Perimeter and Area |

**Measuring Outside and Inside**

Your dad just finished building a deck out back and he wants to put a railing around it, leaving a 2-ft space as an entrance. He also needs a large piece of yellow canvas to cover a square corner of the rectangular deck, so there will be a shady spot. He asks you to help him calculate what length of railing and how much canvas to buy for the project. So you draw this diagram and insert the measures shown:

The length of the **railing** is **2 feet less** than the **perimeter** or **distance around** the deck.

**2 sides** measure **10 feet**, **one** side is **15 feet** and the other is **13 feet** (2 foot gap).

**10 + 10 + 15 + 13 = 48** feet of railing is needed.

The yellow canvas shade cover is a **square** with **side = 5 feet** so we can " ** square it off** " into

The **area** of canvas for the shade is **5 feet × 5 feet = 25 square feet** (ft²).

**Perimeter: Measuring Around the Outside**

The prefix *peri-* means "**around** or **surrounding**" -- as in ** peri**scope -- the device a submarine captain uses to check what's happening at the surface, when the sub is under water. And "meter" means "measure" -- so

When we need to find the **perimeter** of an **irregular figure**, one with no equal sides, we just **add** the lengths of **all the sides**. If we have a **square** or **rectangle**, we can take a shortcut, because -- with a square, **all sides** are the **same length**, and with a **rectangle**, we've got **2 pairs of sides** that are the **same length**. For these shapes, we develop **formulas**.

Because a **rectangle** has **2 pairs of equal sides**, instead of adding their lengths separately to find perimeter, we use a formula. We can see that if we **double the sum of the length and width**, we'll have the perimeter. So, we "name" the length and the width *l* and *w* -- then we write the formula.

**P = 2 l + 2 w = 2 ( l + w )**

and for a

The **length and width** of a rectangle are also **called** it's **base and height**.

**Examples:**

1) Find the **Perimeter** of:

a) A **parallelogram** (//gm) with top/bottom = **7 inches** and sides = **9 inches**

Perimeter = **2 (7 + 9)** = 2 (16) = **32 inches**.

b) A **rectangle** with length (*l*) = **14 ft**. and width (*w*) = **10 ft**.

Perimeter = **2 (14 + 10)** = 2 (24) = **48 feet.**

c) A **Square** with side (*s*) = **12 yards**.

Perimeter = **4 (12)** = **48 yards**.

**BEWARE!!** of **mixed units!!** If one measure is **inches**, the other is **feet**, we have to **convert** in order to have **units of the same size** to add and multiply. **Answers must include units!!**

**Example:**

Find the Perimeter of a **parallelogram** (//gm) with top/bottom = **2 feet,** side = **9 inches**.

**Solution:** We could change feet to inches: **2 feet =** 2 × 12 = **24 inches
**so the

Or, we could realize that 9 inches = ¾ of a foot, (¾ of 12 = 9),

so the

.

**Area: Covering the Surface**

The things we use -- like paint, wallpaper and carpets -- to cover surfaces such as walls and floors are always **measured and priced** in **square units**. The information on every paint can **label** **includes** the number of **square feet** or square meters of wall space, the **paint will cover**.

The **measure of the surface** covered by a closed figure is called **AREA**.

We measure area in **SQUARE UNITS** -- square inches, square feet, square miles --

because, to measure the area of a figure, we "** square it off**" and then count the squares.

We see, that to completely cover the pink square ( 5 × 5 ), with grey squares ( 1 × 1 ),

we would use 5 rows of 5 squares each -- so we'd need 25 squares in all.

Area of a square that is 5 by 5 = 25 units².

Area of a square that is 9 inches per side = 81 in² (*square inches*).

Area Formulas | ||

figure |
Area Formula |
Comments |

Square | A = s² |
all sides = s |

Rectangle | A = l × w |
length = l, width = w |

Triangle | A = ½ ( b × h ) |
base = b, perpendicular height = h |

.

Now get a pencil, an eraser and a note book, copy the questions,

do the practice exercise(s), then check your work with the solutions.

If you get stuck, review the examples in the lesson, then try again.

**Practice Exercises**

1) For each image in the diagram, find the **perimeter**, and **area**.

(be sure to **write the formula** used for both perimeter and area)

2) **Draw a diagram**, then **solve** each of these questions:

a) The Area of a rectangle = 36 in². The length *l* = 9 in, find the width *w *and Perimeter.

b) The Perimeter of a Square = 44 in. Find the length of the side *s* and the Area.

c) The Perimeter of a rectangle is 96 feet. If the **base** is **twice** as long as the **height**,

find the dimensions and the area.

3) The wooden frame around an 8 × 10 inch photo is 2 inches wide on all sides.

Find a) the perimeter of the frame; and b) the area of the wooden frame.

**Solutions**

1)

Perimeter |
Area |

a) P = sum of sides = 28.8 inches |
rectangle area: = 9 × 5 = A = b × h45 in²triangle area: (A = ½) = ½ (9 × 2) = b × h9 in²total area = 45 in² + 9 in² = 54 in² |

b) P = 2( = 8.48 feetl + w) |
= 3.18 × 1.06 = 3.37 ft²A = l × w |

c) P = 428 cms = |
= 49 cm²A = s² |

d) P = sum of sides = 29.3 yds |
The figure is a 4 by 4 square and 2 triangles:square area: = 16 yd²A = s²left triangle: = ½ 4² = 8 yd² A = ½bhright triangle: = ½ (5.5 × 4) = 11 yd² A = ½bh total area = 16 yd² + 8 yd² + 11 yd² = 35 yd ² |

2) a) Area = 36 in². *l* = 9 in, *w* = **36 ÷ 9 = 4 inches**.

b) Perimeter = 44 in. 4*s* = 44 so ** s = 11 in.** Area = 11² =

c)

3) Since the frame is 2 inches wide all around, we add 4 inches to the dimensions of the photo.

The frame's length = 14 inches, the width = 12 inches

so **P = 2( l + w)** =

Area of the picture is **80 in²**, so the area of just the frame is **168 – 80 = 88 in²**.

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