Composite areas · word problems

Composite Area Heuristics

10 question types · PSLE Paper 1 and 2 · one figure that follows the deduction

01

The Leaf in a Square

propertyTwo quadrants overlap in the leaf: leaf = two quadrants − square

ABCD is a square of side 14 cm. Two quadrants are drawn, one with centre A and one with centre C, each of radius 14 cm. They overlap in a leaf shape. Take π = 227. Find the area of the leaf and the total area of the two corner regions outside it.

ABCD14 cm
Two quadrants of radius 14, one from A and one from C: 154 cm² each.
One quadrant: 14 × 227 × 14 × 14 = 154 cm2.
step 1 of 4

Each quadrant fills the square except for one corner region, and the two together cover the leaf twice. So adding the two quadrants counts the square once and the leaf once more.

  1. One quadrant: 14 × 227 × 14 × 14 = 154 cm2.
  2. Two quadrants together: 308 cm2. They cover the whole square, 196 cm2, with the leaf covered twice.
  3. Leaf = 308 − 196 = 112 cm2.
  4. The two corner regions are what the leaf leaves of the square: 196 − 112 = 84 cm2.

answer112 cm2; 84 cm2

Common pitfalls

  • Taking the leaf as half the square by eye.
  • Subtracting one quadrant from the square, 196 − 154 = 42, which is one corner region, not the leaf.
02

The Four-Petal Rose

propertyFour semicircles on the sides make two circles; petals = two circles − square

ABCD is a square of side 28 cm. A semicircle is drawn inside the square on each of its four sides. The overlaps make four petals. Take π = 227. Find the total area of the four petals and the area of the square not covered by any petal.

28 cm14
Each semicircle has radius 14: 308 cm².
Each semicircle: radius 14 cm, area 12 × 227 × 14 × 14 = 308 cm2.
step 1 of 5

Each semicircle has radius 14 cm. The four semicircles together fill the square exactly once with every petal covered twice, because each petal is the overlap of two neighbouring semicircles.

  1. Each semicircle: radius 14 cm, area 12 × 227 × 14 × 14 = 308 cm2.
  2. Four semicircles: 4 × 308 = 1232 cm2, the same as two full circles.
  3. They cover the square, 28 × 28 = 784 cm2, with the petals covered twice.
  4. Petals = 1232 − 784 = 448 cm2.
  5. Not covered by a petal: 784 − 448 = 336 cm2.

answer448 cm2; 336 cm2

Common pitfalls

  • Using 28 cm as the radius of each semicircle; the side is the diameter.
  • Halving 1232 − 784 because there are two circles: the subtraction already counts each petal once.
03

Cut from One Side, Added to the Other

propertyMoving a piece keeps the area; the perimeter changes by the edges it swaps

A rectangle is 20 cm by 14 cm. A semicircle with the left 14 cm side as diameter is cut out, and an identical semicircle is attached to the right side. Take π = 227. Find the area and the perimeter of the new shape.

20 cm14 cm
The semicircle cut from the left is the semicircle added on the right.
The semicircle cut out and the semicircle added are identical: area unchanged, 20 × 14 = 280 cm2.
step 1 of 4

The piece taken away is the piece added back, so the area is the rectangle's. In the outline, both 14 cm sides have been replaced by arcs.

  1. The semicircle cut out and the semicircle added are identical: area unchanged, 20 × 14 = 280 cm2.
  2. Each arc: radius 7 cm, length 12 × 2 × 227 × 7 = 22 cm.
  3. The outline is the two 20 cm sides and two arcs; neither 14 cm side remains.
  4. Perimeter = 20 + 20 + 22 + 22 = 84 cm.

answer280 cm2; 84 cm

Common pitfalls

  • Adding the semicircle's area once and forgetting to subtract it once.
  • Keeping the 14 cm sides in the perimeter; the arcs replaced them.
04

The Propeller

propertyMoving the cut pieces into the gaps keeps the area; every side becomes an arc

A square has sides of 20 cm. A semicircle of radius 10 cm is cut out from each of two opposite sides, and two identical semicircles are attached to the other two sides. Take π = 3.14. Find the area and the perimeter of the shape.

20 cm
Two semicircles cut from the top and bottom, two the same added left and right.
Two semicircles out, two semicircles in: area = 20 × 20 = 400 cm2.
step 1 of 5

What is cut is what is added, so the area is the square's. Each semicircle spans a whole side, so the outline is four arcs and nothing straight.

  1. Two semicircles out, two semicircles in: area = 20 × 20 = 400 cm2.
  2. Each semicircle has diameter 20 cm, the full side, so every side of the square is replaced by an arc.
  3. One arc = 12 × 2 × 3.14 × 10 = 31.4 cm.
  4. Four arcs: 4 × 31.4 = 125.6 cm.
  5. Perimeter = 125.6 cm.

answer400 cm2; 125.6 cm

Common pitfalls

  • Adding the attached semicircles' area without subtracting the cut ones.
  • Keeping any of the square's 80 cm perimeter: every straight edge has been replaced by an arc.
05

The Segment Cut Off by a Chord

propertySegment = sector − triangle; two segments make a leaf

OAB is a quadrant of radius 14 cm and AB is the chord joining the ends of its arc. Take π = 227. Find the area between the chord and the arc, and the area of the leaf made by two such pieces placed back to back along AB.

OAB14 cm
Quadrant of radius 14: 154 cm². The chord AB cuts it in two.
Quadrant = 14 × 227 × 14 × 14 = 154 cm2.
step 1 of 4

The chord cuts the quadrant into a right-angled isosceles triangle and a curved segment. The segment is what is left of the quadrant after the triangle; two of them make the leaf.

  1. Quadrant = 14 × 227 × 14 × 14 = 154 cm2.
  2. Triangle OAB is right-angled at O with legs 14 cm: 12 × 14 × 14 = 98 cm2.
  3. Segment = 154 − 98 = 56 cm2.
  4. Two segments back to back along the chord: leaf = 2 × 56 = 112 cm2.

answer56 cm2; 112 cm2

Common pitfalls

  • Using the chord AB as the base of the triangle with 14 cm as height; the legs OA and OB are the base and height.
  • Taking the segment as a quarter of the quadrant by guesswork.
06

Square Inside a Circle

propertyThe square's diagonal is the circle's diameter, and a square is half the product of its diagonals

A square is drawn with all four corners on a circle of radius 10 cm. Take π = 3.14. Find the area of the square and the area inside the circle but outside the square.

10
Each diagonal passes through the centre: a diameter of 20 cm.
Each diagonal is a diameter: 2 × 10 = 20 cm.
step 1 of 5

Each diagonal of the square passes through the centre, so it is a diameter. The two diagonals cut the square into four right-angled triangles with legs equal to the radius, which gives the area without any side length.

  1. Each diagonal is a diameter: 2 × 10 = 20 cm.
  2. The diagonals cut the square into four triangles, each with two legs of 10 cm meeting at the centre at right angles.
  3. Square = 4 × 12 × 10 × 10 = 200 cm2.
  4. Circle = 3.14 × 10 × 10 = 314 cm2.
  5. Inside the circle, outside the square: 314 − 200 = 114 cm2.

answer200 cm2; 114 cm2

Common pitfalls

  • Taking the square's side as 10 cm or 20 cm; it is neither, and it is not needed.
  • Quartering the circle to get the square.
07

Circle Inside a Square

propertyThe four corner pieces together are square − circle; one corner is a quarter of that

A circle of radius 7 cm is drawn inside a square, touching all four sides. Take π = 227. Find the total area of the four corner regions outside the circle, and the area of one corner region.

7 cm
The circle touches all four sides: the square is 14 cm across.
Side of the square = 2 × 7 = 14 cm; square = 196 cm2.
step 1 of 4

The circle touches all four sides, so the square's side is the circle's diameter. Everything in the square outside the circle is the four corners, and by symmetry they are equal.

  1. Side of the square = 2 × 7 = 14 cm; square = 196 cm2.
  2. Circle = 227 × 7 × 7 = 154 cm2.
  3. Four corners together = 196 − 154 = 42 cm2.
  4. One corner = 42 ÷ 4 = 10.5 cm2.

answer42 cm2; 10.5 cm2

Common pitfalls

  • Using 7 cm as the side of the square.
  • Treating a corner region as a triangle and using a formula for it; it has a curved side, and only subtraction reaches it.
08

Sliding the Apex Between Parallel Lines

propertySame base, same parallel lines, same height: the area does not change

AB is a 12 cm segment on one of two parallel lines 5 cm apart. C and D are two points on the other line. Find the area of triangle ABC and the area of triangle ABD.

ABCD12 cm5 cm
Both apexes sit on the upper line, 5 cm above AB. Slide C along it.
The height of a triangle is the perpendicular distance from the apex to the base line; for any point on the second parallel, that is 5 cm.
step 1 of 4

Both triangles stand on the same base AB and reach up to the same parallel line, so both have height 5 cm. Where the apex sits along that line makes no difference.

  1. The height of a triangle is the perpendicular distance from the apex to the base line; for any point on the second parallel, that is 5 cm.
  2. Triangle ABC: 12 × 12 × 5 = 30 cm2.
  3. Triangle ABD has the same base and the same height: 30 cm2.
  4. Sliding the apex along the parallel line changes the shape but not the area.

answer30 cm2; 30 cm2

Common pitfalls

  • Using the slanted side as the height when the apex is far along the line.
  • Believing the more stretched triangle must be bigger.
09

Trapezium to Rectangle

propertySlice off the triangle on one flank and slide it to the other: a rectangle of the average length

A trapezium has parallel sides of 10 cm and 16 cm and a height of 6 cm, with equal slanted sides. A right-angled triangle is cut from one flank and moved to the other to make a rectangle. Find the length of that rectangle and the area of the trapezium.

10 cm16 cm633
The base overhangs the top by 6 cm: 3 cm on each flank.
The longer side overhangs the shorter by 16 − 10 = 6 cm, 3 cm on each flank.
step 1 of 4

The overhang on each side is half the difference of the parallel sides. Moving one flank's triangle across evens the two sides out to the average length, and the height stays.

  1. The longer side overhangs the shorter by 16 − 10 = 6 cm, 3 cm on each flank.
  2. Cut the 3 cm wide triangle from the right flank and slide it to the left: the top gains 3 cm and the bottom loses 3 cm.
  3. Both sides are now 13 cm: a rectangle 13 cm by 6 cm.
  4. Area = 13 × 6 = 78 cm2, the same as 12 × (10 + 16) × 6.

answer13 cm; 78 cm2

Common pitfalls

  • Multiplying 16 × 6 or 10 × 6: the rectangle's length is the average of the two parallel sides.
  • Using the slanted side as the height.
10

Shaded Sectors of a Circle

propertyEqual sectors share the circle equally; a sector's perimeter is two radii and its arc

A circle of radius 14 cm is cut into 8 equal sectors, and 3 of them are shaded. Take π = 227. Find the shaded area and the perimeter of one sector.

14
Eight equal sectors of a circle of 616 cm².
Circle = 227 × 14 × 14 = 616 cm2.
step 1 of 4

Eight equal sectors each take an eighth of the area and an eighth of the circumference. The shaded part is three eighths, and one sector's outline is its two radii and its arc.

  1. Circle = 227 × 14 × 14 = 616 cm2.
  2. Shaded = 38 × 616 = 231 cm2.
  3. Circumference = 2 × 227 × 14 = 88 cm; one arc = 88 ÷ 8 = 11 cm.
  4. Perimeter of one sector = 14 + 14 + 11 = 39 cm.

answer231 cm2; 39 cm

Common pitfalls

  • Taking the shaded area as 3 × 14 or as three eighths of the circumference.
  • Giving the arc alone as the sector's perimeter.
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