Question

directions: find the area of each figure. round to the nearest hundredth where necessary. 1. 21 m 33 m 21 m 556.5m² 2. 11 in 7 in 8.1 in 15 in 36in² 3. 2.9 ft 26.4ft² 4. 16 cm 20.9 cm 27 cm 176cm² 5. 35 m 38 m 32.7 m 13 m 304.76m² 35 m 6. 17.6 mm 75.12in² 7. 10 yd 19 yd 19 yd 7.8 yd 10 yd 149.88in² 8. 44 in 9. 11.8 ft 16.4 ft 10. 29.1 in 14.6 in 20.4 in 11.

Explanation:

Step1: Recall area formulas

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

Step2: Solve for each figure

Figure 1 (parallelogram)

Base $b = 21$ m, height $h=33$ m. $A = 21\times33=693$ m$^{2}$ (the given answer $556.5$ m$^{2}$ seems incorrect).

Figure 2 (trapezoid)

$a = 11$ in, $b = 15$ in, $h = 7$ in. $A=\frac{(11 + 15)\times7}{2}=\frac{26\times7}{2}=91$ in$^{2}$ (the given answer $36$ in$^{2}$ seems incorrect).

Figure 3 (circle)

$r = 2.9$ ft. $A=\pi\times(2.9)^{2}\approx3.14\times8.41 = 26.42$ ft$^{2}$ (close to the given $26.4$ ft$^{2}$).

Figure 4 (triangle)

$b = 27$ cm, $h = 16$ cm. $A=\frac{1}{2}\times27\times16=216$ cm$^{2}$ (the given answer $176$ cm$^{2}$ seems incorrect).

Figure 5 (trapezoid)

$a = 13$ m, $b = 38$ m, $h = 32.7$ m. $A=\frac{(13 + 38)\times32.7}{2}=\frac{51\times32.7}{2}=833.85$ m$^{2}$ (the given answer $304.76$ m$^{2}$ seems incorrect).

Figure 6 (square)

Side $s=17.6$ mm. $A=(17.6)^{2}=309.76$ mm$^{2}$ (the given answer $75.12$ in$^{2}$ seems incorrect as units don't match and value is wrong).

Figure 7 (rhombus)

Diagonals $d_1 = 7.8$ yd and assume the other diagonal can be found using Pythagorean - like approach for the rhombus parts. But if we use $A=\frac{1}{2}d_1d_2$, assume the figure is composed of two congruent triangles with base and height related to the given sides. However, if we consider the area formula for a general quadrilateral with given side lengths is more complex. If we assume it's a parallelogram - like approach with base $b = 19$ yd and height $h = 7.8$ yd, $A=19\times7.8 = 148.2$ yd$^{2}$ (close to the given $149.8$ yd$^{2}$).

Figure 8 (semi - circle)

$d = 44$ in, $r = 22$ in. $A=\frac{1}{2}\pi r^{2}=\frac{1}{2}\times3.14\times22^{2}=\frac{1}{2}\times3.14\times484 = 759.88$ in$^{2}$.

Figure 9 (rectangle)

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

Figure 10 (triangle)

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

Answer:

1. $693$ m$^{2}$
2. $91$ in$^{2}$
3. $26.42$ ft$^{2}$
4. $216$ cm$^{2}$
5. $833.85$ m$^{2}$
6. $309.76$ mm$^{2}$
7. $148.2$ yd$^{2}$
8. $759.88$ in$^{2}$
9. $193.52$ ft$^{2}$
10. $149.88$ in$^{2}$