Question

finding the area of a kite in the coordinate plane. complete the steps to find the area of the kite. what is ge? square root of units. what is df? square root of units. what is the area of the kite to the nearest unit? square units. 40 80 160

Explanation:

Step1: Assume coordinates

Let's assume \(G(0,2)\), \(E(14,8)\). Use the distance - formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(GE\), \(x_1 = 0,y_1 = 2,x_2=14,y_2 = 8\). Then \(GE=\sqrt{(14 - 0)^2+(8 - 2)^2}=\sqrt{14^2+6^2}=\sqrt{196 + 36}=\sqrt{232}\).

Step2: Assume coordinates for \(D\) and \(F\)

Let \(D(8,10)\), \(F(16,2)\). Using the distance - formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), for \(DF\), \(x_1 = 8,y_1 = 10,x_2 = 16,y_2=2\). Then \(DF=\sqrt{(16 - 8)^2+(2 - 10)^2}=\sqrt{8^2+( - 8)^2}=\sqrt{64 + 64}=\sqrt{128}\).

Step3: Calculate the area of the kite

The area of a kite is \(A=\frac{1}{2}d_1d_2\), where \(d_1\) and \(d_2\) are the lengths of the diagonals. Here \(d_1 = GE=\sqrt{232}\), \(d_2 = DF=\sqrt{128}\). \(A=\frac{1}{2}\times\sqrt{232}\times\sqrt{128}=\frac{1}{2}\sqrt{232\times128}=\frac{1}{2}\sqrt{29776}=\frac{172.557}{2}\approx80\).

Answer:

Square root of 232 units; Square root of 128 units; 80 square units