Why perimeter cannot stand in for area

What to assume they know

• Area as a count of unit squares, and the rectangle formula — Area as a count of unit squares
• Perimeter of a rectangle, and that it depends on both sidelengths
• Multiplication and addition of one- and two-digit numbers, and of simple decimals
• That a statement about "all" shapes is destroyed by one counter-instance
• Reading and drawing shapes on squared paper

What they should be able to do

• State what a measure of area is required to do that boundary length fails to do
• Produce two rectangles with equal perimeters and unequal areas, and verify both claims by computation
• Produce two regions in which the larger perimeter belongs to the smaller area, and say which of the two comparisons each pair of numbers settles
• Hold a perimeter fixed and describe how the area varies as a rectangle is made thinner, including what it tends toward
• Hold an area fixed and show that the perimeter has no upper limit
• Construct a non-rectangular pair in which the comparison can be settled by looking rather than by arithmetic
• Recognise the same independence in three other figures of this chapter and say what stays fixed in each
• State the one genuine relation between the two measures, and why it is an inequality rather than a formula

Where it usually goes wrong

• "Bigger boundary, bigger area." The 1 by 12 rectangle beats the 5 by 5 on boundary by 6 cm and loses on area by 13 cm². One pair kills the rule.
• "Same boundary, same area." The chapter's own two rectangles are 22 cm around apiece and differ by 4 cm². Students find this one the hardest to give up, because the boundary is what the eye follows.
• "Perimeter and area are the same thing measured in different units." They cannot be, because one of them can be changed while the other is held fixed.
• "Doubling the sides doubles both." The boundary doubles; the area quadruples. Part II p.152's fifth item is built on exactly this.
• "If they are independent then no comparison between them is possible." The square is extreme in both directions. Independence rules out a formula, not a bound.
• "The counter-instance needs strange shapes." Two rectangles are enough, which is precisely why the chapter asks for rectangles first and odd shapes second.
• **"A long thin shape only looks like it has less area."** Its unit-square count really is smaller. Overlay the grid and count.
• "Area cannot be squeezed to nothing while the boundary stays put." At 9.9 cm by 0.1 cm the boundary is still 20 cm and the area is under 1 cm². Push the example far enough that the trend is undeniable.

Questions to check understanding

• Given two rectangles, state which has the greater perimeter and which the greater area, and say whether the two answers agree
• Construct two rectangles with equal perimeter and unequal area, with whole-number sides
• Construct two rectangles in which the greater perimeter belongs to the smaller area
• Given a fixed perimeter, find the rectangle with whole-number sides of greatest area, and say why
• Given a fixed area, show that no rectangle has the greatest perimeter
• Decide whether a stated rule linking the two measures is always, sometimes or never true, and support the verdict with a specific pair
• Explain in a sentence why area is measured by counting squares rather than by measuring the boundary — the competency-style item this subheading is written for

Examples worth working on the board

Items marked printed are stated or worked on the page; items marked not in the book are an added construction or arithmetic added here and must not be presented as something the chapter says.

• The question, as the chapter puts it (Part II p.150, under the unnumbered bold subheading ("Why Can't Perimeter be a Measure of Area?")). Printed: three questions in a row — why count squares at all, could the boundary length not serve instead, and does equal boundary length force equal area or does a bigger boundary force a bigger area. Printed: the chapter's answer is that boundary length carries no reliable information about area, because two regions can agree on one and differ on the other, in both directions.
• The chapter's formal demand (Part II p.150). Printed: it asks for two regions in which Region 1's perimeter exceeds Region 2's while Region 1's area is the smaller of the two, and it sets the task of finding two rectangles that do this, pointing the student to the grid paper at the end of the book. A second, Math Talk question asks for a pair of non-rectangular regions with the same property, and requires that the property be obvious to look at.
• The pair the book has already supplied (Part II p.149). The 7 cm by 4 cm and 8 cm by 3 cm rectangles from the rangoli question. Printed: their areas are 28 cm² and 24 cm². Not in the book: their perimeters are both 22 cm. So the first half of the claim — equal boundary, unequal area — is settled by a figure the student met one page earlier, and section 3 should point that out rather than build something new.
• A pair for the harder demand (not in the book; the chapter sets the task and gives no example). Region 1 is 1 cm by 12 cm: perimeter 26 cm, area 12 cm². Region 2 is 5 cm by 5 cm: perimeter 20 cm, area 25 cm². Region 1's boundary is 6 cm longer and its area is 13 cm² smaller. Both fit comfortably on the grid paper the chapter points to, and both are rectangles, which is what the item asks for.
• Fix the perimeter, vary the shape (not in the book). Take 20 cm of boundary and hold it fixed. A 5 by 5 rectangle gives 25 cm²; 6 by 4 gives 24 cm²; 8 by 2 gives 16 cm²; 9 by 1 gives 9 cm²; 9.9 by 0.1 gives 0.99 cm². The area falls away toward zero while the boundary never changes at all. The square is the largest of the family and there is no smallest.
• Fix the area, vary the shape (not in the book). Take 36 cm² and hold it fixed. A 6 by 6 rectangle has perimeter 24 cm; 9 by 4 has 26 cm; 18 by 2 has 40 cm; 36 by 1 has 74 cm; 72 by 0.5 has 145 cm. This direction is the more striking of the two, because the boundary has no ceiling — for any length you name, a thin enough rectangle of area 36 cm² beats it.
• The visually obvious non-rectangular pair (not in the book; the chapter's Math Talk asks for exactly this and supplies nothing). A 6 cm by 6 cm square has perimeter 24 cm and area 36 cm². Against it set a comb cut from a 12 cm by 1 cm strip with nine narrow slits removed: the area is under 12 cm² and the boundary runs to about 40 cm, because every slit adds two long edges. Keep the figure reachable — the bare strip's perimeter is 2(12 + 1) = 26 cm, and a slit cut in from the long edge adds roughly twice its depth, so nine slits 0.9 cm deep add about 16 cm and land near 42 cm. The depth cannot exceed the strip's 1 cm width, so nine slits cannot reach 60 cm whatever their depth (the ceiling is 44 cm); it takes about twenty slits to pass 60. The comparison works handsomely as it is: about 42 cm of boundary on under 12 cm² against the square's 24 cm on 36 cm². A student does not have to compute anything — the comb is plainly less material and plainly more edge. Slits are the cheapest way to add boundary without adding area, and saying so out loud is what makes the pair explanatory rather than merely convincing.
• The same independence, three more times in this chapter. All three are printed figures and all three belong to later topics, so treat them as forward pointers rather than as the explanation's material.
  • Part II p.152, Figure it Out 5. A square cut by two segments into three numbered regions; the item asks by how much each region's area increases when the square's sidelength is doubled. Not in the book: the boundary doubles and every area quadruples, so the two measures do not even scale alike. Units for real areas, and why converting between them squares the length factor owns this item.
  • Part II p.156. All the triangles standing on one base with their apex on a fixed parallel line. Not in the book: every one of them has the same area, while their perimeters differ and have a minimum. The chapter asks both questions and answers only the perimeter one. Triangles with the same base and height have the same area owns this.
  • Part II p.162, Figure it Out 1, and Part II p.163, Figure it Out 4. Seven parallelograms with the same base and height, and then a rectangle set against a parallelogram whose two sidelengths are also 5 cm and 4 cm. Not in the book: the seven all have one area and seven different perimeters; the rectangle and the parallelogram have the same perimeter, 18 cm, and the rectangle has the larger area. Area of a parallelogram, by turning it into a rectangle owns both.
• Where the two measures do connect (not in the book; the chapter states nothing of the kind). Among rectangles of a fixed perimeter, the square holds the most area; among rectangles of a fixed area, the square has the shortest boundary. So there is a real relation — but it is a one-sided bound, not a formula, and it only ever tells you which shape is extreme, never what the other quantity is.

Figures to have open

• The two rangoli rectangles with both measures shown (Part II p.149). The printed figure labels only the sidelengths; the explanation needs the boundary traced and the unit grid added. Schematic.
• A rectangle of fixed perimeter shown at five thinnesses, drawn to scale on one baseline so that the shrinking interior is visible. Standard schematic and the key visual of the topic.
• The same family with area fixed instead, again to scale, so the fifth rectangle runs off the frame. Standard schematic; letting it overflow the frame is the point.
• The slitted comb beside a square of the same footprint. Standard schematic. The comb must be drawn with enough slits that no arithmetic is needed, which is what the chapter's Math Talk asks for.
• The three forward-pointer figures (Part II pp.152, 156, 162) may be reproduced as thumbnails only; each is owned and drawn properly by another topic.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 7, "Area", §7.1 "Rectangle and Squares", under the unnumbered bold subheading ("Why Can't Perimeter be a Measure of Area?"), Part II p.150. It occupies the middle of that page; the Figure it Out block begins below it.
• Part II p.149 for the two rectangles that already serve as the equal-perimeter counter-instance, and for the unit-square definition this topic relies on.
• Part II p.150 for the pointer to grid paper at the end of the book — the "LEARNING MATERIAL SHEETS" section, Part II pp.172–176, of which Part II p.175 is headed "Isometric Grid".
• Forward pointers, each owned elsewhere: Part II p.152 item 5 (owned by Units for real areas, and why converting between them squares the length factor); Part II p.156 questions (i) and (ii) (owned by Triangles with the same base and height have the same area); Part II p.162 item 1 and Part II p.163 item 4 (both owned by Area of a parallelogram, by turning it into a rectangle).
• Backward pointer: Area as a count of unit squares for the count-of-unit-squares definition, which is what section 2 of the explanation argues for.