CDS Mathematics · Mensuration 2D

Equal Perimeters & Re-bent Wires

A wire keeps its length when it is bent into a new shape, so the perimeter carries over; and for a fixed perimeter the rounder shape encloses more area.

All Mensuration 2D notes CDS Mathematics strategy

Why this matters

A family CDS returns to in most years, usually as a quick question: find the length, then the new shape's side or radius. The comparison version (circle against square against triangle) is worth memorising as a ranking, because it needs no working at all.

Concept 1 of 3: The wire keeps its length

Re-bending changes the shape, never the length. So every question has the same two steps: turn the first shape into a length (its perimeter), then turn that length into the second shape.

Definition

Circle: length $2\pi r$ . Square: $4s$ . Equilateral triangle: $3a$ . Rhombus: $4a$ .
Semicircle: $\pi r + 2r$ , because the diameter is part of the boundary.
Sector: arc plus two radii.
A wire cut into two pieces: the two perimeters add to the total length.

Length is conserved

\text{perimeter of the old shape} = \text{perimeter of the new shape}

Worked example

A wire bent into a circle of radius

21

cm is re-bent into a square. Find the area of the square.

(\pi = \tfrac{22}{7})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (I) 2017 — Elementary Mathematics · Q83Moderate

Example 1 · Mensuration 2D · Equal Perimeters and Re-bent Wires

A copper wire when bent in the form of a square encloses an area of

121 \text{ cm}^2

. If the same wire is bent in the form of a circle, it encloses an area equal to

A semicircle's boundary includes the diameter

A wire bent into a semicircle forms the arc AND the straight diameter:

\pi r + 2r

. Using

\pi r

alone gives a larger radius. When no printed option matches either reading, 'None of the above' is the answer the paper wants.

Drill 7 more on the wire keeps its length

Concept 2 of 3: Same perimeter: the rounder shape wins

For a fixed length of boundary, the circle encloses the most area, then the square, then the equilateral triangle. The ratios all follow from writing each area in terms of the common perimeter

P

Definition

With a common perimeter $P$ :

Equilateral triangle: $\dfrac{\sqrt3P^2}{36} \approx 0.048P^2$ .
Square: $\dfrac{P^2}{16} = 0.0625P^2$ .
Circle: $\dfrac{P^2}{4\pi} \approx 0.080P^2$ .

So triangle $<$ square $<$ circle. Circle : square $= 4 : \pi = 14 : 11$ with $\pi = \tfrac{22}{7}$ . Turned round, for equal areas the circle has the smaller perimeter: circle : square $= \sqrt\pi : 2$ .

Equal perimeter

\frac{\text{circle}}{\text{square}} = \frac{4}{\pi}, \qquad \frac{\text{triangle}}{\text{square}} = \frac{4\sqrt3}{9}

Worked example

A square and a regular hexagon have the same perimeter

P

. Which has the larger area?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (II) 2018 — Elementary Mathematics · Q82Moderate

Example 2 · Mensuration 2D · Equal Perimeters and Re-bent Wires

An equilateral triangle, a square and a circle have equal perimeter. If T, S and C denote the area of the triangle, area of the square and area of the circle respectively, then which one of the following is correct ?

Equal perimeter and equal area flip the answer

Equal perimeters: the circle has the MORE area. Equal areas: the circle has the LESS perimeter. Both are the same fact, so read which quantity the stem fixes before choosing.

Drill 5 more on same perimeter: the rounder shape wins

Concept 3 of 3: Same perimeter: the square beats every rectangle

A rectangle and a square with the same perimeter have the same

l + b

. The square's area exceeds the rectangle's by exactly the square of half the difference of the sides.

Definition

If $4s = 2(l + b)$ , then $s = \dfrac{l + b}{2}$ and $s^2 - lb = \left(\dfrac{l - b}{2}\right)^2$ .

So the square always has the larger area, and the rectangle of largest area for a given perimeter is the square.
If the areas differ by $q$ , then $(l - b)^2 = 4q$ .

Square minus rectangle

s^2 - lb = \left(\frac{l-b}{2}\right)^2

Worked example

A square and a rectangle have the same perimeter, and the square's area is

9

^2

more. By how much does the rectangle's length exceed its breadth?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (II) 2022 — Elementary Mathematics · Q89Moderate

Example 3 · Mensuration 2D · Equal Perimeters and Re-bent Wires

A square and a rectangle have same perimeter. They differ in areas by 1 square cm. The length of the rectangle exceeds its breadth by

Drill 2 more on same perimeter: the square beats every rectangle

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The wire keeps its length
Length is conserved
$\text{perimeter of the old shape} = \text{perimeter of the new shape}$
Same perimeter: the rounder shape wins
Equal perimeter
$\frac{\text{circle}}{\text{square}} = \frac{4}{\pi}, \qquad \frac{\text{triangle}}{\text{square}} = \frac{4\sqrt3}{9}$
Same perimeter: the square beats every rectangle
Square minus rectangle
$s^2 - lb = \left(\frac{l-b}{2}\right)^2$

Watch out for (2)

A semicircle's boundary includes the diameter
→ The wire keeps its length
Equal perimeter and equal area flip the answer
→ Same perimeter: the rounder shape wins

Test yourself on Mensuration 2D

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Mensuration 2D chapter test Drill all 17 questions