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.

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

==> DE = \sqrt{KD^2-EK^2} ==> DE = \sqrt{96}

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

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]
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]
05 Oct 2012, 01:19

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 6090 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]
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]
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 5975 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]
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]
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.