ANSWER
A. 15 units^2
EXPLANATION
The area of kite is half the product of the diagonals.
The first diagonal has vertices at,
T(0,0) and R(5,5).
The length of this diagonal is
[tex]TR = \sqrt{ {5}^{2} + {5}^{2} } [/tex]
[tex]TR = \sqrt{25 + 25} [/tex]
[tex]TR = \sqrt{50} =5 \sqrt{2} [/tex]
The other diagonal has vertices at;
Q(0,3) and S(3,0).
The length of this diagonal is
[tex]QS = \sqrt{ {3}^{2} + {3}^{2} } [/tex]
[tex]QS = \sqrt{ 9+ 9 } [/tex]
[tex]QS = \sqrt{18} = 3 \sqrt{2} [/tex]
The area of the kite is
[tex] = \frac{1}{2} \times 3 \sqrt{2} \times 5 \sqrt{2} [/tex]
[tex] = 15 \: {units}^{2} [/tex]
