Overlapping shapes · word problems

Overlap Heuristics

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

01

Identical Shapes, Equal Unshaded Parts

propertyTake the common overlap away from two equal areas and what is left is equal

Two identical squares of side 10 cm overlap. The overlapping region has an area of 15 cm2. Find the unshaded area of each square outside the overlap, and the total area covered by the two squares.

101015
Each square is 100 cm²; the red overlap belongs to both.
Each square: 10 × 10 = 100 cm2.
step 1 of 4

The overlap belongs to both squares. Each square's unshaded part is its own area with that same overlap removed, so the two parts are equal, and the whole figure counts the overlap once.

  1. Each square: 10 × 10 = 100 cm2.
  2. Unshaded part of one square = 100 − 15 = 85 cm2; the same for the other, since the overlap taken away is the same region.
  3. Total covered = 100 + 100 − 15 = 185 cm2: the overlap is counted once, not twice.
  4. Check: 85 + 85 + 15 = 185.

answer85 cm2; 185 cm2

Common pitfalls

  • Adding the two squares to 200 cm2 for the figure: the overlap is one region, covered by both.
  • Subtracting the overlap twice from the total.
02

Different Shapes, a Fixed Difference

propertyUnshaded A − unshaded B = area A − area B, whatever the overlap

A square of side 12 cm and a rectangle 15 cm by 8 cm overlap. Find the difference between the unshaded part of the square and the unshaded part of the rectangle. If the unshaded part of the rectangle is 90 cm2, find the unshaded part of the square.

1212158
Square 144 cm², rectangle 120 cm²; the same overlap sits in both.
Square = 144 cm2; rectangle = 15 × 8 = 120 cm2.
step 1 of 4

Both unshaded parts lose the same overlap, so their difference is the difference of the whole shapes. That difference does not depend on how big the overlap is.

  1. Square = 144 cm2; rectangle = 15 × 8 = 120 cm2.
  2. Unshaded square = 144 − overlap; unshaded rectangle = 120 − overlap.
  3. Difference = 144 − 120 = 24 cm2, the overlap cancelling.
  4. With the rectangle's unshaded part 90 cm2: overlap = 120 − 90 = 30 cm2, so unshaded square = 144 − 30 = 114 cm2; or simply 90 + 24 = 114.

answer24 cm2; 114 cm2

Common pitfalls

  • Trying to find the overlap first when the question asks only for the difference.
  • Subtracting 24 from 90 instead of adding: the square is the bigger shape, so its unshaded part is the bigger one.
03

Total Area of Two Overlapping Shapes

propertyTotal = A + B − overlap; so overlap = A + B − total

A rectangle 20 cm by 12 cm and a square of side 10 cm overlap. When the overlap is 30 cm2, find the total area covered. Later the square is moved so that the total area covered is 290 cm2; find the new overlap.

20121030
Rectangle 240 and square 100: 340 cm² with the overlap counted twice.
Rectangle = 240 cm2; square = 100 cm2; together 340 cm2 with the overlap counted twice.
step 1 of 4

Adding the two shapes counts the overlap twice, so take it away once. Turn the same equation round to find the overlap from the total.

  1. Rectangle = 240 cm2; square = 100 cm2; together 340 cm2 with the overlap counted twice.
  2. Total covered = 340 − 30 = 310 cm2.
  3. For a total of 290 cm2: overlap = 340 − 290 = 50 cm2.
  4. Check: 240 + 100 − 50 = 290.

answer310 cm2; 50 cm2

Common pitfalls

  • Subtracting the overlap from each shape and adding, which takes it away twice.
  • Reading 290 as the overlap.
04

Three Squares in a Row

propertyAdd all three, subtract each pair's overlap; check that the outer two do not meet

Three identical squares of side 10 cm are placed in a row with their bases on one line. Each square overlaps the next by a strip 3 cm wide. Find the total length of the row and the total area covered.

1033
Three squares of 10, each sharing a 3 cm strip with the next.
Length: the first square is 10 cm; each of the next two adds 10 − 3 = 7 cm.
step 1 of 5

Each new square adds its full side less the strip it shares with the previous one. The area works the same way, once it is checked that the first and third squares do not overlap each other.

  1. Length: the first square is 10 cm; each of the next two adds 10 − 3 = 7 cm.
  2. Total length = 10 + 7 + 7 = 24 cm.
  3. Each overlap strip is 3 × 10 = 30 cm2, and there are two of them.
  4. The first and third squares are 14 cm apart at their left edges, more than a side, so they do not overlap each other.
  5. Area = 3 × 100 − 2 × 30 = 240 cm2; or 24 × 10 = 240 cm2.

answer24 cm; 240 cm2

Common pitfalls

  • Subtracting three overlaps for three squares; there are only two places where neighbours meet.
  • Giving the length as 30 − 3 = 27 cm, taking off one strip instead of two.
05

Sliding One Rectangle Over Another

propertyOverlap = (length − slide) × width, as long as the slide is less than the length

Two identical rectangles, 12 cm by 5 cm, lie exactly on top of each other. The top one is slid sideways by x cm. Find the overlap when x = 4. Find x when the overlap is 25 cm2.

125x = 440
Slide the top rectangle: the overlap is (12 − x) long and still 5 wide.
After sliding x cm, the overlap is (12 − x) cm long and still 5 cm wide.
step 1 of 4

Sliding along the length leaves an overlap that is a rectangle of the same width and a shorter length. The overlap shrinks by 5 cm2 for every centimetre of slide.

  1. After sliding x cm, the overlap is (12 − x) cm long and still 5 cm wide.
  2. For x = 4: (12 − 4) × 5 = 40 cm2.
  3. For an overlap of 25 cm2: length of overlap = 25 ÷ 5 = 5 cm.
  4. x = 12 − 5 = 7 cm.

answer40 cm2; 7 cm

Common pitfalls

  • Taking the overlap to be x × 5: the overlap is what is left, not what was moved.
  • Forgetting that once x reaches 12 cm there is no overlap at all.
06

A Square Turning on Another's Centre

propertyA square with a corner at another equal square's centre always covers a quarter of it

Two identical squares have sides of 10 cm. One corner of the second square is fixed at the centre of the first, and the second square is turned about that point. Find the area of the overlap, and the total area covered by the two squares.

10Ooverlap 25.0 cm² at 25°
The dotted quarter is a quarter of the first square: 25 cm².
Draw the two lines from the centre O of the first square to the midpoints of two neighbouring sides: they cut off a quarter, 25 cm2.
step 1 of 4

The two lines of the overlap's boundary meet the first square's sides at points equally far from its centre, and the overlap is made of a triangle and a piece that together are exactly a quarter of the square, whatever the angle.

  1. Draw the two lines from the centre O of the first square to the midpoints of two neighbouring sides: they cut off a quarter, 25 cm2.
  2. Turning the second square by any angle moves a triangle out of that quarter on one side and an identical triangle into it on the other, because the sides of the first square are symmetric about O.
  3. So the overlap is always a quarter of the square: 100 ÷ 4 = 25 cm2.
  4. Total covered = 100 + 100 − 25 = 175 cm2.

answer25 cm2; 175 cm2

Common pitfalls

  • Thinking the overlap changes with the angle and cannot be found without it.
  • Giving 50 cm2, half the square, because the turned square looks large.
07

The Six-Pointed Star

propertyEach triangle splits into nine small ones; the hexagon is six of them and the star is twelve

Two identical equilateral triangles, each of area 36 cm2, overlap to make a six-pointed star with a regular hexagon in the middle. Find the area of the hexagon and the area of the whole star.

Each big triangle is nine small ones of 4 cm².
Each big triangle is nine small equilateral triangles of 36 ÷ 9 = 4 cm2.
step 1 of 4

Lines through the points where the triangles cross divide each big triangle into nine equal small triangles. The hexagon takes six of them; each point of the star is one more.

  1. Each big triangle is nine small equilateral triangles of 36 ÷ 9 = 4 cm2.
  2. The hexagon in the middle is six of them: 6 × 4 = 24 cm2.
  3. The star is the hexagon plus six points, one small triangle each: 24 + 6 × 4 = 48 cm2.
  4. Check: the two triangles add to 72 cm2, and 72 − 24 = 48 cm2, the overlap counted once.

answer24 cm2; 48 cm2

Common pitfalls

  • Taking the hexagon as half of a triangle: it is two thirds.
  • Adding both triangles for the star without removing the overlap.
08

A Fraction of One Is a Fraction of the Other

propertyThe overlap is the same region, so the two fractions describe one number of units

Shapes P and Q overlap. The overlap is 13 of P and 14 of Q. The area of P is 60 cm2. Find the area of Q and the total area covered by the two shapes.

P = 60Qthe overlap is ⅓ of P and ¼ of Q
A third of P: 60 ÷ 3 = 20 cm² of overlap.
Overlap = 13 × 60 = 20 cm2.
step 1 of 3

Both fractions name the same overlap. Find it from P, then it is one quarter of Q, which gives Q; the total counts the overlap once.

  1. Overlap = 13 × 60 = 20 cm2.
  2. That 20 cm2 is 14 of Q: Q = 4 × 20 = 80 cm2.
  3. Total covered = 60 + 80 − 20 = 120 cm2.

answer80 cm2; 120 cm2

Common pitfalls

  • Making 13 of P and 14 of Q two different regions.
  • Taking Q = 34 × 60 by mixing the fractions.
09

Perimeter of Two Shapes Pushed Together

propertyEdges that meet disappear from the outline; the outline is what is left

Two identical squares of side 6 cm are pushed together side by side so that they overlap in a strip 2 cm wide and 6 cm tall. Find the perimeter and the area of the combined figure.

6626
Two squares of 6 sharing a strip 2 wide, full height.
The combined figure is a rectangle 6 cm tall and 6 + 6 − 2 = 10 cm long.
step 1 of 4

Pushed together along their full height, the two squares make one rectangle. Its outline is the outer edges only; its area is the two squares less the strip they share.

  1. The combined figure is a rectangle 6 cm tall and 6 + 6 − 2 = 10 cm long.
  2. Perimeter = 2 × (10 + 6) = 32 cm.
  3. Area = 36 + 36 − 2 × 6 = 60 cm2; or 10 × 6 = 60 cm2.
  4. The two squares' own perimeters total 48 cm; the 16 cm lost is the edges now inside the figure.

answer32 cm; 60 cm2

Common pitfalls

  • Adding the two perimeters, 48 cm, as if the squares only touched at a point.
  • Subtracting the overlap once from the perimeter, which mixes lengths and areas.
10

A Frame of Even Width

propertyFrame = outer rectangle − inner rectangle; the outer sides grow by twice the width

A picture 20 cm by 14 cm sits in a frame that is 3 cm wide all round. Find the area of the frame and the perimeter of its outer edge.

2014332620
The frame adds 3 cm on each side: outer 26 by 20.
Outer length = 20 + 3 + 3 = 26 cm; outer width = 14 + 3 + 3 = 20 cm.
step 1 of 4

The frame adds its width on both sides of each dimension. Its area is the outer rectangle with the picture taken out.

  1. Outer length = 20 + 3 + 3 = 26 cm; outer width = 14 + 3 + 3 = 20 cm.
  2. Outer rectangle = 26 × 20 = 520 cm2; picture = 20 × 14 = 280 cm2.
  3. Frame = 520 − 280 = 240 cm2.
  4. Outer perimeter = 2 × (26 + 20) = 92 cm.

answer240 cm2; 92 cm

Common pitfalls

  • Adding the width once to each dimension: the frame runs along both sides.
  • Finding the frame as 3 cm times the picture's perimeter, which misses the four corner squares.
Mr. Chalk Read the guide