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Percentage change, similarity and re-bent shapes

⏱ 4 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

When a shape grows by a scale factor k, every length multiplies by k but every area multiplies by k squared. For percentage changes the master formula is a + b + ab/100. The same wire bent into different shapes encloses different areas, and the circle always wins.

Lengths scale once, areas twice

Similar figures with a length ratio $k$ have an area ratio $k^2$ . Perimeters are lengths, so they take $k$ too.

Perimeters $3:5$ mean areas $9:25$ . If the larger area is $200$ sq cm, the smaller is $200\times\dfrac{9}{25}=72$ sq cm. Multiply by the ratio; never subtract a guessed difference.

Going from areas back to lengths means a square root: areas $9:25$ give sides $3:5$ . Two circles with areas $154$ and $616$ sq cm are in ratio $1:4$ , so their radii are in ratio $1:2$ .

Rule: Square going from lengths to areas; take the root going back. This single line is half the subtopic.

The percentage change master formula

For two successive percentage changes $a$ and $b$ :

\text{net}=a+b+\frac{ab}{100}

Side $+20\%$ : area $=20+20+\dfrac{400}{100}=+44\%$ . Side $-10\%$ : area $=-10-10+\dfrac{100}{100}=-19\%$ .

Reverse it by factoring: area $+21\%$ means the side grew by $10\%$ , because $1.21=1.1^2$ .

Watch: A $20\%$ side increase is a $44\%$ area increase, not $40\%$ . The cross term is what the options try to hide.

Costs and tiles scale with area

A square room of side $5$ m costs Rs $600$ to carpet. A similar room of side $15$ m has $k=3$ , so area is $9$ times and cost is $600\times9=$ Rs $5400$ .

Tiles ask the same: divide the floor area by one tile's area, then multiply by the rate. A floor of $6$ m by $4$ m carries $24$ sq m $=240{,}000$ sq cm; tiles of $20\times20$ cm cover $400$ sq cm each, so $600$ tiles.

Numbers to trust: side ratio $2:3$ gives cost ratio $4:9$ . A Rs $500$ job grows to Rs $1125$ at that scale, since $500\times\dfrac{9}{4}=1125$ .

Tip: Never scale money by the side ratio. Money follows area, so it takes $k^2$ .

Maps, models and enlargements

A map scale $1:5000$ makes lengths $5000$ times bigger, so areas grow by $5000^2=25{,}000{,}000$ . A $4$ sq cm plot is $10^8$ sq cm in truth, which is $10{,}000$ sq m ( $1$ sq m $=10^4$ sq cm).

Remember: Square the map scale for areas, then convert units once at the end.

Same wire, different shapes

A $44$ cm wire bent into a circle gives $r=\dfrac{44}{2\pi}=7$ cm, area $154$ sq cm. Bent into a square, side $11$ cm, area $121$ sq cm.

For a fixed perimeter, the circle encloses the most area; among rectangles, the square wins. Equilateral beats every other triangle. Order the options by shape generosity when the question says 'maximum area'.

Watch: Compare areas only after converting the perimeter into each shape's own dimensions.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Linear scale to area scale

How to spot it:

Two similar figures with a perimeter or side ratio, an area asked.

Method

Read off the length ratio $k$ .
Square it for the area ratio.
Apply the ratio to the given area.

Why it works:

Squaring once converts any length comparison into an area comparison.

Try this

The perimeters of two similar figures are in the ratio 3 : 5. If the larger area is 200 sq cm, the smaller area is:

Show solution

Area ratio $=\dfrac{9}{25}$ .
Smaller $=200\times\dfrac{9}{25}=72$ sq cm.

Answer

72 sq cm

Type 2common2 practice Q

Cost, tiles and carpet scaling

How to spot it:

A cost for one size, asked for a similar larger size.

Method

Find the linear scale $k$ between the shapes.
Square it: cost multiplies by $k^2$ .
Multiply the given cost.

Why it works:

Cost per unit area is fixed, so total cost behaves exactly like area.

Try this

Carpeting a square room of side 5 m costs Rs 600. The cost of carpeting a square room of side 15 m is:

Show solution

$k=\dfrac{15}{5}=3$ , so area factor $=9$ .
Cost $=600\times9=$ Rs 5400.

Answer

Rs 5400

Type 3very common2 practice Q

Percentage change in side to change in area

How to spot it:

A side percent change, area percent asked (or the reverse).

Method

Put $a=b=$ the percent change.
Net $=a+b+\dfrac{ab}{100}$ .
Reverse direction: take the square root of the factor.

Why it works:

The master formula handles both directions and dodges the missing cross term.

Try this

If each side of a square is increased by 20%, the percentage increase in its area is:

Show solution

$20+20+\dfrac{400}{100}$ .
$=44\%$ .

Answer

44%

Type 4common2 practice Q

Maps, models and photo enlargements

How to spot it:

A map or model with a scale, a true area asked.

Method

Note the linear scale.
Square it for the area factor.
Multiply the map area and convert units.

Why it works:

Areas need the squared scale, and one clean unit conversion finishes it.

Try this

On a map of scale 1 : 5000, a plot measures 4 sq cm. Its actual area is:

Show solution

Area factor $=5000^2=25{,}000{,}000$ .
$4\times25{,}000{,}000=10^8$ sq cm $=10{,}000$ sq m.

Answer

10,000 sq m

Type 5occasional2 practice Q

Same perimeter: square vs circle

How to spot it:

One wire or fence bent into different shapes.

Method

Split the perimeter into each shape's dimensions.
Compute each area.
Compare; the circle encloses the most.

Why it works:

A fixed perimeter is shared, so the comparison isolates the shape effect alone.

Try this

A wire 44 cm long is bent into a circle. The area enclosed is (take pi = 22/7):

Show solution

$r=\dfrac{44\times7}{2\times22}=7$ cm.
$K=\dfrac{22}{7}\times7^2=154$ sq cm.

Answer

154 sq cm

Formula sheet

Similar figures

\frac{a_1}{a_2}=k\Rightarrow\frac{K_1}{K_2}=k^2

Lengths take k once; areas take k squared.

Successive percentage change

\text{net}=a+b+\frac{ab}{100}

Works for two changes in a row, like both dimensions.

Reverse percentage area change

1+\frac{x}{100}=\left(1+\frac{y}{100}\right)^2

Area factor is the side factor squared.

Map areas

\text{true area}=\text{map area}\times(\text{scale})^2

Square the linear scale before converting units.

Shortcuts that save time

⚡ a + b + ab/100 in one line

Both dimensions change by the same percent: plug once, no quadratics, no decimals.

Example

If each side of a square is increased by 20%, the percentage increase in its area is:

Show solution

$a=b=20$ .
$20+20+\dfrac{20\times20}{100}$ .
$=44\%$ .

Answer

44%

⚡ Reverse the square root

Area up 21%? Factor $1.21=1.1^2$ , so the side rose 10%. Recognise perfect squares of decimals.

Example

The area of a square increases by 21%. Each side increased by:

Show solution

$1.21=1.1^2$ .
Side factor $=1.1$ .
Increase $=10\%$ .

Answer

10%

⚡ Wire into shapes: circle wins

Same perimeter, compare areas. Circle beats square beats any rectangle, so guess before computing.

Example

A wire 44 cm long is bent into a circle. The area enclosed is (take pi = 22/7):

Show solution

$2\pi r=44\Rightarrow r=\dfrac{44\times7}{44}=7$ cm.
$K=\dfrac{22}{7}\times49=154$ sq cm.

Answer

154 sq cm

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Multiplying area by $k$ instead of $k^2$ .

Side ratio $3:5$ gives area ratio $9:25$ . Square the length ratio.

Mistake 02

Adding $20\%+20\%=40\%$ for an area change.

Use $a+b+\dfrac{ab}{100}$ : the answer is $44\%$ .

Mistake 03

Scaling cost by the side ratio.

Cost follows area: side $5\to15$ is $k=3$ , so cost multiplies by $9$ .

Mistake 04

Applying the map scale once to areas.

Square it: $1:5000$ multiplies areas by $25{,}000{,}000$ .

Mistake 05

Cutting the area change in half to get the side change.

Take the square root: $+21\%$ area is $+10\%$ side, not $+10.5\%$ .

Quick revision

Read this the night before the exam.

Length ratio $k$ gives area ratio $k^2$ ; perimeters take $k$ once.
Two successive changes: $a+b+\dfrac{ab}{100}$ ; $+20\%$ side means $+44\%$ area.
Area to side: take the square root of the factor ( $1.21\to1.1$ ).
Money, tiles and paint scale with area, never with a side.
Map areas: square the linear scale, convert units once.
Fixed perimeter: circle encloses the most area; square beats other rectangles.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.

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