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Kite $ABCD$ (a quadrilateral with two pairs of adjacent equal sides) has coordinates $A\ (0,7),\ B\ (1,0),\ C\ (12,-2),$ and $D\ (7,8).$ What is the area of $ABCD,$ given that the area of a kite is equal to half the product of its diagonals?

 Aug 3, 2022

Best Answer 

 #1
avatar+113 
+2

A bit of vector algebra. But really only the pythagorean theorem.

We can find the vectors for each diagonal by subtracting startpoint from endpoint, so

AC = (12,-2) - (0,7) = (12-0, -2-7) = (12,-9)

BD = (7,8) - (1,0) = (7-1, 8-0) = (6,8)

We can find the length of each of these by taking the square of each vector component, add

them togehter and then taking the square root of this sum.

 

So the length of AC, denoted |AC| = \(\sqrt{12\times12 + (-9)\times(-9)} = \sqrt{144 + 81} = \sqrt{225} = 15\)

Similarly, the length of BD,  |BD| = \(\sqrt{6\times6 + (8)\times(8)} = \sqrt{36 + 64} = \sqrt{100} = 10\)

 

The rule for finding the Area of the kite is 1/2 x |AC| x |BD| = 1/2 x 15 x 10 = 1/2 x 150 = 75

Hope this helps.

 Aug 3, 2022
edited by tuffla2022  Aug 3, 2022
 #1
avatar+113 
+2
Best Answer

A bit of vector algebra. But really only the pythagorean theorem.

We can find the vectors for each diagonal by subtracting startpoint from endpoint, so

AC = (12,-2) - (0,7) = (12-0, -2-7) = (12,-9)

BD = (7,8) - (1,0) = (7-1, 8-0) = (6,8)

We can find the length of each of these by taking the square of each vector component, add

them togehter and then taking the square root of this sum.

 

So the length of AC, denoted |AC| = \(\sqrt{12\times12 + (-9)\times(-9)} = \sqrt{144 + 81} = \sqrt{225} = 15\)

Similarly, the length of BD,  |BD| = \(\sqrt{6\times6 + (8)\times(8)} = \sqrt{36 + 64} = \sqrt{100} = 10\)

 

The rule for finding the Area of the kite is 1/2 x |AC| x |BD| = 1/2 x 15 x 10 = 1/2 x 150 = 75

Hope this helps.

tuffla2022 Aug 3, 2022
edited by tuffla2022  Aug 3, 2022

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