Primary & Middle School Mathematics · Grades 1–9

34Perimeter and Area

Grade 4 counted grid squares (Chapter 28); this chapter earns the first area formula — length times width for the rectangle — and learns the units cm $^2$ and m $^2$ . The formulas for triangles and other shapes ripen in Chapter 43 and Chapter 53.

34.1 Two different measures

Example 34.1 (Border vs surface)

A gardener needs perimeter to buy the fence and area to buy the grass seed. The two do not follow each other: stretching a rectangle thinner and longer can keep its area while its perimeter grows — compare a $6 \times 2$ and a $4 \times 3$ rectangle (same area $12$ , perimeters $16$ and $14$ ).

34.2 The rectangle formula

Proposition 34.2 (Area of a rectangle)

A rectangle of length $L$ and width $w$ (in the same unit) has area

A = L \times w ,

in square units: cm $^2$ (squares of side $1$ cm), m $^2$ (squares of side $1$ m), …

Why, by counting. Cover the rectangle with unit squares: $w$ rows of $L$ squares each, so $L \times w$ squares in total — exactly the multiplication rectangle of Chapter 16. ∎

Three rows of six centimeter-squares: the formula L × w is the counting, written once and for all. — Three rows of six centimeter-squares: the formula $L \times w$ is the counting, written once and for all.

Example 34.3

A rug measures $2.5$ m by $2$ m: area $2.5 \times 2 = 5$ m $^2$ . A stamp measures $3$ cm by $2.4$ cm: area $3 \times 2.4 = 7.2$ cm $^2$ . Same formula, any unit — as long as both sides use the same one.

Example 34.4 (Square units convert by 100)

$1$ m $= 100$ cm, but $1$ m $^2 = 100 \times 100 = 10\,000$ cm $^2$ : a square meter is a $100$ -by- $100$ grid of square centimeters. Area units jump by hundreds, not tens.

34.3 Composite figures

Method 34.5 (Cut, add, subtract)

For a figure made of rectangles (an L, a T, a frame):

cut it into rectangles, or complete it into a big rectangle;
compute each rectangular area with the formula;
add the pieces — or subtract the hole from the big rectangle;
check against a rough count of grid squares.

Example 34.6

An L-shaped room: a $6$ m $\times$ $4$ m rectangle with a $2$ m $\times$ $2$ m corner missing.

\text{Area} = 6 \times 4 - 2 \times 2 = 24 - 4 = 20 \text{ m}^2 .

Or cut the L into a $6 \times 2$ strip and a $4 \times 2$ strip: $12 + 8 = 20$ m $^2$ — two roads, one answer.

Example 34.7 (Half a rectangle)

Cutting a rectangle along a diagonal gives two triangles of the same area: each is half the rectangle. A right triangle with legs $6$ cm and $4$ cm therefore has area $\frac{6 \times 4}{2} = 12$ cm $^2$ — a picture worth remembering for Chapter 43, where it becomes a formula.

The diagonal cuts the 6 × 4 rectangle into two equal triangles of 12 squares each. — The diagonal cuts the $6 \times 4$ rectangle into two equal triangles of $12$ squares each.

34.4 Exercises

Exercise 34.1 ★

Compute the area and the perimeter of a rectangle $8$ cm $\times$ $5$ cm. Which answer is in cm, which in cm $^2$ ?

Solution

Solution of Exercise 34.1.

Area: $8 \times 5 = 40$ cm $^2$ . Perimeter: $2 \times (8 + 5) = 26$ cm.

Exercise 34.2 ★

Compute the areas: a square of side $9$ cm; a rectangle $12$ m $\times$ $7$ m; a rectangle $4.5$ cm $\times$ $6$ cm.

Solution

Solution of Exercise 34.2.

$9 \times 9 = 81$ cm $^2$ ; $12 \times 7 = 84$ m $^2$ ; $4.5 \times 6 = 27$ cm $^2$ .

Exercise 34.3 ★

A rectangle has area $63$ cm $^2$ and length $9$ cm. Find its width, then its perimeter.

Solution

Solution of Exercise 34.3.

Width: $63 \div 9 = 7$ cm. Perimeter: $2 \times (9 + 7) = 32$ cm.

Exercise 34.4 ★

Draw two different rectangles with area $24$ squares on grid paper, and compute both perimeters. Same area — same perimeter?

Solution

Solution of Exercise 34.4.

For instance $6 \times 4$ (perimeter $20$ ) and $8 \times 3$ (perimeter $22$ ): same area $24$ , different perimeters.

Exercise 34.5 ★

Convert: $3$ m $^2$ in cm $^2$ ; $50\,000$ cm $^2$ in m $^2$ . (Remember Example 34.4: by hundreds!)

Solution

Solution of Exercise 34.5.

$3$ m $^2 = 30\,000$ cm $^2$ ; $50\,000$ cm $^2 = 5$ m $^2$ .

Exercise 34.6 ★

A T-shaped figure is made of a $8 \times 2$ horizontal bar on top of a $2 \times 5$ vertical bar (in cm). Compute its area by adding two rectangles.

Solution

Solution of Exercise 34.6.

Bar: $8 \times 2 = 16$ cm $^2$ ; stem: $2 \times 5 = 10$ cm $^2$ ; total $26$ cm $^2$ .

Exercise 34.7 ★

A picture frame: a $30$ cm $\times$ $20$ cm rectangle with a $24$ cm $\times$ $14$ cm rectangular window cut out. What area of wood does the frame use?

Solution

Solution of Exercise 34.7.

$30 \times 20 - 24 \times 14 = 600 - 336 = 264$ cm $^2$ of wood.

Exercise 34.8 ★

Compute the area of a right triangle with legs $8$ cm and $6$ cm (half a rectangle, Example 34.7).

Solution

Solution of Exercise 34.8.

Half the $8 \times 6$ rectangle: $48 \div 2 = 24$ cm $^2$ .

Exercise 34.9 ★

A rectangular vegetable patch measures $7$ m by $4$ m. Seed costs $2$ per square meter, and fencing $3$ per meter. Compute the cost of the seed, then the cost of the fence.

Solution

Solution of Exercise 34.9.

Seed: area $7 \times 4 = 28$ m $^2$ , cost $28 \times 2 = 56$ . Fence: perimeter $2 \times (7 + 4) = 22$ m, cost $22 \times 3 = 66$ .

Exercise 34.10 ★★

A corridor floor is $12$ m long and $2$ m wide, and must be covered with square tiles of side $50$ cm. How many tiles are needed? (Convert first, or count tiles along each direction.)

Solution

Solution of Exercise 34.10.

Two tiles of $50$ cm make a meter: along the $12$ m length, $24$ tiles; along the $2$ m width, $4$ tiles. Total: $24 \times 4 = 96$ tiles.

Exercise 34.11 ★★

Double the sides of a $4$ cm $\times$ $3$ cm rectangle. What happens to its perimeter? To its area? (Compute both before and after — the two answers differ!)

Solution

Solution of Exercise 34.11.

Before: perimeter $2 \times (4 + 3) = 14$ cm, area $12$ cm $^2$ . After (sides $8$ and $6$ ): perimeter $28$ cm, area $48$ cm $^2$ . The perimeter doubled; the area was multiplied by four ( $2 \times 2$ ): lengths and areas do not scale the same way.