This is the marking copy and it carries every answer. Her sheet is the other file — same numbering, same margin codes, so you can mark straight down the page.
The figure is made up of 5 identical squares arranged as a cross of diamonds, each outer square joined to the middle one along a full side. The perimeter of the figure is 132 cm. What is the area of each square?
/\ /\
/ \ / \
/ \ / \
\ \/ /
\ /\ /
\ / \ /
\/ \/
/\ /\
/ \ / \
/ \/ \
\ /\ /
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5 identical squares · the middle one shares all 4 of its sides
⚠ the drawing above does NOT show this — go by the words, not the pictureanswer 121
method Five squares have 5 × 4 = 20 sides altogether. The middle square shares each of its 4 sides with an outer square, so 4 + 4 = 8 of those sides are tucked inside the figure and do not show. That leaves 20 − 8 = 12 sides on the outside. So 12 sides = 132 cm, one side = 132 ÷ 12 = 11 cm, and the area is 11 × 11 = 121 cm².
watch Dividing 132 by 20 (all the sides) or by 4 (one square). Only the sides on the OUTSIDE make up the perimeter — count how many are hidden first.
In the diagram, ABEF is a square and BCDE is a rectangle. The length of AF is twice the length of BC and AB = 56 cm. Find the length of ED.
A B C +--------------------+---------+ | | /| | square | rect / | | ABEF | / | +--------------------+---------+ F E D AB = 56 cm · AF = 2 x BC
answer 28
method ABEF is a square, so all four of its sides are equal: AF = AB = 56 cm. We are told AF is twice BC, so BC = 56 ÷ 2 = 28 cm. In rectangle BCDE, ED is the side opposite BC, and opposite sides of a rectangle are equal — so ED = 28 cm.
watch Doubling instead of halving, giving 112. "AF is twice BC" makes BC the SMALLER one.
A rectangular piece of paper is folded as shown below, with both top corners folded down. The flat top that is left is 7 cm across, each folded corner covers 6 cm, and the height of the figure is 23 cm. What is the area of the piece of paper at first?
- - - - - +-------------+ - - - - - -+ | /| 7 cm |\ | | | / | | \ | | +----/----+-------------+----\----+ | |<-6cm->| |<-6cm->| 23 cm | | | | | | | | | +---------------------------------+ -+ the dashed lines show where the corners came from
answer 437
method The dashed lines show the paper's original top edge, which runs the full width: 6 + 7 + 6 = 19 cm. The folds only turned the corners down — they did not shorten the sheet — so the height is still 23 cm. Area = 19 × 23 = 437 cm².
watch Using 7 cm as the width because that is the top edge you can see. The folded corners are part of the same sheet, so their 6 cm each still counts.
ABCD is a square of side 8 cm, cut out of a 28 cm by 12 cm rectangle. What is the area of the shaded part (the rectangle outside the square)?
|<----------------- 28 cm ----------------->| +###########################################+ -+ ###########+-----------+##################### | ###########| A B |##################### | ###########| 8 cm |##################### 12 cm ###########| D C |##################### | ###########+-----------+##################### | +###########################################+ -+
answer 272
method Find the whole rectangle: 28 × 12 = 336 cm². Find the square: 8 × 8 = 64 cm². The shaded part is everything except the square, so 336 − 64 = 272 cm².
watch Taking 8 cm as the square's AREA instead of its side, or forgetting to subtract the square at all.
The figures show a square and a rectangle. Square X has sides of 10 cm. Rectangle Y is 2 cm wide. The area of Square X is 4 times the area of Rectangle Y. Find the perimeter of Rectangle Y.
Square X Rectangle Y +----------+ +--+ | | | | | 10 cm | | | 2 cm wide | | | | +----------+ +--+
answer 29
method Square X has area 10 × 10 = 100 cm². Rectangle Y is a quarter of that: 100 ÷ 4 = 25 cm². Its width is 2 cm, so its length is 25 ÷ 2 = 12.5 cm. Perimeter = 2 × (12.5 + 2) = 2 × 14.5 = 29 cm.
watch Multiplying by 4 instead of dividing. Square X is the BIGGER one, so Rectangle Y's area must come out smaller.
Find the perimeter of the I-shaped figure below. All lines meet at right angles. The figure is 14 cm across and 16 cm tall; the top and bottom bars are each 5 cm deep, and the waist is set in 3 cm on the left and 4 cm on the right.
|<-------- 14 cm -------->|
+-------------------------+ -+
| | | 5 cm
+----+---------------+----+ -+
|3cm | | 4cm|
| | 16 cm
+----+---------------+----+
| | | 5 cm
+-------------------------+ -+answer 74
method For a shape like this with no overhangs, start with the rectangle that just surrounds it: 2 × (14 + 16) = 60 cm. Then each notch cut into the side adds twice its depth, because you walk in and back out again: the left notch adds 2 × 3 = 6 cm and the right notch adds 2 × 4 = 8 cm. Perimeter = 60 + 6 + 8 = 74 cm.
watch Assuming the notches make the perimeter SMALLER. Cutting a bite out of the side removes no edge — it replaces one straight edge with three, so the perimeter grows.
The figure is made up of 4 identical rectangles: two standing upright side by side (22 cm tall, 14 cm across the pair) resting centrally on two more lying end to end to form the base. Find the length of XY, the part of the base sticking out to the right of the upright pair.
|<- 14 cm ->|
+-----+-----+ -+
| | | |
| | | 22 cm
| | | |
+-----+-----+-----+----+----+ -+
| | X----Y |
+-----------+---------------+ -+answer 15
method The two uprights together are 14 cm across, so each rectangle is 14 ÷ 2 = 7 cm wide, and each is 22 cm long. The base is two of the same rectangles laid end to end, so it is 22 + 22 = 44 cm long. The upright pair sits centrally, so the base sticks out equally at both ends: (44 − 14) ÷ 2 = 30 ÷ 2 = 15 cm.
watch Forgetting that all four rectangles are IDENTICAL, so the base's length comes from the uprights' 22 cm. Or taking the whole 30 cm overhang instead of halving it between the two sides.
A picture frame is made up of 4 identical rectangular pieces joined together as a pinwheel. The area of each rectangular piece is 69 cm² and its breadth is 3 cm. A square picture fits exactly in the centre of the frame. What is the area of the picture?
+--+------------------+
| | |
| +---------------+ |
| | | |
| | picture | |
| | | |
| +---------------+ |
| | |
+------------------+--+
|<- 3 cm ->| (the frame's width)answer 400
method Each frame piece is 69 cm² with breadth 3 cm, so its length is 69 ÷ 3 = 23 cm. In a pinwheel frame each piece lies along one side of the picture, overlapping the next at the corner: the picture's side is the piece's length minus its breadth, 23 − 3 = 20 cm. So the picture's area is 20 × 20 = 400 cm².
watch Taking the picture's side as the full 23 cm. Look at one side of the hole: the neighbouring piece's 3 cm width eats into it at one end.
A rectangular piece of paper was folded differently on both ends as shown below. What is the area of the rectangular piece of paper before it was folded? Give your answer in cm².
|<- 4 cm ->|
+----------+ -+ 1 cm
+-----------------------+ / -+
/ | /
/ 9 cm | /
+------------------------- +------+
|
3 cm |
+------------
both creases are at 45 degreesanswer 84
method Each crease is at 45°, so the slanted edge runs sideways exactly as far as it drops — and that distance is the paper's WIDTH. The 4 cm across the top right is that same distance, so the paper is 4 cm wide. Now unfold each end: the left flap reaches 3 + 4 = 7 cm, and the right flap reaches 1 + 4 = 5 cm. Laid flat, the length is 7 + 9 + 5 = 21 cm. Area = 21 × 4 = 84 cm².
watch Adding up the visible pieces as if nothing were hidden. A fold tucks paper underneath, so the flat sheet is always LONGER than the folded shape.
A rectangular piece of paper was cut into 3 smaller rectangles with dimensions 9 cm by 12 cm, 9 cm by 2 cm, and 9 cm by 5 cm. Find the length of the paper before it was cut. Give your answer in cm.
+----------+---+------+ | | | | the three pieces, side by side | 12 cm |2cm| 5 cm | every piece is 9 cm tall +----------+---+------+ |<--------- ? -------->|
answer 19
method Every piece is 9 cm on one side, so 9 cm is the paper's breadth and the cuts ran straight across it. The other sides — 12, 2 and 5 — are the pieces of the length, so the length was 12 + 2 + 5 = 19 cm.
watch Adding all six numbers, or adding the 9s as well. The 9 is the side the three pieces SHARE, so it is counted once and is not part of the length.
The figure below is made up of 7 identical rectangles. The perimeter of the whole figure is 340 cm. Find the breadth of 1 such rectangle. Give your answer in cm.
+----------+----------+ -+ | | | | breadth +--+--+--+--+--+------+ -+ | | | | | | | | | | | | length | | | | | | +--+--+--+--+--+ top row: 2 rectangles lying down bottom row: 5 rectangles standing up
answer 20
method The two rows are the same width, so 2 lengths = 5 breadths. The figure's width is 2 lengths and its height is 1 breadth + 1 length, so the perimeter is 2 × (2 lengths + breadth + length) = 2 × (3 lengths + 1 breadth) = 340, giving 3 lengths + 1 breadth = 170. From 2 lengths = 5 breadths, one length is 2½ breadths, so 3 lengths is 7½ breadths. Then 7½ + 1 = 8½ breadths = 170, and one breadth = 170 ÷ 8.5 = 20 cm.
watch Adding the seven rectangles' perimeters. Only the OUTSIDE edge counts — the lines where rectangles touch are inside the figure.
The figure is made up of 2 identical squares overlapping each other. The unshaded rectangle where they overlap has an area of 10 cm². The total area of the shaded parts is 142 cm². Find the length of the square. Give your answer in cm.
+----------+
| shaded |
| +---+------+
| | | |
+------+---+ |
| shaded |
+----------+
the small middle box is the overlap (unshaded), 10 cm2answer 9
method Each square is made of its shaded part plus its half of the overlap. Both squares contain the whole overlap, so the two squares together = shaded + overlap + overlap = 142 + 10 + 10 = 162 cm². That is two squares, so one square is 162 ÷ 2 = 81 cm². A square of area 81 has side 9 cm, since 9 × 9 = 81.
watch Subtracting the overlap once instead of adding it twice. The overlap belongs to BOTH squares, so it has to be counted twice to rebuild them.
A carpet is laid on a rectangular floor measuring 20 m by 13 m, leaving a border of 3 m around it. Find the area of the floor that is not covered by the carpet. Give your answer in square metres.
|<------------- 20 m ------------->| +----------------------------------+ -+ | 3 m | | | +--------------------------+ | | |3m | Carpet | | 13 m | +--------------------------+ | | | | | +----------------------------------+ -+
answer 162
method The whole floor is 20 × 13 = 260 m². The border is 3 m on EVERY side, so it takes 3 m off each end of both measurements: the carpet is (20 − 3 − 3) by (13 − 3 − 3) = 14 × 7 = 98 m². Uncovered = 260 − 98 = 162 m².
watch Taking off 3 m once instead of twice, giving a 17 × 10 carpet. The border runs all the way round, so each dimension loses 3 m at BOTH ends.
The figure shows a rectangular garden 12 m by 16 m with a footpath (shaded) running along the top, down the left side, and part-way down the right side. The width of the footpath is 2 m. The area of the garden not covered by the footpath is 124 m². Find the length of AB, the part of the right-hand edge below the footpath. Give your answer in m.
|<------- 12 m ------->| +######################+ -+ #######################| | ##+----------------+###| | ##| |###| | ##| |###| | 16 m ##| |###+ A | ##| | | | ##| | | | ##+----------------+---+ B -+ # = the 2 m footpath
answer 6
method The garden is 12 × 16 = 192 m², so the footpath covers 192 − 124 = 68 m². Now build the footpath from its parts, taking care not to count the corners twice. The left strip is 2 × 16 = 32 m². The top strip, not counting the corner already used, is (12 − 2) × 2 = 20 m². That leaves 68 − 32 − 20 = 16 m² for the right-hand strip, which is 2 m wide, so it is 16 ÷ 2 = 8 m long below the top strip — reaching 2 + 8 = 10 m down from the top. AB is the rest of that edge: 16 − 10 = 6 m.
watch Adding the three strips as 2×16 + 2×12 + 2×h and double-counting the two corner squares where they meet.
The figure below is made up of 2 squares of different sizes. AG is 8 cm and DE is 6 cm. What is the length of AD? Give your answer in cm.
|<------------- ? ------------->| A B +---------------+---C-------+ D | | | | | | | 8 cm | | 6 cm | | | | | | +---------------+-----------+ G F E
answer 14
method AG = 8 cm is the side of the big square, so AB = 8 cm too. DE = 6 cm is the side of the small square, so BD (the part of the top line beyond B) = 6 cm. AD runs along the top from A to D: 8 + 6 = 14 cm.
watch Trying to find AD from the slanted-looking picture instead of from the two square sides. In a square all four sides are equal — that is the only fact this question needs.
A piece of paper is cut straight down into Square A and Rectangle B. The area of A is twice the area of B. The breadth of Rectangle B is 14 cm. What is the area of Square A?
+----------------+---------+
| | |
| A : B | the cut is straight down,
| (square) : | so A and B are the same height
| | |
+----------------+---------+
|<- 14 cm ->|(A) 28 cm² (B) 112 cm² (C) 336 cm² (D) 784 cm²
answer (D) — 784 cm²
method The cut is straight down, so A and B are the same height — call it h. A is a SQUARE, so its width is h too and its area is h × h. B is h tall and 14 cm wide, so its area is 14 × h. Now use "A is twice B": h × h = 2 × 14 × h, so h = 28 cm. The area of Square A = 28 × 28 = 784 cm².
watch Answering 784 by luck but 28 by stopping early — 28 is the SIDE, and the question asks for the area. Or using 14 as the square's side.
The figure below is made up of 5 identical squares in an I-shape — two across the top, one in the middle, two across the bottom. The area of each square is 36 cm². What is the perimeter of the figure?
+------+------+
| | |
+------+------+
| |
| |
+------+------+
| | |
+------+------+
5 identical squares, 36 cm2 each(A) 60 cm (B) 72 cm (C) 120 cm (D) 180 cm
answer (B) — 72 cm
method Each square has area 36 cm², so its side is 6 cm (because 6 × 6 = 36). Start with the rectangle that just surrounds the figure: it is 2 squares wide and 3 squares tall, so 12 cm by 18 cm, giving 2 × (12 + 18) = 60 cm. Then each notch cut into the waist adds twice its depth, because you walk in and back out: 2 notches, each half a square (3 cm) deep, add 2 × 3 × 2 = 12 cm. Perimeter = 60 + 12 = 72 cm.
watch Taking 36 cm as the SIDE of each square instead of the area, or adding the five squares' perimeters (5 × 24 = 120 cm), which counts the edges where they join.
Square A and Rectangles B, C and D form the rectangle WXYZ. A sits top-left and is a square of side 7 cm, B is top-right, D is bottom-left and C is bottom-right. The area of Rectangle B is twice the area of Rectangle D. The shaded part (A, B and D together) has an area of 385 cm². Find the area of Rectangle C, in cm².
W X +--------+-----------------------+ | A | B | 7 cm |(square)| | +--------+-----------------------+ | D | C | | | | +--------+-----------------------+ Z Y A is a 7 cm square · B = 2 x D · shaded (A + B + D) = 385 cm2
answer 512
method A is a 7 cm square, so the top row is 7 cm tall and the left column is 7 cm wide. Call the right column's width w and the bottom row's height h. Then B = 7 × w, D = 7 × h, and C = w × h. Since B is twice D: 7w = 2 × 7h, so w = 2h. The shaded area is A + B + D = 49 + 7w + 7h = 385, so 7w + 7h = 336 and w + h = 48. Putting w = 2h in: 3h = 48, so h = 16 and w = 32. Area of C = 16 × 32 = 512 cm².
watch Trying to find C by subtracting 385 from the whole rectangle — you do not know the whole rectangle's area yet. Naming the two unknown lengths and using "B is twice D" is what unlocks it.
Sam used Square X (side 5 cm), Rectangle Y (9 cm by 4 cm) and Rectangle Z (14 cm by 3 cm) — with a second copy of Square X — to form Figure A: Z lies along the bottom, one X sits on Z at the left, Y stands upright beside it, and the other X sits at the right-hand end, level with the bottom of Z. What is the perimeter of Figure A, in cm?
+---+
| |
| Y | X = 5 by 5 (square)
+-----+ | Y = 9 tall, 4 wide
| X | | Z = 14 wide, 3 tall
| | | +-----+
+-----+---+----------+ X |
| Z | |
+--------------------+-----+
Figure A
note the STEP on the right: the square stands 2 cm proud of Zanswer 66
method Set the bottom-left corner of Z at 0. Z runs from 0 to 14 across and 0 to 3 up. The left X sits on Z, from 0 to 5 across and 3 to 8 up. Y stands next to it, from 5 to 9 across and 3 to 12 up. The right X sits beyond Z, from 14 to 19 across and 0 to 5 up. Now walk the outline once: up the left side 8, across the top of X 5, up Y's side 4, across Y's top 4, down Y's far side 9, along Z's top 5, up the step 2, across the right X's top 5, down its side 5, and back along the bottom 19. Adding: 8 + 5 + 4 + 4 + 9 + 5 + 2 + 5 + 5 + 19 = 66 cm.
watch Adding the pieces' perimeters (5 × 4 + 5 × 4 + 26 + 34 = 100 cm). Every edge where two pieces touch is inside the figure and must not be counted — walking the outline once is the only safe way.
A rectangular park measures 54 m by 28 m. A pedestrian path 2 m wide and a cyclist path 3 m wide are built along two sides of the park — the pedestrian path wraps the top and left of the park, and the cyclist path wraps the top and left of that. It costs $18 to construct each square metre of the cyclist path. How much does it cost to construct the cyclist path?
################################ -+ 3 m cyclist
##+--------------------------+## -+ 2 m pedestrian
##| | #
##| Park | # 28 m
##| | #
##+--------------------------+-+
|<-------- 54 m -------->|
# = the two paths, along the TOP and the LEFT onlyanswer 4806
method Build outwards from the park. The park is 54 by 28. Adding the 2 m pedestrian path along the top and the left makes that block 54 + 2 = 56 by 28 + 2 = 30. Adding the 3 m cyclist path along the top and left of THAT makes the whole thing 56 + 3 = 59 by 30 + 3 = 33. The cyclist path is the difference between those two: 59 × 33 − 56 × 30 = 1947 − 1680 = 267 m². At $18 a square metre, the cost is 267 × $18 = $4806.
watch Measuring the cyclist path's strips from the PARK's 54 and 28 instead of from the outside of the pedestrian path. The paths are stacked, so the outer one is longer than the inner one.
The figure below is made up of 4 identical rectangles. Find the perimeter of the figure.
|<----- 10 cm ----->|
+---+
| |
| |
| |
+-------------------+ |
+---------------+---+---+---------------+ -+- 2 cm
| +-------------------+ -+-
| |
| |
| |
+---+
4 identical rectangles, each 10 cm long and 2 cm wide, set as a
pinwheel: turn the figure a quarter turn and it looks the same.(1) 96 cm (2) 80 cm (3) 40 cm (4) 36 cm
answer (2) — 80 cm
method The arrows say each rectangle is 10 cm long and 2 cm wide. The figure is a pinwheel — turn it a quarter turn and it looks exactly the same, so the outline is four identical pieces. One piece is 2 cm (the end of an arm) + 10 cm (the long side of that arm) + 8 cm (the part of the next arm still showing) = 20 cm. Four of them: 4 × 20 = 80 cm.
watch Adding the four rectangles' perimeters (4 × 24 = 96 cm). That counts the edges where the rectangles touch, and those edges are inside the figure, not on its outline.
The figure is made up of 2 identical squares and 3 identical rectangles. What is the length of the unknown side?
+-----------+ -+
| | |
| square | |
| | |
|<-------- ? --------->| | |
+-----------+----------+-----------+ |
| | rectangle | 18 cm
| +----------------------+ |
| square | rectangle | |
| +----------------------+ |
| | rectangle | |
+-----------+----------------------+ -+
|<------------ 26 cm ------------->|answer 17
method Look down the right-hand side: 18 cm is the top square's side plus the three stacked rectangles. Now look at the left: the bottom square sits beside those same three rectangles, so the three rectangles stacked are exactly one square's side tall. That makes 18 cm = one square + one square = 2 squares, so each square has side 9 cm. Along the bottom the whole figure is 26 cm, and the top square (9 cm wide) is lined up at the right-hand end, so the marked length = 26 − 9 = 17 cm.
watch Forgetting the two squares are IDENTICAL. That fact is what lets 18 be split as 9 + 9 — without it there is nothing to pin the size down.
Barry has a garden with an area of 216 m². It is made up of a rectangle and a square. The area of the rectangle is 5 times the area of the square. Barry wants to build a fence around part of his garden as indicated by the dashes in the figure. Given that the breadth of the rectangle is 9 m, how many metres of fence does he need?
- - - - - - - - - - - - - - - - -+
' ' | 9 m
' rectangle ' |
- - - - - - - -+------+- - - - - -+
' '
' squ. '
' '
+- - - +
the dashed line is the fence (the whole outline)answer 70
method The square is 1 share and the rectangle is 5 shares, so the garden is 6 equal shares: 216 ÷ 6 = 36 m² for the square and 5 × 36 = 180 m² for the rectangle. A square of area 36 m² has side 6 m. The rectangle has area 180 m² and breadth 9 m, so its length = 180 ÷ 9 = 20 m. Now walk the outline: 20 (top) + 9 (right side) + 9 (left side) + 14 (the bottom of the rectangle, which is 20 minus the 6 m the square covers) + 6 + 6 + 6 (the square's other three sides) = 70 m.
watch Adding the two shapes' perimeters separately (2×(20+9) + 4×6 = 82 m). Where the square joins the rectangle there is no fence — that edge is inside the garden.
The figure below shows Marcus's backyard with 2 identical 8-m square ponds. What is the remaining area of the empty space around the two ponds in the backyard?
|<-------------- 20 m -------------->| +------------------------------------+ -+ | | | | +----------+ +----------+ | | | | pond | | pond | | | | | 8 m sq | | 8 m sq | | 15 m | +----------+ +----------+ | | | | | +------------------------------------+ -+
answer 172
method Find the whole backyard first: 20 × 15 = 300 m². Each pond is an 8 m SQUARE, so its area is 8 × 8 = 64 m², and there are two: 2 × 64 = 128 m². The empty space is what is left: 300 − 128 = 172 m².
watch Treating "8-m square pond" as an area of 8 m² instead of a square with side 8 m, or subtracting only one pond.
After a rectangular piece of paper was cut into a maximum of 18 squares of sides 4 cm each, an L-shaped strip of paper was left over as shown. Given that the length of the rectangular piece of paper is 26 cm, what is the perimeter of the rectangular piece of paper before it was cut?
+--+------------------------------+ | | | | | | | | the 18 squares of 4 cm | | | | | | | +--+------------------------------+ | strip | -+- 1 cm +---------------------------------+ -+- |<------------ 26 cm ------------>| the leftover L-strip is the narrow band down the left edge plus the 1 cm band along the bottom
answer 78
method Fit the 4 cm squares along the 26 cm length: 26 ÷ 4 = 6 remainder 2, so 6 squares fit across and a 2 cm strip is wasted down one side. With 18 squares in rows of 6, there must be 18 ÷ 6 = 3 rows, using 3 × 4 = 12 cm of the breadth. The leftover strip along the bottom is 1 cm, so the breadth is 12 + 1 = 13 cm. Perimeter = 2 × (26 + 13) = 2 × 39 = 78 cm.
watch Trying to get the breadth by dividing the squares' area (18 × 16 = 288 cm²) by 26. The L-strip is wasted paper, so the squares' area is NOT the paper's area.
The figure is not drawn to scale. It is made up of two identical rectangles overlapping each other, forming Square A. The area of Square A is 9 cm² and the area of each rectangle is 50 cm². The length of the rectangle is twice its breadth. Find the perimeter of the figure.
+---------+
| |
| |
| |
+------------+---+ |
| | A | |
| +---+-----+
| |
+----------------+
Square A is where the two rectangles overlapanswer 48
method Square A has area 9 cm², so its side is 3 cm. Each rectangle has area 50 cm² with length = 2 × breadth, so breadth × 2 × breadth = 50 → breadth × breadth = 25 → breadth = 5 cm and length = 10 cm. One rectangle's perimeter is 2 × (10 + 5) = 30 cm, so the two separately total 60 cm. But where they overlap, two sides of Square A are hidden inside each rectangle — four 3 cm edges in all. Perimeter of the figure = 60 − 4 × 3 = 60 − 12 = 48 cm.
watch Taking Square A's side as 9 cm instead of 3 cm (9 is the AREA), or answering 60 by forgetting that the overlap hides edges.