Rectangle in circle with one of its side subtending 120 degrees on the circumference.

[asvk]

A rectangle is inscribed in circle of radius of 8 cm. A triangle is inscribed in this circle in such a way that one of its sides lies on the side of the ractangle. If angle subtended at the third vertex of the triangle (vertex not COMMON to the rectangle) is 120 degrees, then WHAT CAN BE THE MAXIMUM area of the RECTANGLE ?

Solution :

With the given I was able to calculate the angle between the digonals i.e. 60 degrees.

Area of rectangle = 1/2 * (length of diagonal) * (length of diagonal) * sin(60)
= 1/2 * 16 * 16 * sqrt(3)/2 = 64 * sqrt(3).

Area comes out to be constant FOR a given diagonal. so my question is "Is this the MAXIMUM area THE question WANTS ? ".
however the answer given is 36 * SQRT(3).

Please help.

 

[lookagain]

A rectangle is inscribed in circle of radius of 8 cm. A triangle is inscribed in this circle in such a way that one of its sides lies on > > > the side of the rectangle. < < <

Which side of the rectangle does one of the sides of the triangle lie on it!? A general rectangle has two pairs of different length sides.

 

[soroban]

Hello, asvk!

I agree with your answer.

A rectangle is inscribed in circle of radius of 8 cm.
A triangle is inscribed in this circle in such a way that one of its sides lies on the side of the rectangle.
If angle subtended at the third vertex of the triangle (vertex not COMMON to the rectangle) is 120 degrees,
then what can be the maximum area of the rectangle

First of all, the rectangle is unique.

                P
              * o *
       A  *           *  B
        o - - - - - - - o
       *| *           * |*
        |  8*  120d *8  |
      * |     *   *     | *
      * |       o 60d   | *
      * |     * O *     | *
        |   *8     8*   |
       *| *           * |*
        o - - - - - - - o
       D  *           *  C
              * * *

Circle \(\displaystyle O\!:\:OA = OB = OC = OD = 8\)
Draw chords \(\displaystyle PA\) and \(\displaystyle PB\!:\;\angle APB = 120^o\)
Note that \(\displaystyle P\) can be any point on arc \(\displaystyle \widehat{AB}.\)

We find that: .\(\displaystyle \angle AOB = 120^o,\;\angle BOC = 60^o\)

\(\displaystyle \Delta BOC\) is equilateral: .\(\displaystyle BC = 8\)

Law of Cosines \(\displaystyle (\Delta AOB)\!:\:AB^2 \:=\:8^2+8^2 - 2(8^2)\cos120^o \:=\:192\)
. . Hence: .\(\displaystyle AB \,=\,8\sqrt{3}\)

Therefore: .\(\displaystyle \text{Area} \:=\: (AB)(BC) \:=\: (8\sqrt{3})(8) \:=\:64\sqrt{3}\text{ cm}^2\)

 

[lookagain]

Which side of the rectangle does one of the sides of the triangle lie on it!? A general rectangle has two pairs of different length sides.

There is no "lies on the side of the rectangle," because a rectangle doesn't have just one length side. \(\displaystyle \ \ \ \ \) Also, the maximum area of the rectangle here is a direct consequence of the fact that the inscribed-in-circle hexagon that has the hexagon's maximum area when it is a regular hexagon. And regular hexagons have interior angles of 120 degrees.

 

[stapel]

I'll use the sketch provided earlier:

                P
              * o *
       A  *           *  B
        o - - - - - - - o
       *| *           * |*
        |  8*  120d *8  |
      * |     *   *     | *
      * |       o 60d   | *
      * |     * O *     | *
        |   *8     8*   |
       *| *           * |*
        o - - - - - - - o
       D  *           *  C
              * * *

It may be helpful to note that angle AOB is a central angle of 120*. (I'm using "*" to stand for "degrees".) Let "Q" be the point on the circle which is exactly "opposite" P (that is, Q lies at one end of the diameter whose other endpoint is P). Then angle AQB has to have measure 60*. Since APBQ is a cyclic quadrilateral, then angle APB, being opposite angle AQP, has measure 180* - 60* = 120* also.

Triangle BOC is equilateral because angle BOC has measure 60* and, since |OB| = |OC| = 8, then triangle BOC is isosceles.