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Lesson 1 of 7

Perimeter and Area · Lesson 1 of 7

Measuring the Boundary

“Learn how to measure the complete boundary of a shape and use it to solve everyday length problems.”

Learning Objectives

• Explain perimeter as the distance around a closed shape once. • Derive and use rectangle and square perimeter formulas. • Find a missing side from the total perimeter. • Solve problems involving lace, wire, fencing and repeated rounds.

A journey around the edge

Imagine running a finger around the edge of a photo frame. You travel along every outside side and stop when you return to where you began. The length of that complete journey is the perimeter. It measures the boundary, rather than the region inside it.

Definition
Perimeter

The total distance along the boundary of a closed plane figure, measured once around.

Definition
Polygon

A closed flat shape made from straight line segments. Its perimeter is the sum of the lengths of all its sides.

A closed shape has no opening in its boundary. To find its perimeter, choose a starting corner, follow the sides in order, and include each side once. Length units such as centimetres and metres tell us how long the boundary is. Before adding, make sure the side lengths use the same unit.

Example — A triangle boundary

Problem
A triangle has sides 4 cm, 5 cm and 7 cm. Find its perimeter.

  1. 1.There are three outside sides. All their lengths are in centimetres, so they can be added directly.
  2. 2.Perimeter = 4 + 5 + 7 = 16 cm.
  3. 3.The answer describes one complete trip around the triangle.

Why the rectangle and square shortcuts work

A rectangle has two pairs of equal opposite sides. This lets us group repeated lengths instead of adding four unrelated numbers. A square has even more repetition: all four sides have the same length. Both shortcuts come directly from counting the boundary.

12 cm8 cmABCD12 + 8 + 12 + 8= 2 × (12 + 8)= 40 cmWalk along all four sides and return to A.
One complete rectangle boundary— Opposite sides match. Count the top and bottom, and also the left and right.

Let l stand for a rectangle’s length and w for its width, also called its breadth. Travelling around the four sides gives l + w + l + w. There are two copies of l and two copies of w, so the total is twice the sum of one length and one width.

Perimeter of a rectangleLaTeX
P is perimeter, l is length and w is width. Use the same length unit for l and w; P has that length unit too.

For the 12 cm by 8 cm rectangle, the sum of adjacent sides is 12 + 8 = 20 cm. That is only half the boundary. Doubling gives 40 cm. You can check by adding 12 + 8 + 12 + 8 separately.

Let s stand for the side length of a square. Its four sides contribute s + s + s + s. Multiplying the one side length by four gives the same total.

Perimeter of a squareLaTeX
P is perimeter and s is the length of one side. The answer is in the same length unit as s.
Example — Tape for a frame

Problem
A square photo frame has side 1 m. How much tape covers its boundary once?

  1. 1.The tape follows the four equal sides.
  2. 2.Perimeter = 4 × 1 m = 4 m.
  3. 3.Exactly 4 m covers the boundary in this mathematical model; extra tape for fastening would be a separate practical allowance.
Example — Lace for a tablecloth

Problem
A tablecloth is 3 m long and 2 m wide. Find the length of lace needed around it.

  1. 1.The lace goes along the boundary, so use perimeter.
  2. 2.Add one length and one width: 3 + 2 = 5 m.
  3. 3.Double this sum: 2 × 5 = 10 m of lace.
Common mistake

Multiplying length by width does not give perimeter. A boundary is a length, so its answer uses cm or m. Counting the whole outside edge also means that length + width alone is not enough.

Working backwards to a missing side

Sometimes the total boundary length is known but one side is missing. We can undo the steps used to calculate perimeter. For a rectangle, half the perimeter is one length plus one width. For a square, one side is one quarter of its perimeter.

Example — A missing rectangle length

Problem
A rectangle has perimeter 14 cm and width 2 cm. Find its length.

  1. 1.Half of 14 cm is 7 cm. Therefore length + width = 7 cm.
  2. 2.Remove the known width: length = 7 − 2 = 5 cm.
  3. 3.Check all sides: 5 + 2 + 5 + 2 = 14 cm.
Example — Wire bent into a new shape

Problem
A 5 cm by 3 cm wire rectangle is straightened and bent into a square. Find the square’s side.

  1. 1.The wire length is the original rectangle perimeter: 2 × (5 + 3) = 16 cm.
  2. 2.Bending the same wire, without cutting away any part, keeps its total length 16 cm.
  3. 3.The square has four equal sides, so each side is 16 ÷ 4 = 4 cm.
Example — An unknown triangle side

Problem
A triangle has perimeter 55 cm. Two sides are 20 cm and 14 cm. Find the third side.

  1. 1.The known sides together use 20 + 14 = 34 cm of the boundary.
  2. 2.Subtract them from the whole boundary: 55 − 34 = 21 cm.
  3. 3.Check: 20 + 14 + 21 = 55 cm.

If a 36 cm string makes a square, its four equal sides are each 36 ÷ 4 = 9 cm. If it makes an equal-sided triangle, each of the three sides is 12 cm. If it makes an equal-sided hexagon, a six-sided polygon, each side is 6 cm. The number of equal sides tells us how to share the same boundary length.

Several rounds and the cost of a boundary

One complete round of a track covers its perimeter. Several rounds repeat that same distance. A fencing price per metre works similarly: each metre of boundary contributes the same cost. First find the length that is being repeated or paid for, then multiply.

Example — Three park rounds

Problem
Usha walks three rounds of a square park of side 75 m. Find the distance travelled.

  1. 1.One round is 4 × 75 = 300 m.
  2. 2.Three rounds cover 3 × 300 = 900 m.
  3. 3.The distance travelled is 900 m; the park perimeter remains 300 m.
Example — Buying a fence

Problem
A rectangular park is 150 m by 120 m. Fencing costs ₹40 per metre. Find the cost for one boundary.

  1. 1.Perimeter = 2 × (150 + 120) = 2 × 270 = 540 m.
  2. 2.Every metre costs ₹40, so cost = 540 × 40 = ₹21,600.
  3. 3.The length is measured in metres and the total cost in rupees.
Example — Three ropes around a field

Problem
A farmer’s field is 230 m by 160 m. Three complete rounds of rope are needed. Find the total rope length.

  1. 1.One boundary measures 2 × (230 + 160) = 2 × 390 = 780 m.
  2. 2.Three complete rounds require 3 × 780 = 2,340 m.
  3. 3.Multiply the perimeter by three, rather than multiplying just one side.

Quiz

Quick check

Which measurement tells how far you travel once around a shape?

Quick check

A rectangle is 9 cm by 4 cm. What is its perimeter?

Quick check

A square has perimeter 20 cm. What is one side?

Quick check

A rectangle has perimeter 12 m and length 3 m. What is its width?

Quick check

A fence is 80 m long and costs ₹15 per metre. What is its total cost?

Quick check

A 36 cm string forms a triangle with three equal sides. What is each side?

Practice Problems

Practice Problems
  1. Draw a 7 cm by 5 cm rectangle, label every side and find its perimeter in two ways.
  2. A square has perimeter 52 m. Find its side and check by adding all four sides.
  3. A triangle has perimeter 38 cm and two sides 11 cm and 15 cm. Find the remaining side.
  4. A 12 m by 8 m garden needs fencing at ₹25 per metre. Find the length and cost.
  5. A 6 cm by 4 cm wire rectangle is reshaped into a square. Find the new side length.
  6. Explain why four rounds of a square track are not the same as walking along its four sides just once.

Key Takeaways

Key Takeaways

• Perimeter is the length of one complete boundary. • Add every polygon side once; rectangles have P = 2(l + w) and squares have P = 4s. • Use subtraction or division to find a missing side. • Uncut wire or string keeps its length when reshaped. • Find one perimeter before calculating repeated rounds or cost per metre.

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Next · Lesson 2

Perimeter in Paths and Changing Shapes