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Quadrilateral ABCD is inscribed in circle K. The diameter of

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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.
[Reveal] Spoiler: OA

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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.

Answer: B.

Check this for more: math-triangles-87197.html
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Re: Help! [#permalink]

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New post 06 Feb 2011, 07:27
Great explanation Bunuel! Kudos!
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Re: Help! [#permalink]

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New post 09 Feb 2011, 05:10
Great stuff for enlightening about this axiom :-)

" Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle."
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Re: Help! [#permalink]

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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 = \(20*\sqrt{96}\)

Answer B.

Regards,
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Re: Help! [#permalink]

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New post 04 May 2011, 02:15
a not sufficient.giving no values.

b. triangles CED and CDA are similar.
Hence (CE/CD) = (CD/CA)

CD = 20 * 6 = 120 ^ 1/2

thus area of triangle CDA can be found,which multiplied by 2 will give area of ABCD
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Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

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New post 05 Oct 2012, 01:35
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, "

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

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

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New post 05 Oct 2012, 17:18
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.



THanks,

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

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New post 06 Oct 2012, 01:44
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
KiteInCircle-more.jpg [ 26.73 KiB | Viewed 17840 times ]


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Re: Help! [#permalink]

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New post 15 Dec 2012, 10:47
Bunuel wrote:
Attachment:
Circle.PNG
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.

Answer: B.

Check this for more: math-triangles-87197.html



Bunnel

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

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New post 16 Dec 2012, 05:31
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


Looking forward to your inputs.

Thanks
Mridul
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Quadrilateral ABCD is inscribed in circle K.doc [35.5 KiB]
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Re: Quadrilateral ABCD is inscribed in circle K. The diameter of [#permalink]

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New post 02 May 2013, 12:59
Quick question

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


What am I missing here? How do you get DE^2 = CE * EA??? is there a triangle property that I missed out on?

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

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rsworase wrote:
Quick question

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


What am I missing here? How do you get DE^2 = CE * EA??? is there a triangle property that I missed out on?


It's explained here: quadrilateral-abcd-is-inscribed-in-circle-k-the-diameter-of-108776.html#p865608

Image

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

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New post 10 Dec 2013, 13:31
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.

Please see link for diagram.

http://i.imgur.com/LIkYtR0.png

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

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New post 01 May 2016, 00:38
bhandariavi wrote:
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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.

So answer is
[Reveal] Spoiler:
B

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

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New post 02 May 2016, 22:32
Bunuel wrote:
rsworase wrote:
Quick question

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


What am I missing here? How do you get DE^2 = CE * EA??? is there a triangle property that I missed out on?


It's explained here: quadrilateral-abcd-is-inscribed-in-circle-k-the-diameter-of-108776.html#p865608

Image

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

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

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New post 05 May 2016, 02:43
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...

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

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New post 13 Aug 2016, 00:51
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

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

==> 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
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Re: Quadrilateral ABCD is inscribed in circle K. The diameter of   [#permalink] 05 Oct 2017, 22:33
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Quadrilateral ABCD is inscribed in circle K. The diameter of

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