How to find the maximum area of a quadrilateral created by a circle circumscribed around a triangle

Question

ABC is a triangle with B = 60 degrees. AB = 5, BC = 4. A circle is circumscribed around ABC. D is a point on the circle between A and C (D is a point on the minor arc of the circumcircle bounded by A and C). What is the maximum area of the quadrilateral ABCD? Thanks. :)

So far, I have puzzled out that AC = sqrt21 and the radius of the circle is sqrt7, and that for ADC to be the maximum area, AD should be equal to DC (i think). Any help would be much appreciated, thanks so much!

Answer 1

Make AB the bottom of the triangle, with point C at the top.

Draw in the height H, which intersects the base at point E.

We now have two triangles: BCE and AEC

the upper angle of BCE must be 30 deg, and thus know that BE is 2---and therefore AE = 3.

H = 4 sin 60 = √12

Then AC = √(H² + AE²) = √(12 + 9) = √21

The diameter of the circle is any side divided by the sine of the opposing angle, so

d = AC / sin60 = √21 / (√3 / 2) = 2 * √7

and the radius r = √7

Point D is specified as being on the arc between A and C. The triangle formed is ACD. Call the base AC, and the height the distance from the base to point D. For maximum area, the height must be maximum, meaning that the height must bisect AC.

Since we know the chord length (AC) and the radius, we get the chord angle from:

c = 2r*sin(Ø/2)

Ø/2 = asin (c / 2r)

Ø = 2 asin (c / 2r) = 2 asin √21 / 2√7 = 120 deg.

From this, we get the distance from the center to the chord: d = rcos60 = r/2 = √7 / 2

The height of the ACD triangle is therefore also √7 / 2, and the area is ( √21 * (√7 / 2) ) / 2 = √147 / 4

The main triangle area is 5 * √12 / 2

Total area of the quadrilateral: √147 / 4 + 2.5√12

Draw a picture and CHECK THE CALCULATIONS