Question

find the area of each figure, round your answer to the nearest whole number if necessary.

1. 36 cm 36 cm 60 cm

area:

1. 19 ft 10 ft 20 ft

area:

1. 2 m 2 m 4 m

area:

1. 10 in 5 in 10 in

area:

1. 20 yd 2 yd 12 yd 2 yd 12 yd

area:

1. 12 in 16 in 8 in

area:

1. 28 yd 28 yd 14 yd 56 yd

area:

1. 20 cm 18 cm 21 cm 18 cm

area:

1. 8 ft 6 ft 9 ft 6 ft

area:

Explanation:

Step1: Divide figure 1 into two rectangles

The upper rectangle has dimensions 36 cm by 36 cm, and the lower one has dimensions 60 cm by 36 cm.
The area of the upper rectangle $A_1 = 36\times36= 1296$ $cm^2$.
The area of the lower rectangle $A_2=60\times36 = 2160$ $cm^2$.
The total area $A = A_1+A_2=1296 + 2160=3456$ $cm^2$.

Step2: Divide figure 2 into a triangle and a rectangle

The triangle has base 20 ft and height 19 ft, and the rectangle has dimensions 20 ft by 10 ft.
The area of the triangle $A_1=\frac{1}{2}\times20\times19 = 190$ $ft^2$.
The area of the rectangle $A_2=20\times10=200$ $ft^2$.
The total area $A = A_1 + A_2=190+200 = 390$ $ft^2$.

Step3: Divide figure 3 into two rectangles

The vertical - rectangle has dimensions 4 m by 2 m, and the horizontal - rectangle has dimensions 2 m by 2 m.
The area of the vertical rectangle $A_1=4\times2=8$ $m^2$.
The area of the horizontal rectangle $A_2=2\times2 = 4$ $m^2$.
The total area $A=A_1 + A_2=8 + 4=12$ $m^2$.

Step4: Use the trapezoid area formula for figure 4

The trapezoid area formula is $A=\frac{(a + b)h}{2}$, where $a = 5$ in, $b = 10$ in, and $h = 10$ in.
$A=\frac{(5 + 10)\times10}{2}=\frac{15\times10}{2}=75$ $in^2$.

Step5: Divide figure 5 into a rectangle and a triangle

The rectangle has dimensions 20 yd by 12 yd, and the triangle has base 12 yd and height 12 yd.
The area of the rectangle $A_1=20\times12=240$ $yd^2$.
The area of the triangle $A_2=\frac{1}{2}\times12\times12 = 72$ $yd^2$.
The total area $A=A_1+A_2=240 + 72=312$ $yd^2$.

Step6: Divide figure 6 into a rectangle and a triangle

The rectangle has dimensions 16 in by 12 in, and the triangle has base 8 in and height 12 in.
The area of the rectangle $A_1=16\times12=192$ $in^2$.
The area of the triangle $A_2=\frac{1}{2}\times8\times12=48$ $in^2$.
The total area $A=A_1 + A_2=192+48 = 240$ $in^2$.

Step7: Divide figure 7 into two rectangles

The upper rectangle has dimensions 28 yd by 28 yd, and the lower rectangle has dimensions 56 yd by 14 yd.
The area of the upper rectangle $A_1=28\times28=784$ $yd^2$.
The area of the lower rectangle $A_2=56\times14 = 784$ $yd^2$.
The total area $A=A_1+A_2=784+784 = 1568$ $yd^2$.

Step8: Divide figure 8 into two rectangles

The vertical - rectangle has dimensions 21 cm by 20 cm, and the horizontal - rectangle has dimensions 18 cm by 18 cm.
The area of the vertical rectangle $A_1=21\times20 = 420$ $cm^2$.
The area of the horizontal rectangle $A_2=18\times18=324$ $cm^2$.
The total area $A=A_1+A_2=420+324 = 744$ $cm^2$.

Step9: Divide figure 9 into two rectangles

The vertical - rectangle has dimensions 9 ft by 8 ft, and the horizontal - rectangle has dimensions 6 ft by 6 ft.
The area of the vertical rectangle $A_1=9\times8=72$ $ft^2$.
The area of the horizontal rectangle $A_2=6\times6 = 36$ $ft^2$.
The total area $A=A_1+A_2=72+36=108$ $ft^2$.

Answer:

1. 3456 $cm^2$
2. 390 $ft^2$
3. 12 $m^2$
4. 75 $in^2$
5. 312 $yd^2$
6. 240 $in^2$
7. 1568 $yd^2$
8. 744 $cm^2$
9. 108 $ft^2$