8.3
Trapezoids, Rhombuses, Kites, and Other Quadrilaterals
We start by deriving the formula for the area of a trapezoid. This is based easily on the area formula for a triangle.
- Theorem 8.4:
The area of a trapezoid having bases $b_1$ and $b_2$ and altitude $h$
is $\frac{1}{2}(b_1+b_2)$.
Proof:
Subdivide trapezoid $ABCD$ into the two triangles $\triangle ABC$ and
$\triangle ACD$. Using $h$ as the altitude of each one, we have
$$a(\triangle ABC)=\frac{1}{2}b_1\cdot h\text{ and }a(\triangle
ACD)=\frac{1}{2}b_2\cdot h.$$ Adding these, we obtain \begin{align} a(ABCD)&=a(\triangle ABC)+a(\triangle ACD)\\ &=\frac{1}{2}b_1\cdot +\frac{1}{2}\cdot h\\ &=\frac{1}{2}(b_1+b_2)\cdot h \end{align} |
Example:
|
The median of a trapezoid has a
length of 12 and the area of the trapezoid is 60. What is the
altitude of the trapezoid? |
Solution: |
The median of a trapezoid has
length $\frac{1}{2}(b_1+b_2)$. That expression is one of the
factors in the formula for the area of a trapezoid. Hence we can
write \begin{align} a(\text{trapezoid})&=\frac{1}{2}(b_1+b_2)\cdot h\\ 60&=12\cdot h\\ h&=5 \end{align} |
We now turn our attention to the area of a rhombus. Of course, since a rhombus is a special kind of parallelogram, the formula for the area of a parallelogram from the last section is entirely appropriate. However, there is another result that applies to a rhombus. We prove this in a more general situation.
- Theorem 8.5:
If the diagonals of a quadrilateral are perpendicular to each other,
then the area of the quadrilateral is one-half the product of the
diagonals.
Given: In
quadrilateral $ABCD$, $\overline{AC}\perp\overline{BD}$ Prove: $a(ABCD)=\frac{1}{2}(AC)(BD)$ Proof: Let $X$ denot the intersection of the diagonals. Then \begin{align} a(\triangle ABCD)&=a(\triangle ABC)+a(\triangle ACD)\\ &=\frac{1}{2}(AC)(XB)+\frac{1}{2}(AC)(XD)\\ &=\frac{1}{2}(AC)(XB+XD)\\ &=\frac{1}{2}(AC)(BD)\\ \end{align} |
Since the diagonals of a rhombus are perpendicular bisectors of each other, this theorem applies to a rhombus. It also applies to a kite, which has two pairs of distinct consecutive sides congruent. In fact, if the quadrilateral is arrow-shaped, but still has diagonals that (when extended) meet at right angles, the theorem still applies. And it should be noted that a square is a special rhombus, so this theorem can be used to find the area of a square. |
Example:
Solution: |
Determine
the area of a square inscribed in a circle of radius 5. The diagonal of the square is a diameter of the circle, and has length 10. Then the area of the square is $\frac{1}{2}(10)(10)=50$. |
Example:
|
The side of a rhombus is 10 and
one diagonal has length 12. What is the area of the rhombus? |
Solution: | The diagonals are perpendicular
bisectors of each other. That gives us four right triangles with a leg of 6 and a hypotenuse of 10. The other leg is 8 (by the Pythagorean Theorem). Hence the two diagonals have lengths 12 and 16, so the area is $\frac{1}{2}(12)(16)=96$. |
Example:
|
The long diagonal of a kite is $\frac{3}{2}$ times as long as the short diagonal. The area of the kite is 108. How long are the diagonals of the kite? |
Solution: |
We know that one diagonal of a
kite is the perpendicular bisector of
the other. One possible configuration is pictured to the right,
with
the shorter diagonal the one that is bisected. If we let $BD=2x$,
then
$AC=3x$. The area of the kite is then given by
$\frac{1}{2}(2x)(3x)=108$. This leads to $x^2=36$, so
$x=6$. The
diagonals are 12 and 18. There is a second possible configuration, with the longer diagonal the one that is bisected. The kite looks peculiar, but it is still conceivable. The short diagonal is two-thirds as long as the long diagonal, so we have the diagram to the right. The short diagonal is $\frac{4}{3}x$ and the long diagonal is $2x$. Then the area is given by $\frac{1}{2}\cdot (2x)\cdot (\frac{4}{3}x)=108$. This gives $x^2=81$, so $x=9$. The diagonals are again 12 and 18. |