In a cyclic quadrilateral, the diagonal would be the hypotenuse of two right-angled triangles.
Clearly if the hypotenuse , in this case the diagonal, is 4√3, then the two given sides of 4 and 5 cannot be sides of the same triangle.
I made the following sketch
Quad ABCD in a circle, with BD as the diagonal of 4√3
AB = 5 and BC = 4
in triangle ABD
AD^2 + 5^2 = (4√3)^2
AD^2 = 48-25 = 23
AD = √23
It should also be obvious that the other triangle BCD is congruent to the first one.
So total area = 2 (1/2)(5)(√23 = 5√23
Find the area of a cyclic quadrilateral whose 2 sides measure 4 & 5 units, & whose diagonal coincides with a diameter of the circle. Suppose the radius of the circumscribing circle is 2 sq.root of 3 units.
Please include solution. Thanks.
1 answer