Question

2.2.1 to 2.2.3 review
find the area of each.
1)
2)
3)
4)
5)
6)
find the area of the shaded regions and the composited figures.
7)
8)
9)
10)

Explanation:

Step1: Recall area formulas

For a parallelogram, $A = base\times height$. For a triangle, $A=\frac{1}{2}\times base\times height$. For a rectangle, $A = length\times width$. For a trapezoid, $A=\frac{1}{2}(a + b)h$ where $a$ and $b$ are the lengths of the parallel - sides and $h$ is the height. For composite figures, find the area of component shapes and then add or subtract as needed.

Step2: Solve for each figure

1) Parallelogram

Base $b = 11$ in, height $h = 9$ in. Using the formula $A=b\times h$, we have $A=11\times9 = 99$ in².

2) Triangle

Base $b = 9$ ft, height $h = 7.2$ ft. Using the formula $A=\frac{1}{2}\times b\times h$, we get $A=\frac{1}{2}\times9\times7.2=32.4$ ft².

3) Rectangle

Length $l=4x + 3$, width $w = 2x+2$. Using the formula $A=l\times w=(4x + 3)(2x + 2)=8x^{2}+8x+6x + 6=8x^{2}+14x + 6$.

4) Triangle

Base $b = 5$ yd, height $h = 6.5$ yd. Using the formula $A=\frac{1}{2}\times b\times h$, we have $A=\frac{1}{2}\times5\times6.5 = 16.25$ yd².

5) Trapezoid

Parallel - sides $a = 4$ mi and $b = 12$ mi, height $h = 5$ mi. Using the formula $A=\frac{1}{2}(a + b)h=\frac{1}{2}(4 + 12)\times5=\frac{1}{2}\times16\times5=40$ mi².

6) Square

Side length $s = 10$ ft. Using the formula $A=s^{2}$, we get $A = 10\times10=100$ ft².

7) Composite figure

We can split it into two rectangles. One rectangle has dimensions $22$ cm by $25$ cm and the other has dimensions $(45 - 25)$ cm by $30$ cm.
Area of first rectangle $A_1=22\times25 = 550$ cm².
Area of second rectangle $A_2=(45 - 25)\times30=20\times30 = 600$ cm².
Total area $A=A_1+A_2=550 + 600=1150$ cm².

8) Composite figure

The area of the large rectangle is $A_{total}=12\times4 = 48$ cm². The area of the non - shaded triangle with base $b = 9$ cm and height $h = 4$ cm is $A_{triangle}=\frac{1}{2}\times9\times4 = 18$ cm².
The area of the shaded region $A=48-18 = 30$ cm².

9) Composite figure

Area of rectangle $A_{rect}=30\times28 = 840$ in².
Area of triangle $A_{triangle}=\frac{1}{2}\times25\times22=275$ in².
The area of the shaded region $A=840 - 275=565$ in².

10) Composite figure

We can split it into a rectangle and a trapezoid.
Rectangle has dimensions $8$ cm by $7$ cm, so its area $A_1=8\times7 = 56$ cm².
Trapezoid has parallel - sides $a = 7$ cm and $b = 13$ cm, and height $h=(16 - 8)$ cm. Using the trapezoid area formula $A_2=\frac{1}{2}(a + b)h=\frac{1}{2}(7 + 13)\times(16 - 8)=\frac{1}{2}\times20\times8 = 80$ cm².
Total area $A=A_1+A_2=56 + 80=136$ cm².

Answer:

1. $99$ in²
2. $32.4$ ft²
3. $8x^{2}+14x + 6$
4. $16.25$ yd²
5. $40$ mi²
6. $100$ ft²
7. $1150$ cm²
8. $30$ cm²
9. $565$ in²
10. $136$ cm²