Composite Areas: 12 Applications on Leaves, Petals and Cut-and-Paste

The Leaf Where Two Quadrants Overlap in a Square

propertyThe Two Quadrants Cover the Square Once and the Leaf Twice: Leaf = Two Quadrants − Square

ABCD is a square with sides of 14 cm. Two quadrants are drawn inside it, one with center A and one with center C, each with a radius of 14 cm. They overlap in a leaf shape. Take π = 227. (a) Find the area of the leaf. (b) Find the total area of the two parts of the square outside the leaf.

One quadrant of radius 14 cm is 14 × 227 × 14 × 14 = 154 cm².

One quadrant is a quarter of a circle of radius 14 cm, so its area is 14 × 227 × 14 × 14 = 154 cm².

step 1 of 5

Each quadrant covers all of the square except one corner part. Together the two quadrants cover the whole square, and they cover the leaf twice, because the leaf is in both of them. So the two quadrants add up to the square and one more leaf.

All steps

One quadrant is a quarter of a circle of radius 14 cm, so its area is 14 × 227 × 14 × 14 = 154 cm².
The two quadrants together are 2 × 154 = 308 cm². Between them they cover the whole square, and the leaf, which is in both of them, is covered twice.
The square is 14 × 14 = 196 cm². Take it away from the two quadrants, and what is left is the leaf, counted the second time.
(a) The leaf is 308 − 196 = 112 cm².
(b) The two parts outside the leaf are the rest of the square: 196 − 112 = 84 cm². Check: each of them is the square less one quadrant, 196 − 154 = 42 cm², and 42 + 42 = 84.

answer(a) 112 cm²; (b) 84 cm²

techniqueArea of a Circle · Area of a Composite Figure

examsPSLE · SAT · GCSE Higher

Common pitfalls

Answering 196 − 154 = 42 cm² for the leaf. That is the square less one quadrant, which is one of the two corner parts outside the leaf, not the leaf.
Giving the leaf as 308 cm², the two quadrants added together. That is more than the whole square, 196 cm², which cannot be. The quadrants overlap in the leaf, so it has been counted twice; take the square away once: 308 − 196 = 112 cm².

The Four Petals Made by Four Semicircles in a Square

propertyThe Four Semicircles Cover the Square Once and Each Petal Twice: Petals = Four Semicircles − Square

ABCD is a square with sides of 28 cm. A semicircle is drawn inside the square on each of its four sides. Where the semicircles overlap, they make four petals. Take π = 227. (a) Find the total area of the four petals. (b) Find the area of the part of the square that is not in any petal.

Each semicircle has a radius of 28 ÷ 2 = 14 cm and an area of 308 cm².

The side of the square, 28 cm, is the diameter of each semicircle, so each radius is 28 ÷ 2 = 14 cm. Each semicircle is 12 × 227 × 14 × 14 = 308 cm².

step 1 of 5

Each petal is where two neighboring semicircles overlap. Together the four semicircles cover the whole square, and each petal is covered twice. So the four semicircles add up to the square and the petals once more.

All steps

The side of the square, 28 cm, is the diameter of each semicircle, so each radius is 28 ÷ 2 = 14 cm. Each semicircle is 12 × 227 × 14 × 14 = 308 cm².
The four semicircles together are 4 × 308 = 1232 cm².
Between them they cover the whole square, 28 × 28 = 784 cm². Each petal is where two semicircles overlap, so each petal is covered twice.
(a) Take the square away once, and what is left is the petals, counted the second time: 1232 − 784 = 448 cm².
(b) The part of the square not in any petal is 784 − 448 = 336 cm².

answer(a) 448 cm²; (b) 336 cm²

techniqueArea of a Circle · Area of a Composite Figure

examsPSLE · SAT · GCSE Higher

Common pitfalls

Using 28 cm as the radius of each semicircle. 28 cm is the side of the square, which is the diameter of a semicircle, so the radius is half of it, 14 cm.
Halving 1232 − 784 = 448 because each petal is made by two semicircles. Taking the square away once already takes away the first covering of every petal, so what is left counts each petal exactly once.

A Semicircle Cut from One End of a Rectangle and Added to the Other

propertyMoving a Piece Keeps the Area; in the Perimeter, Curved Edges Take the Place of the Straight Sides

A rectangle is 20 cm long and 14 cm wide. A semicircle with the left 14 cm side as its diameter is cut out of the rectangle, and the same semicircle is added onto the right 14 cm side. Take π = 227. (a) Find the area of the new shape. (b) Find the perimeter of the new shape.

The semicircle cut out on the left is the same size as the one added on the right.

The semicircle cut out on the left and the semicircle added on the right are the same size, so the shape loses exactly as much area as it gains.

step 1 of 5

The piece taken away is the same size as the piece added back, so the area does not change. The outline does change: both 14 cm sides are gone, and a curved edge is in the place of each one.

All steps

The semicircle cut out on the left and the semicircle added on the right are the same size, so the shape loses exactly as much area as it gains.
(a) The area is the same as the rectangle's: 20 × 14 = 280 cm².
Each semicircle has a diameter of 14 cm, so its radius is 7 cm. Its curved edge is half the circumference of a circle of radius 7 cm: 12 × 2 × 227 × 7 = 22 cm.
Walk round the new shape. The two 20 cm sides are still there, but both 14 cm sides are gone. In their place are two curved edges, one bending in on the left and one bulging out on the right.
(b) The perimeter is 20 + 22 + 20 + 22 = 84 cm.

answer(a) 280 cm²; (b) 84 cm

techniqueArea of a Composite Figure · Area and Perimeter

examsPSLE · SAT · GCSE Higher

Common pitfalls

Adding the semicircle on the right to the rectangle, 280 + 77 = 357 cm², and forgetting the piece cut out on the left. The same semicircle was taken away first, so the area goes back to 280 cm².
Keeping the two 14 cm sides in the perimeter: 20 + 20 + 14 + 14 + 22 + 22 = 112 cm. Those straight sides were cut away, and the curved edges took their place, so they are no longer part of the outline.

The Propeller: Semicircles Cut from Two Sides of a Square and Added to the Other Two

propertyMoving the Pieces Keeps the Area; Every Side Is Replaced by a Curved Edge

A square has sides of 20 cm. A semicircle of radius 10 cm is cut out of the square on each of two opposite sides, and two semicircles of the same size are added onto the other two sides. Take π = 3.14. (a) Find the area of the new shape. (b) Find the perimeter of the new shape.

Two semicircles are cut out, from the top and the bottom, and two of the same size are added, on the left and the right.

The two semicircles cut out and the two semicircles added are all the same size, so the shape loses exactly as much area as it gains.

step 1 of 5

The pieces cut out are the same size as the pieces added, so the area is the square's. Each semicircle has a whole side as its diameter, so the outline is four curved edges and no straight edge at all.

All steps

The two semicircles cut out and the two semicircles added are all the same size, so the shape loses exactly as much area as it gains.
(a) The area is the same as the square's: 20 × 20 = 400 cm².
Each semicircle has the whole 20 cm side as its diameter, so no straight part of any side is left. The outline is 4 curved edges.
Each curved edge is half the circumference of a circle of radius 10 cm: 12 × 2 × 3.14 × 10 = 31.4 cm.
(b) The perimeter is 4 × 31.4 = 125.6 cm. Check: the 4 curved edges make 2 whole circles of diameter 20 cm, and 2 × 3.14 × 20 = 125.6 cm.

answer(a) 400 cm²; (b) 125.6 cm

techniqueArea of a Composite Figure · Circumference and Diameter

examsPSLE · SAT · GCSE Higher

Common pitfalls

Adding the two semicircles on the outside to the square, 400 + 314 = 714 cm², and forgetting the two cut out. The pieces cut out are the same size as the pieces added, so the area stays 400 cm².
Adding the square's perimeter, 4 × 20 = 80 cm, to the curved edges. Each semicircle takes up a whole side of the square, so no straight edge is left on the outline. The perimeter is the 4 curved edges only.

The Curved Part of a Quadrant Cut Off by a Straight Line

propertyCurved Part = Quadrant − Triangle; Two Curved Parts Make a Leaf

OAB is a quadrant of a circle with center O and radius 14 cm. The straight line AB joins the two ends of its curved edge. Take π = 227. (a) Find the area of the part of the quadrant between the line AB and the curved edge. (b) Two of these parts are placed back to back along AB to make a leaf. Find the area of the leaf.

The quadrant is a quarter of a circle of radius 14 cm: 154 cm².

The quadrant is a quarter of a circle of radius 14 cm, so its area is 14 × 227 × 14 × 14 = 154 cm².

step 1 of 5

The line AB cuts the quadrant into a right-angled triangle and a curved part. The curved part is what is left of the quadrant when the triangle is taken away, and two of them make the leaf.

All steps

The quadrant is a quarter of a circle of radius 14 cm, so its area is 14 × 227 × 14 × 14 = 154 cm².
The line AB cuts the quadrant into triangle OAB and the curved part. The angle at O is a right angle, so OA is the base of the triangle and OB is its height, both 14 cm.
The area of triangle OAB is 12 × 14 × 14 = 98 cm².
(a) The curved part is the quadrant less the triangle: 154 − 98 = 56 cm².
(b) The leaf is two of these parts: 2 × 56 = 112 cm². Check: this is the same leaf that two quadrants make in a 14 cm square, and 2 × 154 − 14 × 14 = 112 cm² too.

answer(a) 56 cm²; (b) 112 cm²

techniqueArea of a Circle · Area of a Triangle

examsPSLE · SAT · GCSE Higher

Common pitfalls

Using AB as the base of the triangle and 14 cm as its height. The 14 cm sides are not at a right angle to AB. The right angle is at O, so the base and the height are OA and OB.
Taking the area of the triangle as 14 × 14 = 196 cm² and forgetting the half. A triangle is half of the rectangle that boxes it in, here the 14 cm square, so it is 98 cm². With 196, the curved part would be 154 − 196, which is less than zero and cannot be.

A Square Drawn Inside a Circle

propertyEach Half of a Diagonal Is a Radius; the Diagonals Cut the Square into Four Right-Angled Triangles

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

The diagonals cross at the center. Each half of a diagonal reaches the circle, so it is a radius, 10 cm.

The diagonals of the square cross at its center, which is also the center of the circle. Each half of a diagonal runs from the center to a corner on the circle, so it is a radius, 10 cm.

step 1 of 5

The diagonals of the square cross at the center of the circle, so each half of a diagonal is a radius. The diagonals cut the square into four right-angled triangles whose shorter sides are radii, which gives the area without knowing the side of the square.

All steps

The diagonals of the square cross at its center, which is also the center of the circle. Each half of a diagonal runs from the center to a corner on the circle, so it is a radius, 10 cm.
The diagonals of a square cross at right angles. So they cut the square into 4 right-angled triangles, each with two shorter sides of 10 cm.
(a) Each triangle is 12 × 10 × 10 = 50 cm², so the square is 4 × 50 = 200 cm².
The area of the circle is 3.14 × 10 × 10 = 314 cm².
(b) The area inside the circle but outside the square is 314 − 200 = 114 cm².

answer(a) 200 cm²; (b) 114 cm²

techniqueArea of a Composite Figure · Area of a Triangle

examsPSLE · SAT · GCSE Higher

Common pitfalls

Taking the side of the square as 10 cm, the radius, and answering 100 cm². The radius runs from the center to a corner, which is half of a diagonal, not a side. A side is longer than 10 cm, so the square is bigger than 100 cm².
Taking the side of the square as 20 cm, the diameter, and answering 400 cm². That square would be bigger than the whole circle, 314 cm², so it cannot fit inside it. 20 cm is the length of a diagonal, not of a side.

A Circle Drawn Inside a Square

propertyThe Side of the Square Is the Diameter; the Four Corners Together Are the Square Less the Circle

A circle of radius 7 cm is drawn inside a square and touches all four sides of the square. Take π = 227. (a) Find the total area of the four corner parts of the square outside the circle. (b) Find the area of one corner part.

The circle touches the left and right sides, so the square is 2 × 7 = 14 cm across: 196 cm².

The circle touches the left side and the right side of the square, so the side of the square is the diameter, 2 × 7 = 14 cm. The square is 14 × 14 = 196 cm².

step 1 of 5

The circle touches all four sides, so the side of the square is the circle's diameter. Everything in the square outside the circle is the four corner parts, and the four are the same size.

All steps

The circle touches the left side and the right side of the square, so the side of the square is the diameter, 2 × 7 = 14 cm. The square is 14 × 14 = 196 cm².
The area of the circle is 227 × 7 × 7 = 154 cm².
(a) The part of the square outside the circle is the four corners: 196 − 154 = 42 cm².
The circle sits exactly in the middle of the square, so the four corner parts are the same shape and the same size.
(b) One corner part is 42 ÷ 4 = 10.5 cm². Check: 4 × 10.5 = 42, and 154 + 42 = 196.

answer(a) 42 cm²; (b) 10.5 cm²

techniqueArea of a Circle · Area of a Composite Figure

examsPSLE · SAT · GCSE Higher

Common pitfalls

Using 7 cm as the side of the square, which gives a square of 49 cm², smaller than the circle inside it. 7 cm is the radius, from the center to a side. The circle is 14 cm across, and that is the side of the square.
Treating a corner part as a triangle and working out 12 × 7 × 7. One of its edges is curved, so it is not a triangle. Its area can only be found by taking the circle away from the square.

Two Triangles on One Base with Their Top Corners on a Parallel Line

propertySame Base, Top Corners on the Same Parallel Line, So the Same Height and the Same Area

Two parallel lines are 5 cm apart. AB is a line 12 cm long on the lower line, and C and D are two different points on the upper line. (a) Find the area of triangle ABC. (b) Find the area of triangle ABD.

Both triangles stand on the same base, AB = 12 cm.

Where C sits: 6 cm along

Both triangles have the same base, AB = 12 cm.

step 1 of 5

Both triangles stand on the same base, AB, and both top corners are on the upper line. The height of a triangle is measured straight up from its base line to its top corner, so both triangles have the height of the gap between the lines.

All steps

Both triangles have the same base, AB = 12 cm.
The height of a triangle is measured from its top corner straight down to the base line, at a right angle to it. C is on the upper line, and every point of the upper line is 5 cm above the lower line, so the height of triangle ABC is 5 cm.
(a) The area of triangle ABC is 12 × 12 × 5 = 30 cm².
D is on the upper line too, so triangle ABD has the same base, 12 cm, and the same height, 5 cm.
(b) The area of triangle ABD is 12 × 12 × 5 = 30 cm² as well. Sliding the top corner along the upper line changes the shape of the triangle, but not its area.

answer(a) 30 cm²; (b) 30 cm²

techniqueArea of a Triangle · Area of a Parallelogram

examsPSLE · SAT · GCSE Higher

Common pitfalls

Using a slanted side, such as BD, as the height. A slanted side leans over, so it is longer than 5 cm. The height is measured straight up from the base line to the top corner, and for both triangles that is the 5 cm between the lines.
Deciding that the more stretched triangle must have the bigger area. Both triangles have the same base and the same height, and the area of a triangle depends only on those two, so both areas are 30 cm².

A Trapezium Cut and Rearranged into a Rectangle

propertyMove the Triangle from One End to the Other: the Rectangle's Length Is Halfway Between the Parallel Sides

A trapezium has parallel sides of 10 cm and 16 cm, and its height is 6 cm. Its two slanted sides are equal. A right-angled triangle is cut off one end of the trapezium and fitted onto the other end, making a rectangle. (a) How long is the rectangle? (b) What is the area of the trapezium?

The longer side is 16 − 10 = 6 cm longer, so it sticks out 3 cm at each end.

The longer parallel side is 16 − 10 = 6 cm longer than the shorter one. The slanted sides are equal, so it sticks out by the same amount at each end: 6 ÷ 2 = 3 cm.

step 1 of 5

The longer side sticks out past the shorter side by the same amount at each end. Cut off the triangle at one end and fit it into the gap at the other end: the shorter side grows and the longer side shrinks until both have the same length, and the height stays the same.

All steps

The longer parallel side is 16 − 10 = 6 cm longer than the shorter one. The slanted sides are equal, so it sticks out by the same amount at each end: 6 ÷ 2 = 3 cm.
Cut straight down from the right end of the shorter side. This cuts off a right-angled triangle 3 cm wide and 6 cm tall.
Turn the triangle over and fit it into the gap beside the slanted side at the left end. It fits exactly, because that gap is also 3 cm wide and 6 cm tall.
(a) The shorter side has grown by 3 cm and the longer side has lost 3 cm, so both are 10 + 3 = 16 − 3 = 13 cm. The rectangle is 13 cm long and 6 cm high.
(b) No piece was added or thrown away, so the trapezium has the same area as the rectangle: 13 × 6 = 78 cm². Check: 12 × (10 + 16) × 6 = 78 cm².

answer(a) 13 cm; (b) 78 cm²

techniqueArea of a Trapezium · Area of a Composite Figure

examsO-Level · SAT · GCSE Higher

Common pitfalls

Multiplying the longer side by the height, 16 × 6 = 96 cm². That is the rectangle drawn round the trapezium, and it includes the two corner triangles outside it. The rectangle made by moving the triangle is only 13 cm long, so the area is 78 cm².
Adding all 6 cm to the shorter side, because the longer side is 6 cm longer, and making the rectangle 16 cm long. The 6 cm sticks out in two parts, 3 cm at each end, and only one part is moved. So the shorter side grows by 3 cm, to 13 cm.

A Circle Cut into Eight Equal Wedges, Three of Them Shaded

propertyEqual Wedges Share the Area and the Circumference Equally; a Wedge's Perimeter Is Two Radii and Its Curved Edge

A circle of radius 14 cm is cut into 8 equal wedges by lines through its center, and 3 of the wedges are shaded. Take π = 227. (a) Find the shaded area. (b) Find the perimeter of one wedge.

The circle is 227 × 14 × 14 = 616 cm².

The area of the circle is 227 × 14 × 14 = 616 cm².

step 1 of 5

Each of the 8 equal wedges takes 18 of the circle's area and 18 of its circumference. The shaded part is 38 of the area. The outline of one wedge is two radii and its curved edge.

All steps

The area of the circle is 227 × 14 × 14 = 616 cm².
The 8 wedges are equal, so each is 18 of the circle, and 3 of them are 38 of it.
(a) 38 of 616 is 616 ÷ 8 × 3 = 77 × 3 = 231 cm².
The circumference is 2 × 227 × 14 = 88 cm, and each wedge has 18 of it as its curved edge: 88 ÷ 8 = 11 cm.
(b) Walk round one wedge: out from the center along a radius, round the curved edge, and back along another radius. Its perimeter is 14 + 11 + 14 = 39 cm.

answer(a) 231 cm²; (b) 39 cm

techniqueArea of a Circle · A Fraction of an Amount

examsPSLE · SAT · GCSE Higher

Common pitfalls

Finding 38 of the circumference, 33 cm, for the shaded area. The circumference is the length round the edge; the shaded area is the space inside, so it is 38 of the circle's area, 616 cm².
Giving the curved edge alone, 11 cm, as the perimeter of a wedge. The perimeter is the whole way round the wedge, and that includes the two straight edges from the center, each a radius of 14 cm.

Eight Rectangles Round a Square Hole

propertyOne side of the big square is a length and two breadths; one side of the hole is a length less two breadths

Square ABCD is made of 8 identical rectangles arranged round a square hole EFGH in the middle. Along each side of the hole, two of the rectangles lie side by side, with their long sides along the side of the hole and along the edge of ABCD, and the four pairs turn at the corners like a pinwheel. The perimeter of ABCD is 92 cm and the area of the hole EFGH is 49 cm². (a) Find the length and the breadth of one rectangle. (b) Find the area of square ABCD. (c) In two of the rectangles a right-angled triangle is shaded; its two shorter sides are the length and the breadth of the rectangle. Find the total area of the 8 rectangles that is not shaded.

Side of ABCD: 92 ÷ 4 = 23 cm. The hole is 49 cm², so its side is 7 cm.

The side of ABCD is 92 ÷ 4 = 23 cm. The hole is a square of area 49 cm², and 7 × 7 = 49, so its side is 7 cm.

step 1 of 6

Walk along one side of ABCD and then along one side of the hole, and name each piece you pass. The two walks pass the same length and the same two breadths, once added and once taken away, so together they give the length.

All steps

The side of ABCD is 92 ÷ 4 = 23 cm. The hole is a square of area 49 cm², and 7 × 7 = 49, so its side is 7 cm.
Along the bottom of ABCD lie the length of one rectangle and then the breadths of the two rectangles of the next pair: length + 2 breadths = 23 cm. The same length runs along the bottom of the hole, but the two breadths of the pair on the left cover part of it: length − 2 breadths = 7 cm.
Adding the two sides, the breadths cancel: 2 × length = 23 + 7 = 30, so the length is 15 cm. Then 2 breadths = 23 − 15 = 8, so the breadth is 4 cm. (a) Each rectangle is 15 cm by 4 cm.
(b) The area of ABCD is 23 × 23 = 529 cm². Check: the 8 rectangles and the hole make 8 × 15 × 4 + 49 = 480 + 49 = 529 cm².
Each shaded triangle has the rectangle's length and breadth as its two shorter sides, so it is half the rectangle: 12 × 15 × 4 = 30 cm². The two triangles are 60 cm².
(c) The 8 rectangles cover 480 cm², so the unshaded part of them is 480 − 60 = 420 cm².

answer(a) 15 cm by 4 cm (b) 529 cm² (c) 420 cm²

techniqueArea of a Composite Figure · Area of a Triangle

examsPSLE · SAT · GCSE Higher

Common pitfalls

Taking the side of ABCD as length + breadth, as in a pinwheel of four rectangles: here two rectangles lie along each side of the hole, so two breadths sit beside the length.
Answering (c) with 529 − 60 = 469 cm²: the hole is not part of any rectangle, so the unshaded area of the rectangles leaves it out.

A Flower Bed Shaped Like a Parallelogram: Its Area and a Fence Around It

propertyThe area is the base times the height at a right angle to it; the slanted side belongs to the fence, not to the area

A flower bed is in the shape of a parallelogram. Its base is 8 m long and each of its slanted sides is 5 m long. Its height, measured at a right angle to the base, is 4 m. (a) What is the area of the flower bed? (b) A low fence goes all the way around the edge of the flower bed. How long is the fence?

The height, 4 m, goes straight down from the top side, at a right angle to the base. The slanted side, 5 m, leans over.

The height is the distance from the top side straight down to the base, at a right angle to it: 4 m. The slanted side, 5 m, leans over, so it is longer than the height.

step 1 of 5

Cut the triangle off one slanted end and slide it to the other end: the parallelogram becomes a rectangle with the same base and the same height. The slanted side leans over, so it is longer than the height and is not used for the area. It is part of the edge, so it is used for the fence.

All steps

The height is the distance from the top side straight down to the base, at a right angle to it: 4 m. The slanted side, 5 m, leans over, so it is longer than the height.
Cut along the height to take the triangle off the left end, and slide it to the right end. The two pieces make a rectangle 8 m long and 4 m high. No piece was added or thrown away, so the area has not changed.
(a) The area of the flower bed is base × height = 8 × 4 = 32 m².
The fence runs along all four edges: two sides of 8 m and two slanted sides of 5 m.
(b) The fence is 8 + 5 + 8 + 5 = 26 m long. The height is a line inside the flower bed, not an edge, so it is not part of the fence.

answer(a) 32 m²; (b) 26 m

techniqueArea of a Parallelogram

examsO-Level · SAT · GCSE Higher

Common pitfalls

Multiplying the base by the slanted side: 8 × 5 = 40 m². The slanted side leans over, so it is longer than the height. The rectangle that the cut and the slide make is only 4 m high, so the area is 8 × 4 = 32 m².
Using the height for the fence: 8 + 4 + 8 + 4 = 24 m. The fence goes along the edges of the flower bed, and the edges at the two ends are the slanted sides of 5 m, so the fence is 26 m long.