How to find the area of a kite - ISEE Upper Level Mathematics Achievement

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Question

Cassie is making a kite for her little brother. She has two plastic tubes to use as the skeleton, measuring inches and inches. If these two tubes represent the diagnals of the kite, how many square inches of paper will she need to make the kite?

Answer

To find the area of a kite, use the formula $\frac{1}{2}ab$ , where represents one diagnal and represents the other.

Since Cassie has one tube measuring inches, we can substitute for . We can also substitute the other tube that measures inches in for .

$\frac{1}{2}(7)(12)=42\ in^{2}$

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Question

Two diagonals of a kite have the lengths of $10\ in$ and $6\ in$ . Give the area of the kite.

Answer

The area of a kite is half the product of the diagonals, i.e.

$Area=\frac{d_{1}d_{2}}{2}$ ,

where $d_{1}$ and $d_{2}$ are the lengths of the diagonals.

$Area=\frac{d_{1}d_{2}}{2}=\frac{6\times 10}{2}=30\ in^2$

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Question

In the following kite, $a=10\ cm$ , $b=15\ cm$ and $\angle C=120^{\circ}$ . Give the area of the kite. Figure not drawn to scale.

Answer

When you know the length of two unequal sides of a kite and their included angle, the following formula can be used to find the area of a kite:

$Area=ab\times sinC$ ,

where __are the lengths of two unequal sides, is the angle between them and is the sine function.

$Area=ab\times sinC=10\times 15\times sin120^{\circ}=10\times 15\times \frac{\sqrt{3}}{2}\Rightarrow Area=75\sqrt{3}\ cm^2$

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Question

Find the area of a kite with one diagonal having length 18in and the other diagonal having a length that is half the first diagonal.

Answer

To find the area of a kite, we will use the following formula:

$A = \frac{pq}{2}$

where p and q are the lengths of the diagonals of the kite.

Now, we know the length of one diagonal is 18in. We also know the other diagonal is half of the first diagonal. Therefore, the second diagonal has a length of 9in.

Knowing this, we can substitute into the formula. We get

$A = \frac{18\text{in} \cdot 9\text{in}}{2}$

$A = \frac{162\text{in}^2}{2}$

$A = 81\text{in}^2$

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