Quadrilateral ABCD is inscribed in circle K. The diameter of

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