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ISEE Upper Level Math : Kites

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Example Questions

Example Question #43 : Quadrilaterals

Cassie is making a kite for her little brother. She has two plastic tubes to use as the skeleton, measuring $\displaystyle 7$ inches and $\displaystyle 12$ inches. If these two tubes represent the diagnals of the kite, how many square inches of paper will she need to make the kite?

Possible Answers:

$\displaystyle 82$

$\displaystyle 42$

$\displaystyle 84$

$\displaystyle 40$

$\displaystyle 84$

Correct answer:

$\displaystyle 42$

Explanation:

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

Since Cassie has one tube measuring $\displaystyle 7$ inches, we can substitute $\displaystyle 7$ for $\displaystyle a$. We can also substitute the other tube that measures $\displaystyle 12$ inches in for $\displaystyle b$.

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

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Example Question #2 : How To Find The Area Of A Kite

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

Possible Answers:

$80\ in^2$ $\displaystyle 80\ in^2$

$45\ in^2$ $\displaystyle 45\ in^2$

$30\ in^2$ $\displaystyle 30\ in^2$

$60\ in^2$ $\displaystyle 60\ in^2$

$36\ in^2$ $\displaystyle 36\ in^2$

Correct answer:

$30\ in^2$ $\displaystyle 30\ in^2$

Explanation:

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

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

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

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

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Example Question #3 : How To Find The Area Of A Kite

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

Kite

Possible Answers:

$75\ cm^2$ $\displaystyle 75\ cm^2$

$75\sqrt{2}\ cm^2$ $\displaystyle 75\sqrt{2}\ cm^2$

$150\ cm^2$ $\displaystyle 150\ cm^2$

$150\sqrt{3}\ cm^2$ $\displaystyle 150\sqrt{3}\ cm^2$

$75\sqrt{3}\ cm^2$ $\displaystyle 75\sqrt{3}\ cm^2$

Correct answer:

$75\sqrt{3}\ cm^2$ $\displaystyle 75\sqrt{3}\ cm^2$

Explanation:

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:

$Area=ab\times sinC$ $\displaystyle Area=ab\times sinC$,

where $\displaystyle a,b$are the lengths of two unequal sides, $\displaystyle C$ is the angle between them and sin $\displaystyle sin$ is the sine function.

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

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Example Question #4 : How To Find The Area Of A Kite

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.

Possible Answers:

$81\text{in}^2$ $\displaystyle 81\text{in}^2$

$76\text{in}^2$ $\displaystyle 76\text{in}^2$

$162\text{in}^2$ $\displaystyle 162\text{in}^2$

$108\text{in}^2$ $\displaystyle 108\text{in}^2$

$54\text{in}^2$ $\displaystyle 54\text{in}^2$

Correct answer:

$81\text{in}^2$ $\displaystyle 81\text{in}^2$

Explanation:

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

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

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

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

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

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

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Example Question #1 : Kites

A kite has the area of $60\ in^2$ $\displaystyle 60\ in^2$. One of the diagonals of the kite has length $15\ in$ $\displaystyle 15\ in$. Give the length of the other diagonal of the kite.

Possible Answers:

$6\ in$ $\displaystyle 6\ in$

$8\ in$ $\displaystyle 8\ in$

$7\ in$ $\displaystyle 7\ in$

$5\ in$ $\displaystyle 5\ in$

$4\ in$ $\displaystyle 4\ in$

Correct answer:

$8\ in$ $\displaystyle 8\ in$

Explanation:

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

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

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

$Area=\frac{d_{1}d_{2}}{2}=\frac{{15}\times {d_{2}}}{2}=60\Rightarrow d_{2}=\frac{60\times 2}{15}\Rightarrow d_{2}=8\ in$ $\displaystyle Area=\frac{d_{1}d_{2}}{2}=\frac{{15}\times {d_{2}}}{2}=60\Rightarrow d_{2}=\frac{60\times 2}{15}\Rightarrow d_{2}=8\ in$

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ISEE Upper Level Math : Kites

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #43 : Quadrilaterals

Example Question #2 : How To Find The Area Of A Kite

Example Question #3 : How To Find The Area Of A Kite

Example Question #4 : How To Find The Area Of A Kite

Example Question #1 : Kites

All ISEE Upper Level Math Resources

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