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mixed practice on perimeter and area 1) perimeter = 2) perimeter = find…

Question

mixed practice on perimeter and area
1)

perimeter =

2)

perimeter =

find the perimeter (or circumference) and the area of each figure.
3)

perimeter =
area =

4)

circumference =
area =

Explanation:

1. Hexagon Perimeter:

Step1: Identify side lengths

Hexagon has 3 sides of \(4\frac{1}{8}\) in and 3 sides of \(7\frac{3}{4}\) in.

Step2: Calculate perimeter
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2. Trapezoid Perimeter:

Step1: Sum all sides

Sides are \(8\) in, \(6\) in, \(9\) in, \(6\) in.

$$P = 8+6+9+6=29\text{ in}$$

3. Rectangle Perimeter and Area:

Perimeter:
Step1: Use perimeter formula \(P = 2(l + w)\)

\(l = 12\) ft, \(w = 5\) ft

$$P=2(12 + 5)=2\times17 = 34\text{ ft}$$
Area:
Step1: Use area formula \(A=l\times w\)
$$A=12\times5 = 60\text{ ft}^2$$

4. Circle Circumference and Area:

Circumference:
Step1: Use formula \(C = 2\pi r\)

\(r = 8\) in

$$C=2\times\pi\times8=16\pi\approx16\times3.14 = 50.24\text{ in}$$
Area:
Step1: Use formula \(A=\pi r^2\)
$$A=\pi\times8^2=64\pi\approx64\times3.14 = 200.96\text{ in}^2$$

Answer:

  1. \(35\frac{5}{8}\text{ in}\)
  2. \(29\text{ in}\)
  3. Perimeter \(34\text{ ft}\), Area \(60\text{ ft}^2\)
  4. Circumference \(50.24\text{ in}\), Area \(200.96\text{ in}^2\)