Kite Within a Square - Problem With Solution

A problem on finding the sine of the angle of a kite within a square is presented along with detailed solution.

Problem : In the figure below, ABCD is a square whose side has a length equal to 2 units. Points M and N are on the sides BC and CD such that x is the length of BM and DN. Express sin(a) as a function of x.

Solution to Problem :

  • The right triangles ABM and ADN are congruent. The area of one of them is given by

    (1/2)(2)(x) = x

  • We now subtract the areas of the two right triangles ABM and ADN from the area of the square to obtain the area of the kite AMCN.

    area of AMCN = (2)(2) - 2 x = 4 - 2x

  • The area of the kite AMCN can also be found as the area of the isosceles triangle AMN plus the right triangle MCN

    area of isosceles triangle AMN = (1/2) sin(a) (AM)(AN) = (1/2) sin(a) AM2 = (1/2) sin(a) (x2 + 4)

    area of MCN = (1/2)(x - 2)2

    area of AMCN = (1/2) sin(a) (x2 + 4) + (1/2)(x - 2)2

  • We now equate the two expressions for the area of the kite AMCN found above to obtain

    (1/2) sin(a) (x2 + 4) + (1/2)(x - 2)2 = 4 - 2x

  • Multiply all terms of the above equation by 2 to obtain

    sin(a) (x2 + 4) + (x - 2)2 = 2(4 - 2x)

  • Solve for sin(a) to obtain.

    sin(a) = (4 - x2) / (4 + x2)

  • As an exercise, put x = 0 in the expression of sin(a) obtained and find angle a. Check your answer using the geometrical figure shown above. Set x = 0 and check your answer. Any contradiction in your answers?


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