*20=10*DE (AC is perpendicular to BD means DE is the height of CDA), so area of ABCD=2*CDA=20*DE. All we need to find is the lengths of the line segment DE.

Alternately as ABCD is a kite (triangles CDA and CBA are congruent then CB=CD and AB = AD --> ABCD is a kite) then its area equals to \(\frac{d_1*d_2}{2}\) and as one diagonal is a diameter CA, which equals to 20, then we need the second diagonal BD which equals to 2*DE.

Next, you should know the following properties to solve this question: • A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So, as CA is a diameter then angles CDA and CBA are right angles.

• Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular DE divides right triangle CDA into two similar triangles CDE and DAE. Now, as in similar triangles, corresponding sides are all in the same proportion then CE/DE=DE/EA --> DE^2=CE*EA.

(1) AB = AD --> we knew this ourselves (from the fact that triangles CDA and CBA are congruent). Not sufficient.

(2) The length of CE is 8 --> EA=CA-CE=20-8=12 --> DE^2=CE*EA=8*12=96. Sufficient.

I always have confusion about taking which sides into consideration when we are equating corresponding corresponding sides, can you please throw some light into this...i.e how to write the equation CE/DE=DE/EA
_________________

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

16 Dec 2012, 04:31

EvaJager wrote:

There are two important properties that can help in answering this question:

1) In any quadrilateral with perpendicular diagonals, the area is given by half the product of the diagonals. 2) A diameter perpendicular to a chord bisects that chord. In addition, any point on the diameter, is equally distant from the endpoints of the chord. In the attached drawing, it means that necessarily AB = AD and BC = DC. AC being the perpendicular bisector of BD means that BE = ED.

Now to our question:

(1) AB = AD doesn't help. It follows from the property of the diameter perpendicular to a chord. In addition, think of moving the chord BD parallel to itself, it will remain perpendicular to the diameter AC, AB will remain equal to AD, but the area of the quadrilateral will change. Not sufficient.

(2) Since CE = 8 and CK is radius, then EK = 10 - 8 = 2. From the right triangle KED we can compute ED (using Pythagoras's) and having ED in fact we have BD = 2ED. Then the area of the quadrilateral can be computed as we have both diagonals. Sufficient.

Answer B.

Hello Eve,

Thanks for the your approach to handle this Q.I thought your approach was better

I wanted your inputs w.r.t Similarity of triangles.

Please refer to the attachment for the same Q. I am confused on how do we define the proportion of sides ie. Which side upon which side is proportional. Is there any general rule that we can follow.

This sometimes confuses me while solving Questions on Similarity of triangles

• Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular DE divides right triangle CDA into two similar triangles CDE and DAE. Now, as in similar triangles, corresponding sides are all in the same proportion then CE/DE=DE/EA --> DE^2=CE*EA.

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

10 Dec 2013, 12:31

Quadrilateral ABCD is inscribed in circle K. The diameter of K is 20. AC is perpendicular to BD. What is the area of ABCD?

(1) AB = AD AB = AD, BE = ED. However, we know nothing about BE = ED length. Depending on the vertical height of BD in relation to it's position up and down CA, the area of the quadrilateral can be of greatly different area. Insufficient.

(2) The length of CE is 8.

If the length of CE = 8 then EA = 12. Furthermore, we can derive the height of this triangle. We know that the diameter is 20 which means the center is at 10. If CE = 8 then EK = 2. We also know that the radius = 10 so we can find the length of ED. We now have the length of the base and the height which allows us to find area. Because CA bisects and is perpendicular to BD, we know that CD = CB and that DA = BA so the triangles are similar which means that they both have the same area. Sufficient.

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

03 May 2015, 07:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.
_________________

Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

30 Apr 2016, 23:38

bhandariavi wrote:

Quadrilateral ABCD is inscribed in circle K. The diameter of K is 20. AC is perpendicular to BD. What is the area of ABCD?

(1) AB = AD (2) The length of CE is 8.

Here is my 2 cents. We can solve it using a simple circle property and much faster/easier way. Also explained the property.

Statement 1 is clearly insufficient bcz we are not given the lengths of any sides or the other diagonal.

Remember this The perpendicular from the center of the circle to a chord bisects the chord. (why well create triangle by joining the two ends of a chord from center(2 radii). So we have an isosceles triangle. Now the perpendicular drawn from unequal angle vertex to the opposite side is the median, angle bisector and perpendicular bisector of that side. )

Statement2. So we know ED=BD. AC =20 so KC= 10 and KE = 2 and CE=8. Now we can easily find ED using pythagoras formula. So BE = 2 ED. now we can find the area of ABCD in 2 way. Let's say we dont know the area of kite = d1*d2/2. Still we can find area of triangle CBD and ABC bcz we know the base BE = sqrt(96) and Altitudes CE and AE correspondingly. Sum of those two area = Area of ABCD.

• Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular DE divides right triangle CDA into two similar triangles CDE and DAE. Now, as in similar triangles, corresponding sides are all in the same proportion then CE/DE=DE/EA --> DE^2=CE*EA.

Hope it's clear.

The ratio of any pair of corresponding sides is the same.

Bunuel, why we have taken CE/DE=DE/EA , why not CE/DE=EA/DE base to base and perpendicular to perpendicular. Please explain

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

05 May 2016, 01:43

is it clear from the question that K is the centre of the circle and the CA passes through the diameter? Option B would only be sufficient if this were true...

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

Show Tags

12 Aug 2016, 23:51

muralimba wrote:

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Well, this is a nice one. Howevr, the given qtn can be easily solved with out using this.

the area of the Quadrilateral ABCD = 2 * area of trianlge ACD.

area of triangle ACD = 1/2*AC * DE = 1/2 * 20 * DE we knw that AC = 20 (i.e. diameter)

==>the area of the Quadrilateral ABCD = 2 * area of trianlge ACD = 2 ( 1/2*AC * DE) = 20*DE

Stmnt1: AB = AD..

We can move the BD line up/down on the diameter (AC) keeping that line (BD) perpendicular to AC and also keeping AB=AD. Well but the area of the Quad will change as the line (BD) moves Up/Down...JUST IMAGINE.... Hence this stmnt is of no use

Stmnt2: CE=8

connect k and D making a line KD that is radius = 20/2 = 10

now we also know that CK (radius) = 10. hence KE = CK-CE = 10-8 = 2

Now we have EKD, the right angle triangle, with EK = 2 and KD = 10

==> the area of the Quadrilateral ABCD = 20*DE = \(20*\sqrt{96}\)

Answer B.

Regards, Murali.

Gr8 explanation : Just one doubt : how do we assume that ABCD is divided into two congruent triangles ABC and ADC ? Why are they mirror images of each other? Or a kite ? Thanks in advance
_________________

You have to dig deep and find out what it takes to reshuffle the cards life dealt you

gmatclubot

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of
[#permalink]
12 Aug 2016, 23:51

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

[rss2posts title=The MBA Manual title_url=https://mbamanual.com/2016/11/22/mba-vs-mim-guest-post/ sub_title=MBA vs. MiM :3qa61fk6]Hey, guys! We have a great guest post by Abhyank Srinet of MiM-Essay . In a quick post and an...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...