The area of a kite is half the product of the diagonals.

\(\displaystyle A=\frac{p \cdot q}{2}\)

The diagonals of the kite are the height and width of the rectangle it is superimposed in, and we know that because the area of a rectangle is base times height.

Therefore our equation becomes:

\(\displaystyle A=\frac{l \cdot w}{2}\).

We also know the area of the rectangle is \(\displaystyle A=l \cdot w=80\). Substituting this value in we get the following:

\(\displaystyle A=\frac{l \cdot w}{2}=\frac{80}{2}=40\)

Thus,, the area of the kite is \(\displaystyle 40\).

The Quadrilateral \(\displaystyle KITE\) is shown below with its diagonals \(\displaystyle \overline{I E}\) and \(\displaystyle \overline{KT}\).

. We call the point of intersection \(\displaystyle X\):

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, \(\displaystyle \overline{KT}\) bisects the \(\displaystyle 60 ^{\circ }\) and \(\displaystyle 90^{\circ }\)angles of the kite. Consequently, \(\displaystyle \bigtriangleup KXI\) is a 30-60-90 triangle and \(\displaystyle \bigtriangleup TXI\) is a 45-45-90 triangle. Also, the diagonal that connects the common vertices of the pairs of adjacent sides bisects the other diagonal, making \(\displaystyle X\) the midpoint of \(\displaystyle \overline{IE}\). Therefore, 

\(\displaystyle IX = \frac{1}{2} \cdot IE = \frac{1}{2} \cdot 24 = 12\).

By the 30-60-90 Theorem, since \(\displaystyle \overline{IX}\) and \(\displaystyle \overline{KX}\) are the short and long legs of \(\displaystyle \bigtriangleup KXI\), 

\(\displaystyle KX = IX \cdot \sqrt{3}= 12 \sqrt{3}\)

By the 45-45-90 Theorem, since \(\displaystyle \overline{IX}\) and \(\displaystyle \overline{XT}\) are the legs of a 45-45-90 Theorem, 

\(\displaystyle XT = IX = 12\).

The diagonal \(\displaystyle \overline{KT}\) has length 

\(\displaystyle KT =XT + KX= 12+ 12\sqrt{3}\).

In order to solve this problem, first observe that the red diagonal line divides the kite into two triangles that each have side lengths of \(\displaystyle 15\) and \(\displaystyle 8.\) Notice, the hypotenuse of the interior triangle is the red diagonal. Therefore, use the Pythagorean theorem: \(\displaystyle a^2+b^2=c^2\), where \(\displaystyle c=\) the length of the red diagonal. 

The solution is: 

\(\displaystyle 8^2+15^2=c^2\)

\(\displaystyle 64+225=c^2\)

\(\displaystyle c^2=289\)

\(\displaystyle c=\sqrt{289}=\sqrt{17\times 17}=17\)

To solve this problem, apply the formula for finding the area of a kite: 

\(\displaystyle Area=\frac{diagonalA\times diagonalB}{2}\)

However, in this problem the question only provides information regarding the exact area. The lengths of the diagonals are represented as a ratio, where 
\(\displaystyle diagonalA:diagonalB=1:2\)

Therefore, it is necessary to plug the provided information into the area formula. Diagonal \(\displaystyle A\) is represented by \(\displaystyle x\) and diagonal \(\displaystyle B=2\)\(\displaystyle x\).

The solution is:

\(\displaystyle 196=\frac{x\times 2x}{2}\)

\(\displaystyle 196\times2=x\times 2x\)

\(\displaystyle 392=2x^2\)

\(\displaystyle x^2=\frac{392}{2}=196\)

\(\displaystyle x=\sqrt{196}=14\)

Thus, if \(\displaystyle x=14\), then diagonal \(\displaystyle B\) must equal \(\displaystyle 2(14)=28\)

This problem can be solved by applying the area formula: 



Since this question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal. 

Thus the solution is: 

\(\displaystyle 60=\frac{8\times diagonal B}{2}\)

\(\displaystyle 60\times2=8\times diagonalB\)

\(\displaystyle 120=8(diagonalB)\)

\(\displaystyle diagonal B=\frac{120}{8}=15\)

This problem can be solved by applying the area formula: 



Since this question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal. 

Thus the solution is: 

\(\displaystyle 45=\frac{18\times diagonal B}{2}\)

\(\displaystyle 45\times2=18\times diagonalB\)

\(\displaystyle 90=18(diagonalB)\)

\(\displaystyle diagonal B=\frac{90}{18}=5\)

First find the length of the missing diagonal before you can find the sum of the two perpendicular diagonals. 

To find the missing diagonal, apply the area formula: 




This question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal. 


\(\displaystyle 6,250=\frac{250\times diagonal B}{2}\)

\(\displaystyle 6250\times2=250\times diagonalB\)

\(\displaystyle 12,500=250(diagonalB)\)

\(\displaystyle diagonal B=\frac{12,500}{250}=50\)

Therefore, the sum of the two diagonals is: 

\(\displaystyle 250+50=300\)

You must find the length of the missing diagonal before you can find the sum of the two perpendicular diagonals. 

To find the missing diagonal, apply the area formula: 




This question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal. 


\(\displaystyle 28=\frac{4\times diagonal B}{2}\)

\(\displaystyle 28\times2=4\times diagonalB\)

\(\displaystyle 56=4(diagonalB)\)

\(\displaystyle diagonal B=\frac{56}{4}=14\)

Therefore, the sum of the two diagonals is: 

\(\displaystyle 14+4=18\)

To find the length of the black diagonal apply the area formula: 



Since this question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal. 

Thus the solution is: 

\(\displaystyle 125=\frac{25\times diagonal B}{2}\)

\(\displaystyle 125\times2=25\times diagonalB\)

\(\displaystyle 250=25(diagonalB)\)

\(\displaystyle diagonal B=\frac{250}{25}=10\)

This problem can be solved by applying the area formula: 



Since this question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal. 

Thus the solution is: 

\(\displaystyle 297=\frac{22\times diagonal B}{2}\)

\(\displaystyle 297\times2=22\times diagonalB\)

\(\displaystyle 594=22(diagonalB)\)

\(\displaystyle diagonal B=\frac{594}{22}=27\)