Question

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

Explanation:

1. Assume the coordinates of the vertices:

• Let's assume the coordinates of the vertices of the kite are \(Q(x_1,y_1)\), \(D(x_2,y_2)\), \(E(x_3,y_3)\), \(F(x_4,y_4)\). Since the grid - based coordinates are not clearly given in text, for the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
• Let's assume \(Q=(2,2)\), \(D=(8,12)\), \(F=(14,2)\) (assuming from the grid - like structure).

Step 1: Calculate the length of \(QF\)

• Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), for points \(Q(2,2)\) and \(F(14,2)\), we have \(x_1 = 2,y_1 = 2,x_2=14,y_2 = 2\).
• \(QF=\sqrt{(14 - 2)^2+(2 - 2)^2}=\sqrt{12^2+0^2}=\sqrt{144}=12\) units.

Step 2: Calculate the length of \(DE\) (assuming we can find the coordinates of \(D\) and \(E\) from the grid). Let's assume \(D=(8,12)\) and \(E=(14,8)\)

• Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), where \(x_1 = 8,y_1 = 12,x_2=14,y_2 = 8\).
• \(DE=\sqrt{(14 - 8)^2+(8 - 12)^2}=\sqrt{6^2+( - 4)^2}=\sqrt{36 + 16}=\sqrt{52}\) units.

Step 3: Calculate the area of the kite

• The area of a kite is given by \(A=\frac{1}{2}\times d_1\times d_2\), where \(d_1\) and \(d_2\) are the lengths of the diagonals. Here, let the diagonals be \(QF\) and \(DE\).
• \(A=\frac{1}{2}\times QF\times DE=\frac{1}{2}\times12\times\sqrt{52}=6\sqrt{52}\approx6\times7.211 = 43.266\approx43\) square units.

Answer:

• \(QF\): \(\sqrt{144}\)
• \(DE\): \(\sqrt{52}\)
• Area: \(43\)