8.2
Parallelograms and Triangles
We now turn our attention to some other figures whose areas can be computed based on our work in the last section. In a parallelogram, an altitude is the distance between a vertex and the line containing the opposite side. In the figure below, given that $ABCD$ is a parallelogram, $\overline{AP}$ and $\overline{BQ}$ are both altitudes to $\overline{CD}$. In fact, if any point $P$ is chosen on side $\overline{AB}$, the distance from that point to side $\overline{DC}$ will be the same as the length of $\overline{AX}$, so that segment could be considered an altitude to $\overline{DC}$.
We now develop the area of a parallelogram based on the area formula for a rectangle.
- Theorem 8.2: The area of a parallelogram is the product of the length of one side with the length of the altitude drawn to that side.
Given:
Parallelogram $ABCD$ with $\overline{DX}\perp\overline{AB}$.
$AB=b$ and $DX=h$. Prove: $a(ABCD)=b\cdot h$ |
Proof: Extend $\overline{AB}$ and drop a perpendicular $\overline{CY}$ to that extended segment. $\triangle ADX\cong\triangle BCY$ (by Hypotenuse-Leg). Noting that $XYCD$ is a rectangle, we compute the area of the parallelogram as follows. \begin{align} a(ABCD)&=A(\triangle AXD)+a($XBCD)\\ &=a(\triangle BYC)+a(XBCD)\\ &=a(XYCD)\\ &=b\cdot h \end{align} |
There is a great tendency to think of an altitude of a parallelogram as having to go vertically. It is important to recognize that altitudes can be drawn to each side of a parallelogram and that the area of a parallelogram can be computed as the product of the length of any side with the length of the altitude drawn to that side.
Example: In parallelogram $ABCD$, $AB=8.2$, $AD=2.7$, and $\angle A$ is a $27^\circ$ angle. Determine the length of the altitude from $B$ to $\overline{AD}$. Solution: With the parallelogram as pictured to the right, we add the altitude $\overline{BX}$ from $B$ to $\overline{AD}$. Since the known angle is not one of our familiar angles (for which we know the ratios of sides of a right triangle), we turn to trigonometry to get the answer. In right $\triangle ABX$, we have $\sin \angle A=\frac{BX}{AB}$. We know $m\angle A=27$ and $AB=8.2$. Then $BX=AB\cdot \sin 27$. With a calculator, we obtain $BX=3.723$. |
With the area of a parallelogram taken care of, we can deal easily with the area of a triangle. A triangle is basically half od a parallelogram, giving us the following theorem.
- Theorem 8.3: The area of a triangle is one-half the product of the length of one side with the length of the altitude drawn to that side.
Given: In $\triangle ABC$, $\overline{CX}\perp\overline{AB}$. $AB=b$ and $CX=h$.
Prove: $a(\triangle ABC)=\frac{1}{2}b\cdot h$
Proof: Draw the line through $C$ that is parallel to $\overline{AB}$ and the line through $B$ that is parallel to $\overline{AC}$. Let the intersection of those lines be point $D$. (Check the boxes in the figure above.) Then $ABDC$ is a parallelogram, by definition. Its area is $b\cdot h$. But that parallelogram is made up of two congruent triangles, $\triangle ABC$ and $\triangle DCB$. Since the triangles are congruent, their areas are equal. Since they have no interior points in common, each must be half of the entire parallelogram. Hence $$a(\triangle ABC)=\frac{1}{2}b\cdot h.$$
Example: |
Determine the area of the triangle pictured to the right. |
Solution:
|
Draw the altitude to the base, creating both a 30-60-90 triangle and a 45-45-90 triangle. The short leg of the 30-60-90 triangle is 6, so the long leg is $6\sqrt{3}$. Then the second leg of the 45-45-90 triangle is also $6\sqrt{3}$, and the base of the triangle is $6+6\sqrt{3}$. The area is $\frac{1}{2}(6+6\sqrt{3})6\sqrt{3}=18\sqrt{3}+54$. |
An important corollary to Theorem 8.3 is the following, having to do with what happens when a median is drawn in a triangle.
- Corollary: A median of a triangle divides the triangle into two regions with the same area.
If $\overline{AM}$ is a median of $\triangle ABC$, then $MB=MC$. Using the altitude from vertex $A$ to compute the areas of the triangles $\triangle ABM$ and $\triangle ACM$, we see that these two areas are equal. |
Example: |
In the figure to the right, $M$ is the midpoint of $\overline{AB}$ and $P$ is the midpoint of $\overline{CM}$. If $a(\triangle ABC)=24$, then what is $a(\triangle APC)$? |
Solution: | Since $M$ is the midpoint of $\overline{AB}$, $a(\triangle AMC)$ is half the area of the whole triangle. In the same way, with $P$ the midpoint of $\overline{CM}$, $a(\triangle APC)$ is half the area of $\triangle AMC$. That means that $a(\triangle APC)$ is one-quarter the area of $\triangle ABC$. Hence, $a(\triangle APC)=6$. |
There are some important things to remember as you work on area problems. One is that every triangle has three altitudes, one from each vertex. This allows the area of the triangle to be computed in three ways. And since the area of the triangle doesn't change when we compute it a second way, we can make use of a strategy we'll call "area equals area."
Example: Determine the altitude to a leg of an isosceles triangle with sides 10, 10, and 12.
Solution 1: |
A
common algebraic approach to this problem is to set up two equations in
two unknowns. Let denote the desired length, and
let denote one of the lengths on the leg of length 10 cut
off by the altitude. The we can use the Pythagorean Theorem in two
separate triangles. \begin{cases} h^2+(10-x)^2&=12^2 \\ h^2+x^2&=10^2 \end{cases} Subtracting the second equation from the first gives $100-20x=44$, so $x=\frac{14}{5}=2.8$. We can then substitute this in the second equation and obtain $h^2=100-\frac{196}{25}=\frac{2304}{25}$, so $h=\frac{48}{5}$. |
There is a much easier approach, using "area equals area."
Solution 2: |
If we drop an altitude from the vertex angle to the base of the isosceles triangle, it gives us two identical right triangles with hypotenuse 10 and one leg of 6. The other leg — the altitude to the base of the isosceles triangle — is then 8. Then the area of the triangle is $\frac{1}{2}\cdot 12\cdot 8=48$. We set that equal to the area using the altitude $h$, and we have $\frac{1}{2}\cdot 10\cdot h=48$, and clearly $h=\frac{48}{5}$. |
An example of the use of "area equals area" in a proof is the next problem.
Example:
Given: Prove: |
Point $P$ is any
point on the base $\overline{BC}$ of isosceles $\triangle ABC$.
Perpendiculars $\overline{PM}$ and $\overline{PN}$ are drawn to the
legs of the triangle. Altitude $\overline{BD}$ is drawn from base
$\angle ABC$. $PM+PN=BD$ |
Proof:
|
$\triangle ABC$ can be thought
of as being made up of two smaller triangles, $\triangle ABP$ and
$\triangle ACP$. With altitudes $\overline{PM}$ and
$\overline{PN}$ in those triangles, we get the area of the entire
$\triangle ABC$ as $$\dfrac{1}{2}\cdot PM\cdot AB+\dfrac{1}{2}\cdot
PN\cdot AC.$$ $\overline{AB}$ and $\overline{AC}$ are he legs of
the isosceles triangle, so $AB=AC$. Hence the area of the whole
triangle is $$\frac{1}{2}\cdot PM\cdot AC+\frac{1}{2}\cdot PN
\cdot AC=\frac{1}{2}(PM+PN)\cdot AC$$ But the area of the whole triangle is also $\frac{1}{2}\cdot BD\cdot AC$. Equating these two results ("area equals area") gives $PM+PN=BD$. |
As we saw earlier, it is not hard to find the altitude to the base of an isosceles triangle whose sides are known. We conclude this section with an "area equals area" application having to do with the inscribed circle of an isosceles triangle.
Example: |
Determine the radius of the
inscribed circle in an isosceles triangle with sides 29, 29, and 42. |
Solution:
|
Working as we did in Solution 2
above, we find that the altitude to the base of the triangle has length
20. Thus the area of the triangle is $$\frac{1}{2}\cdot 20 \cdot
42=420.$$ We now look at the triangle as being made up of three
non-overlapping parts — $\triangle ABI$, $\triangle BCI$, and
$\triangle ACI$. In each case, the altitude drawn to the side of
the triangle that is also a side of the big is a radius of
the inscribed circle. Thus, \begin{align} a(\triangle ABC)&=\frac{1}{2}\cdot 29 \cdot r+\frac{1}{2}\cdot 42 \cdot r\\ &=\frac{1}{2}(29+29+42)\cdot r\\ &=50r \end{align} Equating the two results for the area of the triangle ("area equals area"), we find $r=\frac{42}{5}$. |