Question

homework 1.6: area of plane figures
this is a 2 - page document!
directions: find the area of each figure. round to the nearest hundredth where necessary.
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Explanation:

Step1: Recall area formulas

For parallelogram: $A = base\times height$. For trapezoid: $A=\frac{(a + b)h}{2}$ where $a$ and $b$ are bases and $h$ is height. For circle: $A=\pi r^{2}$. For triangle: $A=\frac{1}{2}\times base\times height$. For rectangle: $A = length\times width$. For semi - circle: $A=\frac{1}{2}\pi r^{2}$.

Step2: Solve for each figure

Figure 1 (parallelogram)

Base = 21 m, height = 33 m. $A=21\times33 = 693m^{2}$

Figure 2 (trapezoid)

$a = 11$ in, $b = 15$ in, $h = 7$ in. $A=\frac{(11 + 15)\times7}{2}=\frac{26\times7}{2}=91in^{2}$

Figure 3 (circle)

$r = 2.9$ ft. $A=\pi\times(2.9)^{2}=\pi\times8.41\approx26.41ft^{2}$

Figure 4 (triangle)

Base = 27 cm, height = 16 cm. $A=\frac{1}{2}\times27\times16=216cm^{2}$

Figure 5 (trapezoid)

$a = 13$ m, $b = 38$ m, $h = 32.7$ m. $A=\frac{(13 + 38)\times32.7}{2}=\frac{51\times32.7}{2}=833.85m^{2}$

Figure 6 (square)

Side = 17.6 mm. $A=(17.6)^{2}=309.76mm^{2}$

Figure 7 (parallelogram)

Base = 10 yd, height = 7.8 yd. $A=10\times7.8 = 78yd^{2}$

Figure 8 (semi - circle)

$d = 44$ in, so $r = 22$ in. $A=\frac{1}{2}\pi\times(22)^{2}=\frac{1}{2}\pi\times484\approx759.88in^{2}$

Figure 9 (rectangle)

Length = 16.4 ft, width = 11.8 ft. $A=16.4\times11.8 = 193.52ft^{2}$

Figure 10 (triangle)

Base = 20.4 in, height = 14.6 in. $A=\frac{1}{2}\times20.4\times14.6=149.88in^{2}$

Figure 11 (triangle)

Base = 26 cm, height can be found using Pythagorean theorem. Let the base be bisected into two parts of length 13 cm. The hypotenuse is 30 cm. $h=\sqrt{30^{2}-13^{2}}=\sqrt{900 - 169}=\sqrt{731}\approx27.04$ cm. $A=\frac{1}{2}\times26\times27.04 = 351.52cm^{2}$

Answer:

1. $693m^{2}$
2. $91in^{2}$
3. $26.41ft^{2}$
4. $216cm^{2}$
5. $833.85m^{2}$
6. $309.76mm^{2}$
7. $78yd^{2}$
8. $759.88in^{2}$
9. $193.52ft^{2}$
10. $149.88in^{2}$
11. $351.52cm^{2}$