# Find the area of the kite PQRS with P(5,0), Q(8,6), R(5,8) and S(2,6)?

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To find the area of the kite PQRS with vertices P(5,0), Q(8,6), R(5,8), and S(2,6), you can divide the kite into two triangles, PQS and QRS, and then use the formula for the area of a triangle, which is half the base times the height.

First, find the lengths of the diagonals of the kite.

For diagonal PR: Using the distance formula: [PR = \sqrt{(5-5)^2 + (8-0)^2} = \sqrt{0 + 64} = \sqrt{64} = 8]

For diagonal QS: Using the distance formula: [QS = \sqrt{(8-2)^2 + (6-6)^2} = \sqrt{36 + 0} = \sqrt{36} = 6]

Next, find the lengths of the two segments PQ and RS, which are half the lengths of the diagonals.

[PQ = \frac{PR}{2} = \frac{8}{2} = 4] [RS = \frac{QS}{2} = \frac{6}{2} = 3]

Now, find the coordinates of the intersection point of the diagonals, which is the midpoint of PR and QS.

The midpoint of PR: [M\left(\frac{5+5}{2}, \frac{0+8}{2}\right) = (5, 4)]

The midpoint of QS: [N\left(\frac{8+2}{2}, \frac{6+6}{2}\right) = (5, 6)]

The height of the kite, which is the distance between M and N: [MN = |6 - 4| = 2]

Now, calculate the area of each triangle:

For triangle PQS: [A_{PQS} = \frac{1}{2} \times PQ \times MN = \frac{1}{2} \times 4 \times 2 = 4]

For triangle QRS: [A_{QRS} = \frac{1}{2} \times RS \times MN = \frac{1}{2} \times 3 \times 2 = 3]

Finally, add the areas of the two triangles to find the total area of the kite:

[A_{\text{kite}} = A_{PQS} + A_{QRS} = 4 + 3 = 7]

So, the area of the kite PQRS is 7 square units.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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