In the figure shown, adjacent sides of the parallelogram are $4$ cm and $6$ cm. The indicated angle is $25^\circ$. What is the number of square centimeters in the area of the parallelogram?
Holy smoke! What a lot of math! If the sides are a and b, and b is the base, then
A = bh = b * (a sinθ) = 6 * 4sin25° = 10.14
That's correct, but please note that it is not necessary to use the sine function in this case. The area of a parallelogram can also be calculated by multiplying the base by the corresponding height, which is the perpendicular distance between the base and the opposite side. In this case, the height is 4 cm, so the area is indeed $6 \times 4 = 24$ square cm.
To find the area of the parallelogram, we need to find the length of the base (the side opposite the indicated angle).
Since the opposite sides of a parallelogram are equal in length, the base of the parallelogram is also 6 cm.
To find the height of the parallelogram, we can use trigonometry.
The height can be found by taking the sine of the indicated angle, since the opposite side of the angle is the height of the parallelogram.
Using the formula:
height = adjacent side × sine(angle)
height = 4 cm × sine(25°)
Calculating this using a calculator, we find:
height ≈ 4 cm × 0.4226 ≈ 1.6904 cm
Now that we have the base and height of the parallelogram, we can find its area by multiplying the base and height.
Area = base × height
Area = 6 cm × 1.6904 cm
Calculating this, we find:
Area ≈ 10.1424 cm^2
Therefore, the area of the parallelogram is approximately 10.1424 square centimeters.
We drop an altitude from the vertex at the bottom to the opposite side. [asy]
size(3cm);
pair A=(0,0),B=(4,0),C=dir(25)*6,D=A+C,E=B+C;
draw(A--B--E--D--A^^D--C^^rightanglemark(D,C,A,6)^^C--A);
label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,SW); label("$E$",E,NE);
draw(shift((7,0))*A--shift((7,0))*B--shift((7,0))*E--shift((7,0))*D--shift((7,0))*A^^shift((7,0))*D--shift((7,0))*C^^shift((7,0))*C--shift((7,0))*A);
label("$A$",shift((7,0))*A,SW); label("$B$",shift((7,0))*B,SE); label("$C$",shift((7,0))*C,NE); label("$D$",shift((7,0))*D,SW); label("$E$",shift((7,0))*E,NE);
draw((0,-1)--(7,-1),EndArrow); draw((2.5,-1)--(2.5,0.5),EndArrow);
label("$4$",(2,-1)); label("$6$",(5,-1));
[/asy] Then $AD$ is the shorter base of the parallelogram, and $BC$ is the corresponding longer base. Since $\angle EDC=75^\circ$ and $\angle CDE=25^\circ$, then $\angle CED=180^\circ-75^\circ-25^\circ=80^\circ$. Since opposite angles of a parallelogram are equal, we know $\angle BDC = 80^\circ$. Thus $\triangle BDC$ and $\triangle ABD$ are similar right triangles. If we denote $\frac{AD}{BD}=\frac{CD}{BC}=r$, then $\tan(80^\circ)=\tan(\angle BDC)=\frac{CD}{BD}=\frac{r}{1}=r$, so $AD=rBD$. Since $AD=\overline{DC}=4\text{ cm}$ and $BD=6\text{ cm}$, we can see that $r=\frac43$.
[asy]
size(5cm);
pair A=(0,0),B=(4,0),C=dir(25)*6,D=A+C;
draw(A--B--D--A^^D--C^^rightanglemark(D,C,A,6));
label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,SW);
[/asy]
By dropping a perpendicular from $C$ to $\overline{BD}$, we can create a rectangle whose height is $AD=4\text{ cm}$ and whose width is $rBD=\frac{4\cdot6}{3}=8\text{ cm}$. The area of the rectangle is $4\text{ cm}\times8\text{ cm} = 32\text{ cm}^2$, so the area of the parallelogram is twice this value, or $\boxed{64}$ square centimeters.