If the radius of a circle is reduced by 50%, its area will be reduced by

Correct — D, (d) 75%.

THE STEP THAT DECIDES THIS QUESTION is that AREA GOES WITH THE SQUARE OF A LENGTH. The area of a circle is πr², so if the radius is multiplied by some factor, the area is multiplied by the SQUARE of that factor.

Reducing the radius by 50 per cent means halving it, so the multiplier on the radius is 1/2. The multiplier on the area is therefore

(1/2)² = 1/4

The new area is a QUARTER of the old one. A quarter remains, so three quarters have gone, and the area has been reduced by 75 per cent. That is option (d).

WITH NUMBERS, IF THAT IS EASIER TO TRUST. Take a circle of radius 10, whose area is 100π. Halve the radius to 5 and the area becomes 25π. The area has fallen from 100π to 25π, a fall of 75π, which is 75 per cent of what it was.

THE GENERAL FORMULA, worth carrying because these questions recur with different percentages. If the radius falls by p per cent, the remaining fraction of the area is (1 − p/100)², and the percentage reduction in area is

100 [ 1 − (1 − p/100)² ] = 2p − p²/100

Check it at p = 50: 100 − 25 = 75. At p = 10 it gives 19 per cent; at p = 20 it gives 36 per cent; at p = 1 it gives 1·99 per cent. Notice two things about that expression. First, for small reductions the area falls by very nearly TWICE the linear reduction, which is a useful mental rule. Second, the formula is exactly the successive-percentage rule applied twice over — two reductions of p per cent compounding into 2p − p²/100 — because reducing a length reduces both of the dimensions that make up an area.

WHY THE ANSWER MUST EXCEED 50 PER CENT, which is worth establishing before any calculation. Area is a two-dimensional quantity, so a cut in the radius bites twice: the circle becomes narrower AND shorter, so to speak, and the two effects compound. Any reduction in the radius must therefore produce a LARGER percentage reduction in the area. A candidate who knows only that much has already ruled out the options at or below 50 per cent.

BACK-SUBSTITUTION FINISHES THE JOB. Each option can be tested by asking what cut in the radius it would require:

a reduction of 30% leaves 70% of the area, so the radius would fall to √0·70 ≈ 0·837 of its value — a cut of about 16 per cent a reduction of 50% leaves 50% of the area, so the radius would fall to √0·50 ≈ 0·707 — a cut of about 29 per cent a reduction of 60% leaves 40% of the area, so the radius would fall to √0·40 ≈ 0·632 — a cut of about 37 per cent a reduction of 75% leaves 25% of the area, so the radius would fall to √0·25 = 0·5 — a cut of exactly 50 per cent

Only the last matches the stem. It is worth noticing from that table how slowly the radius has to change to produce a large change in area, which is the same fact seen from the other side.

WHAT ELSE CHANGES, AND BY HOW MUCH. The circumference, being a length, is proportional to r and falls by exactly 50 per cent — the same as the radius. The diameter likewise. Only the area, being two-dimensional, falls by 75 per cent. Had the question been about a sphere, the surface area would fall by 75 per cent and the VOLUME, being three-dimensional, by 87·5 per cent, since (1/2)³ = 1/8.

The percentages are printed with no space before the sign, and the stem ends on the words "reduced by" with no punctuation, as this booklet sets them.

Why the others are wrong

• (a)30% — Thirty per cent is the smallest reduction on the list and it is far too small. A fall of 30 per cent would leave 70 per cent of the area, which would require the radius to fall to the square root of 0·70, about 0·837 of its original value — a cut of roughly sixteen per cent, not fifty. The option cannot be reached by any correct handling of the relationship between radius and area, and it can be rejected by the general argument alone: since the area depends on the SQUARE of the radius, a cut in the radius always produces a larger percentage cut in the area, so any option below 50 per cent is impossible before the arithmetic begins. That single observation removes half the list. It is worth adding that the reverse question — by what percentage must the radius fall to halve the area — has the answer about 29 per cent, and a candidate who has that figure in mind from another problem may be reaching for a small number here for the wrong reason.
• (b)50% — Fifty per cent is the trap this question was built for, and it is the answer of a candidate who transfers the change in the radius directly to the area. The mistake is to treat area as though it were proportional to the radius, when it is proportional to the radius SQUARED. Halving a length halves everything one-dimensional about the circle — the diameter, the circumference — but it quarters the area, because both of the dimensions that make up the area have been halved. Tested backwards, a 50 per cent fall in area would leave half the area, which requires the radius to fall to 1/√2, about 0·707 of its value, or a cut of only about 29 per cent. The habit that prevents this error is to ask, of every scaling question, what DIMENSION the quantity has: lengths scale as k, areas as k², volumes as k³. Getting that right is nearly the whole of the topic.
• (c)60% — Sixty per cent is the near miss on the high side, and it is the option for a candidate who has understood that the area must fall by more than the radius does but has not computed how much more. It is a real improvement on the naive answer, since it at least respects the direction of the effect, and that makes it more dangerous rather than less. A fall of 60 per cent would leave 40 per cent of the area, requiring the radius to fall to the square root of 0·40, about 0·632 — a cut of some 37 per cent rather than the 50 the stem states. There is no need to estimate the size of the effect when it can be computed exactly in one step: half squared is a quarter, a quarter remains, and three quarters have gone. Qualitative reasoning is valuable for eliminating options and for checking a result, but where an exact multiplier is available it should be used.

SCALE FACTORS AND DIMENSION are the whole of this topic, and the rule is short enough to state in one line: if every length in a figure is multiplied by k, then every LENGTH is multiplied by k, every AREA by k², and every VOLUME by k³.

That rule holds for any shape whatever, not merely for circles. Doubling the side of a square quadruples its area; doubling the radius of a sphere multiplies its surface area by four and its volume by eight; enlarging a photograph by 20 per cent in each direction increases its area by 44 per cent. The exponent is simply the number of dimensions the quantity has.

THE CIRCLE'S OWN FORMULAE make this concrete. Circumference = 2πr, a length, proportional to r. Area = πr², proportional to r². So a change in the radius passes straight through to the circumference and is squared on the way to the area. For a sphere, surface area = 4πr² and volume = (4/3)πr³, so the same change is squared for one and cubed for the other.

THE PERCENTAGE FORM of the rule is what examinations usually want. If a length changes by p per cent, the multiplier is (1 + p/100) with p taken as negative for a reduction, and the area multiplier is its square. Expanding gives the familiar expression

percentage change in area = 2p + p²/100

with the sign of p carried through. For an increase both terms are positive, so the area rises by more than twice the linear increase — a 10 per cent rise in radius raises the area by 21 per cent. For a decrease the first term is negative and the second positive, so the area falls by a little less than twice the linear fall for small changes — a 10 per cent cut in radius cuts the area by 19 per cent — but for large changes the square term matters a great deal, which is why a 50 per cent cut gives 75 rather than 100.

THIS IS THE SUCCESSIVE-PERCENTAGE RULE AGAIN. An area is the product of two lengths, and changing a figure's scale changes both of them, so the area undergoes two successive percentage changes of the same size. The formula a + b + ab/100 with a = b = p gives 2p + p²/100 exactly. Recognising that the same rule governs two discounts on a price, two stages of a chained race and the scaling of an area is worth more than memorising three separate results.

A PRACTICAL CAUTION. Percentage change questions must always be read for their BASE and their DIRECTION. "Reduced by 75 per cent" and "reduced to 25 per cent" describe the same thing in words that look opposite, and an option list will often offer both. Here the stem says "reduced by", so the answer is the amount removed and not the amount remaining.

Key facts

• The area of a circle is πr², so the area is proportional to the SQUARE of the radius and a scale factor on the radius is squared on the way to the area.
• Halving the radius multiplies the area by (1/2)² = 1/4, so a quarter of the area remains and three quarters — 75 per cent — have gone.
• A circle of radius 10 has area 100π; halving the radius to 5 gives 25π, a fall of 75π, which is 75 per cent of the original.
• If the radius falls by p per cent, the percentage reduction in area is 2p − p²/100; at p = 50 that is 100 − 25 = 75.
• Because area is two-dimensional, any percentage cut in the radius produces a LARGER percentage cut in the area, which rules out every option at or below 50 per cent.
• Back-substitution shows what each option would require of the radius: cuts of about 16, 29 and 37 per cent for reductions of 30, 50 and 60 per cent in area.
• The circumference, being a length, falls by exactly the same 50 per cent as the radius; only the area falls by 75.
• For a sphere the same halving would cut the surface area by 75 per cent and the volume by 87·5 per cent, since (1/2)³ = 1/8.
• The general rule is that scaling every length by k scales lengths by k, areas by k² and volumes by k³, whatever the shape.
• "Reduced BY 75 per cent" and "reduced TO 25 per cent" describe the same circle, and reading the wrong one of the two costs the item after correct working.

Common traps

• Carrying the change in the radius straight across to the area. Area goes with the square of the radius, so a half becomes a quarter.
• Knowing the area must fall by more than 50 per cent but guessing how much more. The exact multiplier takes one step and removes the guess.
• Answering with the area that remains rather than the amount removed. A quarter remains; three quarters have gone.
• Applying the area rule to the circumference, which is a length and falls by exactly the same percentage as the radius.
• Forgetting that the same reasoning cubes for a volume, so a halved radius leaves an eighth of a sphere and not a quarter.

• practice — not a real PYQ

  If the radius of a circle is increased by 20 per cent, by what percentage does the area of that circle increase ?

  • (a)20 per cent
  • (b)40 per cent
  • (c)44 per cent
  • (d)48 per cent

  Answer(c) 44 per cent — the radius is multiplied by 1·2, so the area is multiplied by 1·2² = 1·44, an increase of 44 per cent. The formula 2p + p²/100 gives 40 + 4 = 44, and the extra 4 above twice the linear change is the square term.

• practice — not a real PYQ

  If the edge of a cube is reduced to half its original length, the volume of the cube is reduced to what fraction of its original volume ?

  • (a)One half
  • (b)One quarter
  • (c)One sixth
  • (d)One eighth

  Answer(d) One eighth — volume is a three-dimensional quantity, so the scale factor is cubed: (1/2)³ = 1/8. The surface area, being two-dimensional, would fall to one quarter, and the edge itself to one half.