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.

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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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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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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.

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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.

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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

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



• 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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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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