Area, Perimeter & Shape

The diagram shows exactly one third of the shaded area in the original diagram. It is made up of the quadrilateral \(UOPX\), together with a sector of the outer circle, \(POQ\), where \(O\) is the centre of the original circle.
Since \(P\) and \(Q\) are two of the six points equally spaced around the circle, \(\angle POQ = \tfrac{1}{6} \times 360° = 60°\). The outer circle has radius 2 cm, so the area, in cm², of sector \(POQ\) is \(\tfrac{1}{6} \times \pi \times 2^2 = \tfrac{2\pi}{3}\).

Sophia can make a small square by joining two of her isosceles right-angled triangles, as shown in the first diagram. She can then join 4 or 9 or 16 or 25 of these small squares to make larger squares using up to 50 triangles. She can also create a different small square by joining four of her isosceles triangles, as shown in the second diagram. Note that the side-length of this square is \(\sqrt{2}\) times the side-length of the previous small square. She can then join 4 or 9 of this second type of small square to make larger squares using up to 36 triangles. The eight squares described so far all have different side-lengths and it can be shown that there are no other possible sizes of squares. Hence Sophia can make 8 different-sized squares with the triangles she has.

In the diagram, \(SR\) is the diameter of the upper small semicircle and \(P\) and \(Q\) are the centres of the two lower small semicircles.
Note that the line \(SR\) touches the two semicircles with centres \(P\) and \(Q\) at points \(S\) and \(R\) respectively.
So \(\angle SRQ = \angle RSP = 90°\).
Also, \(SR = PQ = 2\) cm. Therefore \(PQRS\) is a rectangle.

First we prove that the four white triangles are congruent. Note that they each have a right angle. Also angles \(HGA\) and \(FGE\) are equal (vertically opposite). The quadrilateral \(ACEG\) is a parallelogram since each of its sides come from the rectangles. Hence angles \(FGE\) and \(GAC\) are equal (corresponding), and angles \(GAC\) and \(ECD\) are equal (corresponding). Therefore each triangle has a right-angle, and an angle equal to \(HGA\), and therefore they each have the same angles. Moreover, they each have a corresponding side of length 3 cm. By labelling triangle \(ABC\) with lengths 3, \(x\) and \(y\), Pythagoras’ Theorem gives \(x^2 + 3^2 = y^2\) [1]. The triangles \(ABC\) and \(CDE\) are congruent, so \(CD = CB = x\). Since \(AD = 9\), we can see \(x + y = 9\) [2].

Equation [2] rearranges to \(x = 9 - y\), so \(x^2 = (9 - y)^2 = 81 - 18y + y^2\). Substituting this into [1] gives \(81 - 18y + y^2 + 3^2 = y^2\), so \(18y = 90\) and \(y = 5\). Thus \(x = 9 - 5 = 4\). The area of each white triangle is \(\tfrac{1}{2} \times 3x = \tfrac{1}{2} \times 3 \times 4 = 6\) cm². Thus the overlap is \(3 \times 9 - 2 \times 6 = 15\) cm².

The diagram shows the top left-hand corner of the original diagram. The centre of the semicircle shown is \(Q\). Also, \(U\) and \(S\) are the points where the edges of the bigger square touch the semicircle shown. Therefore both \(QU\) and \(QS\) are radii of the semicircle and \(\angle TSQ = \angle TUQ = 90°\). Also \(\angle UTS\) is a right angle as it is the corner of a square. Therefore \(UQST\) is a square. Hence \(QS = ST\).

Divide the garden up and let the lengths of the various sides be as shown on the diagram. Since all sides are either parallel or perpendicular, \(a + b + c = 3\). Therefore the perimeter of Sacha’s garden is \(3 + 5 + a + x + b + 4 + c + (4 + (5 - x)) = 21 + a + b + c = 24\).

Let \(u\) be the area of triangle \(JLM\), and \(v\) be the area of triangle \(JMK\). Since \(M\) is the midpoint of \(KL\), the triangle \(JLM\) has half the area of the triangle \(JLK\), so \(u = v = \tfrac{1}{2}S\).

Note that \(\angle QJR = \angle KJM\), \(JR = 4 \times JK\) and \(JQ = 3 \times JM\). Hence, using the formula “Area of a triangle \(= \tfrac{1}{2}ab\sin C\)”, we see that the area of triangle \(JQR = 3 \times 4 \times v = 6S\).

Similarly the area of triangle \(PJQ = 2 \times 3 \times u = 6u = 3S\) and the area of triangle \(JPR = 2 \times 4 \times S = 8S\). Hence the area of triangle \(PQR = 6S + 3S - 8S = S\).

Label the intersection of \(WQ\) and \(XP\) as \(V\) and the midpoint of \(WX\) as \(U\). Let the side-length of the square be 1 unit. The area of triangle \(WXR\) is \(\tfrac{1}{2} \times 1 \times 1\) units² = \(\tfrac{1}{2}\) units². Consider triangle \(WXP\) and triangle \(UXV\). These two triangles have the same angles and hence are similar. Since \(UX\) is half of \(WX\), it follows that \(VU\) is half of \(PW\) and hence has length \(\tfrac{1}{4}\) unit. Therefore the area of triangle \(WXV\) is \(\left(\tfrac{1}{2} \times 1 \times \tfrac{1}{4}\right)\) units² = \(\tfrac{1}{8}\) units². Hence the shaded area is \(\left(\tfrac{1}{2} - \tfrac{1}{8}\right)\) units² = \(\tfrac{3}{8}\) units². Therefore the fraction of the square that is shaded is \(\tfrac{3}{8}\).

Since the total shaded area is half the area of the larger equilateral triangle, the area of that triangle is six times the area of each of the shaded equilateral triangles . Now when two similar figures have side-lengths in the ratio \(1 : k\), the ratio of their areas is \(1 : k^2\). In this case, the ratio of the areas is \(1 : 6\) so the ratio of the side-lengths is \(1 : \sqrt{6}\).
Hence the the length of the sides of the larger triangle is \(\sqrt{6}\) cm.

Let \(O\) be the centre of the circle, and \(P\), \(Q\), \(P'\) and \(Q'\) the vertices of the square. The triangles \(OPQ\) and \(OP'Q'\) are congruent since they are right-angled and have two equal sides (\(PQ = P'Q'\) since they are edges of a square, and \(OP = OP'\) because each is a radius). Hence \(OQ = OQ'\), and \(O\) is thus the midpoint of the edge of the square.

Let the side of the square be \(2x\). And \(OP = 1\) since it is a radius. And by Pythagoras’ Theorem on triangle \(OPQ\) we have \((2x)^2 + x^2 = 1^2\), so \(4x^2 + x^2 = 1\), and \(5x^2 = 1\). This gives \(x^2 = \tfrac{1}{5}\). The area of the square is \((2x)^2 = 4x^2 = 4 \times \tfrac{1}{5} = \tfrac{4}{5}\).

In the diagram shown, \(T\) is the foot of the perpendicular from \(R\) to \(PQ\) produced. Consider triangle \(QTR\). Angle \(QTR = 90°\) and, because the sum of angles on a straight line = 180°, angle \(RQT = (180 - 2 \times 60)° = 60°\). So \(\sin 60° = \dfrac{RT}{QR}\) and hence
\(RT = QR \times \sin 60° = 2 \times \dfrac{\sqrt{3}}{2} = \sqrt{3}\). Also \(\cos 60° = \dfrac{QT}{QR}\), so
\(QT = QR \times \cos 60° = 2 \times \tfrac{1}{2} = 1\). Therefore \(PT = 7\).

By drawing radii and chords as shown, we can see that the triangles are equilateral and therefore each of the angles is 60°. Hence the left and right circles have each lost 120° (one third) of their circumferences, and the central circle has one-third \(\left(\tfrac{1}{6} + \tfrac{1}{6}\right)\) of its circumference remaining. Since \(1 - \tfrac{1}{3} + 1 - \tfrac{1}{3} + \tfrac{1}{3} = \tfrac{5}{3}\), the perimeter of the shape is \(\tfrac{5}{3} \times 2\pi R = \tfrac{10\pi R}{3}\).

The diagram shows that the hexagon is an equilateral triangle of side 11 which has an equilateral triangle of side 1 removed from the bottom left corner, an equilateral triangle of side 2 removed from the bottom right corner and an equilateral triangle of side 6 removed from the top.

Let the area of each of the three squares be \(S\). Then the areas of triangles \(AJI\), \(AJB\), \(BKC\), \(CKD\) and \(GLF\) are all equal to \(\tfrac{1}{2}S\).
Also, the areas of triangles \(DEF\) and \(GHI\) are both equal to \(\tfrac{1}{4}S\).
So the total area inside the rectangle, but outside the three squares, is \(5 \times \tfrac{1}{2}S + 2 \times \tfrac{1}{4}S = 3S\). Therefore the area of the rectangle is \(6S = 6 \times 2 \times 2 = 24\).
(It is left as an exercise for the reader to show that the height and width of the rectangle are \(4\sqrt{2}\) and \(3\sqrt{2}\) respectively and therefore that its area is \(4\sqrt{2} \times 3\sqrt{2} = 24\).)

Let each of the small squares in the grid have side-length \(x\) cm. Remove the shading and divide the dove into regions as shown.
It can be seen that the regions marked A and B combine to make a square of side \(2x\) cm and hence of area \(4x^2\) cm². Similarly, regions C, D and E combine to make a rectangle with sides \(2x\) cm and \(3x\) cm and hence area \(6x^2\) cm². Finally, region F is a rectangle with sides \(2x\) cm and \(x\) cm and hence area \(2x^2\) cm². Since the total area of the dove is 192 cm², we have \(4x^2 + 6x^2 + 2x^2 = 192\) and hence \(12x^2 = 192\). Therefore \(x^2 = 16\) and hence \(x = 4\). Hence the flag has length (6 × 4) cm = 24 cm and height (4 × 4) cm = 16 cm.

Let the dimensions of the box be \(x\) cm by \(y\) cm by \(z\) cm as indicated. From the diagram, we have \(2x + 2y = 26\), \(x + z = 10\) and \(y + z = 7\). When we add the last two of these, we obtain \(x + y + 2z = 17\) and, when we then double this, we obtain \(2x + 2y + 4z = 34\). Therefore \(4z = 34 - 26 = 8\) and hence \(z = 2\). Therefore \(x = 8\) and \(y = 5\) and hence the volume of the box in cm³ is 8 × 5 × 2 = 80.

The diagrams show that all four options could be the shape of the region where two triangles overlap.

In the diagram the square of side-length 4 has been dissected into squares of side-length 1. The small curved pieces have then been moved as indicated by the arrows. The resulting shaded area consists of eight squares each with area 1, and the non-shaded area is the same.

The diagram shows that the hexagon may be divided into twelve congruent triangles, of which five are shaded. Let the area of each of these triangles be \(a\) cm². Then \(5a = 20\), so \(a = 4\). Hence the area of the hexagon, in cm², is \(12a = 12 \times 4 = 48\).

First note that the length of each side of the larger square is 7 cm and that of the smaller square is 5 cm. So their areas are 49 cm² and 25 cm² respectively. Let \(\angle BAC = x°\). As the interior angles of a triangle sum to 180° and \(\angle ABC = 90°\), \(\angle ACB = (180 - 90 - x)° = (90 - x)°\).
The angles on a straight line sum to 180°, so \(\angle ECD = [180 - 90 - (90 - x)]° = x°\).

By joining the vertices of the inner triangle to the centre of the hexagon in both the first and third diagrams, it can be seen that each hexagon has been dissected into six equal parts, three of which are shaded.
Therefore, \(X\), \(Y\) and \(Z\) are each half of the hexagon and hence they are all the same.

The diagram shows part of the diagram in the question. Point \(O\) is the centre of the circle and points \(A\), \(B\), \(C\), \(D\), \(E\), \(F\), are as shown. Consider triangle \(BOF\): \(BF\) is equal in length to the side of the square so \(BF = 2\). Also \(OB = OF = 2\) as they are both radii of the circle. So triangle \(BOF\) is equilateral.
Therefore \(\angle BOF = 60\)°, so the area of sector \(OBF = \frac{60}{360} \times \pi \times 2^2 = \frac{2\pi}{3}\).
By Pythagoras' Theorem: \(OA = \sqrt{2^2 - 1^2} = \sqrt{3}\).
So the area of triangle \(OBF = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3}\). Therefore the area of the segment bounded by arc \(BF\) and line segment \(BF = \frac{2\pi}{3} - \sqrt{3}\). The area of rectangle \(BFEC = BF \times AD = 2 \times (\sqrt{3} - x)\).
The shaded region has area 2, so the area of the above segment + area of rectangle \(BFEC\) = area of the given square minus 2 = 4 − 2 = 2.
Hence \(\frac{2\pi}{3} - \sqrt{3} + 2\sqrt{3} - 2x = 2\). So \(x = \frac{\pi}{3} + \frac{\sqrt{3}}{2} - 1\).

Divide the tablecloth into 25 equal squares as shown. Half of each of the 16 outer squares is coloured black which is equivalent to 8 complete squares. Therefore the percentage of the tablecloth that is coloured black is
\(\dfrac{8}{25} \times 100 = 32\).

Let square \(PQRS\) have side \(3x\) cm. Then, as \(PT : TQ = 1 : 2\), \(PT = x\) cm. Similarly, \(US = x\) cm. In triangles \(PTS\) and \(UST\), \(PT = US\), \(\angle PTS = \angle UST\) (alternate angles) and \(TS\) is common to both. So the triangles are congruent (SAS).
Therefore \(UT = PS = 3x\) cm and \(\angle TUS = \angle SPT = 90\)°.
Hence \(PTUS\) is a rectangle, which has perimeter 40 cm. So \(40 = 2(3x + x) = 8x\). Therefore \(x = 5\) and the area of \(PTUS\), in cm², is 15 × 5 = 75.

Let the area of the equilateral triangle of side 1 be \(A\). If similar figures have sides which are in the ratio \(k : 1\), then the ratio of their areas is \(k^2 : 1\). So the areas of the triangles with sides 2, 3, 4 are \(4A\), \(9A\), \(16A\) respectively.
So, as shown in the diagram, the total shaded area is \(3A + 8A + 15A = 26A\). Therefore \(n = 26\).

The tiled area may be considered to be a tessellation of the figure shown, except for the dotted lines. For every hexagonal tile, there are two triangular tiles. The diagram shows that the area of each hexagonal tile is 24 times the area of each triangular tile. As there are two triangular tiles to each hexagonal tile, the ratio of the fraction of the floor shaded black to that which is shaded grey is 2 : 24 = 1: 12. Therefore, in the repeating pattern of tiles, the fraction which is shaded black is 1/13.
The exact ratios given are for the infinite plane. Since we are dealing with a finite floor, this is approximate since the edges are unpredictable, but close to correct since the numbers involved are large.

Let \(a\) be the side length of the octagon and \(b\) be as shown on the diagram. The square in the centre is \(a\) by \(a\), each rectangle is \(a\) by \(b\) and the triangles are each half of a \(b\) by \(b\) square. Applying Pythagoras' Theorem to a triangle shows that \(a^2 = 2b^2\). So the shaded area is \(b^2 + ab = b^2 + \sqrt{2}b^2 = b^2(1 + \sqrt{2})\).
Similarly the total area of the figure is \(a^2 + 4ab + 2b^2 = 4b^2 + 4\sqrt{2}b^2\)
\(= b^2(4 + 4\sqrt{2})\). Therefore the ratio required is \((1 + \sqrt{2}) : (4 + 4\sqrt{2}) = 1 : 4\).

In the diagram, \(O\) is the centre of the lower semicircle, \(A\) and \(C\) are the points of intersection of the two semicircles and \(B\) is the point at the centre of the rectangle and also of the overlap. Now \(OA\) is a radius of the semicircle so \(OA\) has length 5 cm. Also \(OB\) is half the height of the rectangle so has length 4 cm. Angle \(ABO\) is a right angle. So triangle \(ABO\) is a (3, 4, 5) triangle and hence \(AC = 2 \times 3\) cm = 6 cm.

Since the area of the square is 36 cm², the length of a side of the square is 6 cm. Since the shaded area is 27 cm², the area not shaded is (36 − 27) cm² = 9 cm². Let \(a\) cm, \(b\) cm and \(c\) cm be the lengths of the parts of the sides shown on the diagram. The area of a triangle is \(\tfrac{1}{2}\) × base × height.

Therefore \(\tfrac{1}{2} \times a \times 6 + \tfrac{1}{2} \times b \times 6 + \tfrac{1}{2} \times c \times 6 = 9\) and hence \(a + b + c = 3\). Therefore, since \((a + b + c) + (p + q + r + s)\) is the sum of the lengths of two sides of the square and so is equal to 12 cm, the value of \(p + q + r + s\) is 9 cm.

Let the length of the side of the square be \(2x\) cm. Then, using Pythagoras' Theorem in the triangle shown, \((2x)^2 + x^2 = 1^2\). So \(4x^2 + x^2 = 1\). Therefore \(x^2 = \frac{1}{5}\) and the area of the square is \(4x^2\) cm² = \(\frac{4}{5}\) cm².

By dissecting the triangle into smaller, identical triangles, we see that the shaded area is 22/25 of the larger triangle, which corresponds to 88/100 = 88%.

The area of the whole rectangle is 200 cm². Suppose the rectangle is cut in two by a vertical cut joining the midpoints of its longer edges and the right-hand half is then given a quarter turn about its centre to produce the arrangement as shown. It can then be seen that every grey region has a corresponding white region of the same shape and size. Hence, the total area of the grey regions is half the area of the rectangle and so is 100 cm².

Points \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) on the perimeter of the triangle are as shown. Let \(AD\) have length \(x\) so that \(DB\) has length \(3x\). Let the perpendicular from \(C\) to \(AB\) have length \(4h\). So, by similar triangles, the perpendicular from \(E\) to \(DB\) has length \(h\). The area of triangle \(ABC\) is \(\tfrac{1}{2} \times 4x \times 4h = 8xh\). The area of triangle \(DBE\) is \(\tfrac{1}{2} \times 3x \times h = \tfrac{3}{2}xh\). So the area of triangle \(DBE\) is \(\tfrac{3}{16}\) of the area of triangle \(ABC\).
Similarly, by drawing perpendiculars to \(CB\) from \(A\) and from \(F\), it may be shown that the area of triangle \(FEC\) is \(\tfrac{3}{16}\) of the area of triangle \(ABC\).
So the fraction of the area of the triangle that is shaded is \(1 - \tfrac{3}{16} - \tfrac{3}{16} = \tfrac{5}{8}\).

Let the length of the sides of each small square be \(x\). Then the shaded area is \(24x^2\). Let the perimeter of the square be divided into eight line segments, each of length \(y\), and four line segments of length \(z\). Some of these are labelled in the diagram. By Pythagoras' Theorem in triangle \(ABC\):
\(y^2 + y^2 = (5x)^2\), that is \(2y^2 = 25x^2\).
So \(y = \tfrac{5}{\sqrt{2}}x = \tfrac{5\sqrt{2}}{2}x\). Similarly, in the triangle with hypotenuse \(CD\):
\(x^2 + x^2 = z^2\), that is \(2x^2 = z^2\). So \(z = \sqrt{2}x\). Therefore the length of the side of the large square is \(2y + z = 5\sqrt{2}x + \sqrt{2}x = 6\sqrt{2}x\). So the area of the large square is \((6\sqrt{2}x)^2 = 72x^2\). Hence the required fraction is \(\tfrac{24x^2}{72x^2} = \tfrac{1}{3}\).

Label the corners of the square \(A\), \(B\), \(C\) and \(D\) going anticlockwise from the top left corner. Draw in the lines from each marked point on the diagonal to \(B\) and to \(D\). All the triangles with a base on the diagonal and a vertex at \(B\) or \(D\) have the same perpendicular height. Hence their areas are directly proportional to the length of their bases. The two triangles with \(e\) as their base both have area 4 cm². Hence the two triangles with base \(d\) both have area 5 cm². Similarly the two triangles with base \(a\) have area 2 cm², and so the two triangles with base \(b\) have area 3 cm². Finally, since the area of the square is 30 cm², the two triangles with base \(c\) have area 1 cm². Since the triangles with largest area are those of area 5 cm², the longest base is that part labelled \(d\).

The small numbers in the figure show the lengths in cm of each line segment. The larger numbers inside the figure show the areas in cm² of each square or trapezium. (The area of a trapezium is \(\tfrac{1}{2}(a + b)h\) where \(a\) and \(b\) are the lengths of the parallel sides and \(h\) is the perpendicular distance between them.)
So the area of the shaded portion in cm² is
11 × 11 − (1 + 12 + 4 + 15 + 9 + 14 + 16 + 15) = 35.
(See the extended solutions for a beautifully elegant solution of this problem.)

The stripes are of equal width, so the width of each stripe is (72 ÷ 6) cm = 12 cm. The diagram shows that the difference between the areas of any two adjacent stripes is equal to the area of a rectangle of width 12 cm and height \(h\) cm. By similar triangles, \(\dfrac{h}{12} = \dfrac{24}{72}\). So \(h = \dfrac{12 \times 24}{72} = 4\). So the required area is 12 × 4 cm² = 48 cm².

Three vertical lines have been added to the diagram. These divide the original diagram into 4 rectangles. In each of these, a diagonal divides the rectangle into two triangles of equal area, one shaded and one unshaded. So the total shaded area in the original rectangle is equal to the total unshaded area and is therefore equal to half the area of the original rectangle. So the total shaded area is \(\tfrac{1}{2} \times 3 \times 14\) cm² = 21 cm².

The centre of the top square is directly above the common edge of the lower two squares. Hence a rectangle half the size of the square, and so of area \(\tfrac{1}{2}\) cm², can be added to the diagram to form a right-angled triangle as shown. The area of the shaded region and the added rectangle is equal to \(\left(\tfrac{1}{2} \times 2 \times 1\tfrac{1}{2}\right)\) cm² = \(1\tfrac{1}{2}\) cm². Hence the area of the shaded region, in cm², is \(1\tfrac{1}{2} - \tfrac{1}{2} = 1\).

Let the length of the side of the regular hexagon be \(a\). Then its perimeter is \(6a\). Therefore the perimeter of the equilateral triangle is also \(6a\), so the length of each of its sides is \(2a\). The diagrams show that the equilateral triangle may be divided up into 4 equilateral triangles of side \(a\), whereas the regular hexagon may be divided into 6 such triangles. So the required ratio is 4 : 6 = 2 : 3.

Let each side of the octagon have length \(x\). The octagon may be divided into eight triangles by joining the centre of the circle to the vertices of the octagon. One such triangle is shown. Each of these triangles has one side of length 1 (the radius of the smaller circle), one side of length 2 (the radius of the larger circle) and one side of length \(x\). So all eight triangles are congruent. Therefore \(\angle AOB = 360^\circ \div 8 = 45^\circ\).
Let \(D\) be the foot of the perpendicular from \(B\) to \(AO\). Then triangle \(BDO\) is an isosceles right-angled triangle.
Let \(OD = DB = y\). Applying Pythagoras' Theorem to triangle \(BDO\):
\(y^2 + y^2 = 1\). So \(y = \tfrac{1}{\sqrt{2}}\).
Applying Pythagoras' Theorem to triangle \(ADB\):

So the length of the perimeter is \(8x = 8\sqrt{5 - 2\sqrt{2}}\).
(Note that the length of AB may also be found by applying the Cosine Rule to triangle OAB.)

Let the radius of the circle be \(r\) and let the angle of the sector be \(a\)°.
Then the perimeter of the sector is \(2r + \tfrac{a}{360} \times 2\pi r\).
This equals \(2\pi r\), the circumference of the original circle.
So \(2r + \tfrac{a}{360} \times 2\pi r = 2\pi r\).
Therefore the fraction of the area of the disc removed is
\(\tfrac{a}{360} = \tfrac{2\pi r - 2r}{2\pi r} = \tfrac{\pi - 1}{\pi}\).

\(A\), \(B\), \(C\), \(D\), \(E\) are the centres of the five semicircles. Note that \(AC\) joins the centres of two touching semicircles and therefore passes through the point of contact of the semicircles. So \(AC\) has length 2 + 2 = 4.
This also applies to all of the other sides of triangles \(ACD\) and \(BED\). Hence both triangles are equilateral. So each of the nine arcs which make up the perimeter of the shaded shape subtends an angle of 60° at the centre of a semicircle.
So the length of the perimeter of the shaded figure is 9 × \(\tfrac{1}{6}\) × 2 × π × 2 = 6π.

The large shape consists of 21 small squares, so the required shape is made up of 7 small squares. So A and C may be eliminated. The diagram on the right shows that shape E is as required. It is left to the reader to check that neither B nor D was the shape used.

The length of the side of the triangle is equal to four times the radius of the arcs. So the arcs have radius 2 ÷ 4 = \(\tfrac{1}{2}\).
In the first diagram, three semicircles have been shaded dark grey. The second diagram shows how these semicircles may be placed inside the triangle so that the whole triangle is shaded. Therefore the difference between the area of the shaded shape and the area of the triangle is the sum of the areas of three sectors of a circle. The interior angle of an equilateral triangle is 60°, so the angle at the centre of each sector is 180° − 60° = 120°.
Therefore each sector is equal in area to one-third of the area of a circle. Their combined area is equal to the area of a circle of radius \(\tfrac{1}{2}\). So the required area is \(\pi \times \left(\tfrac{1}{2}\right)^2 = \tfrac{\pi}{4}\).

Let the radius of each semicircle be \(r\). In the left-hand diagram, let the side-length of the square be 2\(x\). By Pythagoras' Theorem, \(x^2 + (2x)^2 = r^2\) and so \(5x^2 = r^2\). So this shaded area is \(4x^2 = \tfrac{4r^2}{5}\). In the right-hand diagram, let the side-length of each square be \(y\). Then by Pythagoras' Theorem, \(y^2 + y^2 = r^2\) and so this shaded area is \(r^2\).
Therefore the ratio of the two shaded areas is \(\tfrac{4}{5}\) : 1 = 4 : 5.

The large square has area 196 = 14². So it has side-length 14. The ratio of the areas of the inner squares is 4 : 1, so the ratio of their side-lengths is 2 : 1. Let the side-length of the larger inner square be 2\(x\), so that of the smaller is \(x\). The figure is symmetric about the diagonal \(AC\) and so the overlap of the two inner squares is also a square which therefore has side-length 1. Thus the vertical height can be written as \(x + 2x - 1\). Hence \(3x - 1 = 14\) and so \(x = 5\). Also, the two shaded rectangles both have side-lengths \(2x - 1\) and \(x - 1\); that is 9 and 4. So the total shaded area is 72.

Let the base \(AB\) of the triangle be the side of length 8 cm and let \(AC\) be the side of length 6 cm. So \(C\) must lie on the circle with centre \(A\) and radius 6 cm as shown. The area of the triangle is to be 7 cm², so the perpendicular from \(C\) to \(AB\) (or to \(BA\) produced) must be of length \(\tfrac{7}{4}\) cm.
The diagram shows the four possible positions of \(C\). However, since\(\angle BAC_1 = \angle BAC_3\) and \(\angle BAC_2 = \angle BAC_4\), these correspond to exactly two possibilities for the length of the third side \(AC\). The diagrams below show the two possibilities.

The diagrams show isosceles triangles \(T\) and \(U\).
The perpendicular from the top vertex to the base divides an isosceles triangle into two congruent right-angled triangles as shown in both \(T\) and \(U\).
Evidently, by Pythagoras' Theorem, \(h_1 = 4\) and \(h_2 = 3\). So both triangles \(T\) and \(U\) consist of two ‘3, 4, 5’ triangles and therefore have equal areas.

The radius of each disc in the figure is equal to half the side-length of the square, i.e. \(\tfrac{1}{\pi}\). Because the corners of a square are right-angled, the square hides exactly one quarter of each disc.
So three-quarters of the perimeter of each disc lies on the perimeter of the figure. Therefore the length of the perimeter is \(4 \times \tfrac{3}{4} \times 2\pi \times \tfrac{1}{\pi} = 6\).

The diagram shows the top-right-hand portion of the square. The shaded trapezium is labelled \(QXYZ\) and \(W\) is the point at which \(ZY\) produced meets \(PQ\).
As \(QXYZ\) is an isosceles trapezium, \(\angle QZY = \angle ZQX = 45^\circ\).
Also, as \(YX\) is parallel to \(ZQ\), \(\angle XYW = \angle WXY = 45^\circ\). So \(WYX\) and \(WZQ\) are both isosceles right-angled triangles. As \(\angle ZWQ = 90^\circ\) and \(Z\) is at the centre of square \(PQRS\), we deduce that \(W\) is the midpoint of \(PQ\). Hence \(WX = XQ = \tfrac{1}{4}PQ\). So the ratio of the side-lengths of similar triangles \(WYX\) and \(WZQ\) is 1 : 2 and hence the ratio of their areas is 1 : 4.
Therefore the area of trapezium \(QXYZ\) = \(\tfrac{3}{4}\) × area of triangle \(ZWQ\) = \(\tfrac{3}{32}\) × area \(PQRS\) since triangle \(ZWQ\) is one-eighth of \(PQRS\). So the fraction of the square which is shaded is 4 × \(\tfrac{3}{32}\) = \(\tfrac{3}{8}\).

Let the length of the shorter sides of the cards be 1 unit. Then, by Pythagoras' Theorem, the length of the hypotenuse of each card is \(\sqrt{1^2 + 1^2} = \sqrt{2}\).
So the lengths of the perimeters of the five figures in order are: \(4\sqrt{2}\); \(4 + 2\sqrt{2}\); \(4 + 2\sqrt{2}\); 6; \(4 + 2\sqrt{2}\). Also, as \(\left(\tfrac{3}{2}\right)^2 = \tfrac{9}{4} = 2\tfrac{1}{4} > 2\) we see that \(\tfrac{3}{2} > \sqrt{2}\). Therefore, \(4\sqrt{2} < 6 < 4 + 2\sqrt{2}\). So figure A has the shortest perimeter.

The figure is itself a quadrilateral. It can be divided into four small quadrilaterals labelled A, B, C, D. There are also four quadrilaterals formed in each case by joining together two of the smaller quadrilaterals: A and B; B and C; C and D; D and A.

We consider the triangular segment of the octagon formed by joining two adjacent vertices, \(P\) and \(Q\) to the centre, \(O\). For convenience, we show this segment, drawn on a larger scale, on the left, where we have added the lines \(RW\), \(ST\), \(TW\) and \(UV\). These lines are parallel to the edges of the triangle \(POQ\), as shown and together with the lines \(RU\) and \(SV\) they divide the triangle \(OPQ\) into 9 congruent triangles, of which 2 are shaded. Thus \(\dfrac{2}{9}\) of the segment is shaded. The same holds for all the other congruent segments of the octagon. So \(\dfrac{2}{9}\) of the whole octagon is shaded.

The diagram shows the original diagram enclosed within a square of side 2\(r\), where \(r\) is the radius of the original circle. The unshaded area of the square consists of four quadrants (quarter circles) of radius \(r\). So the shaded area is \(4r^2 - \pi r^2\) \(= r^2(4 - \pi)\). Therefore the required fraction is

Let triangle \(CEF\) have area \(a\). Note that \(\angle AFD = \angle CFE\) (vertically opposite angles) and \(\angle DAF = \angle ECF\) (alternate angles), so triangles \(ADF\) and \(CEF\) are similar.
Note also that the side \(AD\) is twice the length of the corresponding side \(CE\).
Hence:
(i) triangle \(ADF\) has area 4\(a\) and;
(ii) \(AF\) has twice the length of the corresponding side \(CF\).
View \(AF\) and \(CF\) as bases of the triangles \(ADF\) and \(CDF\) (which then share the same height). Therefore, by (ii), triangle \(ADF\) has twice the area of triangle \(CDF\) (area P), which thus is 2\(a\).
The area of triangle \(ACD\) is 6\(a\); so that of triangle \(ABC\) is also 6\(a\) and that of area Q is 5\(a\). So the required ratio is 2 : 5.

Let 4\(x\) be the length of each side of the regular octagon. The diagram shows part of the figure. The four triangles shown in the diagram are all isosceles right-angled triangles. In such triangles the ratio of the length of the hypotenuse to the length of the shorter sides is \(\sqrt{2} : 1\).
So in the larger triangles which have hypotenuse of length 4\(x\), the length of the shorter sides is \(2\sqrt{2}x\), whilst the smaller triangles with hypotenuse 2\(x\) have shorter sides of length \(\sqrt{2}x\).
Therefore the shaded square in the question has side of length \((4 + 2\sqrt{2})x\).
The length of the side of the outer square is \((4\sqrt{2} + 4)x = \sqrt{2}(4 + 2\sqrt{2})x\).
Therefore the two squares have sides in the ratio \(1 : \sqrt{2}\), which means that their areas have ratio 1 : 2.

As the diagram shows, each equilateral triangle may be divided into four equilateral triangles of side 1, whilst the hexagon may be divided into six equilateral triangles of side 1.
Therefore the fraction of the whole shape which is shaded is
\(\dfrac{6}{6 \times 4 + 6} = \dfrac{6}{30} = \dfrac{1}{5}\).

The L-shape needs to be divided as shown since neither of the pieces is to be a square. Notice that one of the pieces must be turned over. The difference between the areas of the two pieces is 7 − 5 = 2.

Triangles \(A\) and \(C\) are each of area \(\tfrac{1}{2}\) × 5 × 5 cm².
Triangles \(B\) and \(D\) are each of area \(\tfrac{1}{2}\) × 3 × 3 cm².
So the shaded area is [64 − (25 + 9)] cm² = 30 cm².

Let the length of the side of the regular octagon be \(x\) units and let \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) be the points shown. So \(AC = CE = x\).
Now \(\angle ACE\) = 135° (interior angle of regular octagon), so \(\angle ACB\) = 45° and hence triangle \(ABC\) is an isosceles right-angled triangle with \(AB = BC\).
Also, by Pythagoras' Theorem: \(AB^2 + BC^2 = AC^2 = x^2\) so \(AB = BC = \tfrac{\sqrt{2}}{2}x\).
Similarly, \(EF = \tfrac{\sqrt{2}}{2}x\).
Therefore \(BF = \left(\tfrac{\sqrt{2}}{2}x + x + \tfrac{\sqrt{2}}{2}x\right)\) units \(= x(1 + \sqrt{2})\) units.
But we are given that \(BF = (1 + \sqrt{2})\) units so \(x = 1\).
Now the area of the octagon formed by the overlap of the squares is equal to the area of one of these squares minus the sum of the area of four triangles, each of which is congruent to triangle \(CDE\).
Thus, in square units, the required area is

Let the centre of the circle be \(O\) and let \(A\) and \(B\) be corners of one of the shaded squares, as shown. As the circle has area π units², its radius is 1 unit. So \(OB\) is 1 unit long. Let the length of the side of each of the shaded squares be \(x\) units.
By Pythagoras' Theorem: \(OB^2 = OA^2 + AB^2\), that is \(1^2 = (2x)^2 + x^2\).
So \(5x^2 = 1\). Now the total shaded area is \(8x^2 = 8 \times \tfrac{1}{5} = 1\tfrac{3}{5}\) units².

In one hour, the snail can reach points within 1 m of the corner at which it starts. So it can reach some of the points on the three faces which meet at that corner, but none of the points on the other three faces. On each of the three reachable faces, the points which the snail can reach form a quarter of a circle of radius 1 m.
So the required fraction is \(\dfrac{3 \times \frac{1}{4}\pi \times 1 \times 1}{6 \times 1 \times 1} = \dfrac{\pi}{8}\).

Firstly, note that the black squares have side 2 units. The pattern may be considered to be a tessellation of the shape shown on the right. So the ratio of squares to rectangles is 1:2 and hence the fraction coloured black is \(\dfrac{4}{4 + 2 \times 48} = \dfrac{4}{100} = \dfrac{1}{25}\).

If the shaded regions in the top-right and bottom-left corners of the diagram are moved as shown, the area of the shaded region in both the top half and bottom half of the diagram is now that of a 3 × 2 rectangle which has a quarter of a circle of radius 2 removed from it.
So the total shaded area is
2(3 × 2 − \(\tfrac{1}{4}\) × π × 2²) cm² = (12 − 2π) cm².

The small square on top will be in the upper half of the divided figure. Now consider the figure formed by moving this square to become an extra square on the left of the second row, as shown. It may now be seen from the symmetry of the figure that the line \(PX\) splits the new figure in half – with that small square in the upper half. So the line \(PX\) does the same for the original figure.

Let the vertices of the quadrilateral be \(A\), \(B\), \(C\), \(D\) as shown. Then, by Pythagoras' Theorem:
\(AC^2 = AD^2 + DC^2 = (7^2 + 9^2)\) cm² = 130 cm².
Similarly, \(AB^2 + BC^2 = AC^2\), so
\(AB^2 = (130 - 9)\) cm² = 121 cm². Therefore \(AB\) has length 11 cm and the area, in cm², of quadrilateral \(ABCD\) is \(\tfrac{1}{2} \times 9 \times 7 + \tfrac{1}{2} \times 3 \times 11 = 48\).

In each of 6 possible directions, the view of the sculpture is as shown, with the outer square having side 3 units and the inner square having side 1 unit.
So the surface area of the sculpture is 6 × 8 units² = 48 units².

Let the radius of the semicircle be \(r\) and let the perpendicular height of the isosceles triangle be \(h\). Then \(\tan x\)° = \(\tfrac{h}{r}\).
Now the area of the semicircle = \(\tfrac{1}{2}\pi r^2\), whilst the area of the triangle = \(\tfrac{1}{2} \times 2r \times h = rh\). So \(rh = \tfrac{1}{2}\pi r^2\), giving \(\tfrac{h}{r} = \tfrac{\pi}{2}\).

From the symmetry of the figure, the two circles must be concentric. Let their centre be \(O\). Let the radius of the semicircles be \(r\). Then the radius of the outer circle is \(2r\) and, by Pythagoras' Theorem, the radius of the inner shaded circle is \(\sqrt{r^2 + r^2}\), that is \(\sqrt{2}r\).
So the radii of the two circles are in the ratio \(\sqrt{2} : 2\), that is \(1 : \sqrt{2}\), and hence the ratio of their areas is \(1 : 2\).

Let \(r\) be the radius of the circle. Then, in the smaller square:

and in the larger square:

So \(\tfrac{5x}{4} = \tfrac{y}{2}\) and we deduce that \(x : y = 2 : 5\).

Let \(T\) be the centre of the semicircle with diameter \(QR\) and let \(OT\) produced meet the circumference of the larger semicircle at \(U\).
By symmetry, we note that \(OT\) is perpendicular to \(QR\).
As \(TR = TO = TQ\) (radii of semicircle), triangles \(ORT\) and \(OQT\) are both isosceles, right-angled triangles. So \(QOR\) is a right angle. By Pythagoras' Theorem: \(QR^2 = OQ^2 + OR^2 = 2^2 + 2^2 = 8\). So \(QR = \sqrt{8} = 2\sqrt{2}\) and the radius of semicircle \(QOR\) is \(\sqrt{2}\).
The area of the shaded region is equal to the area of semicircle \(QOR\) plus the area of the quadrant bounded by \(OQ\), \(OR\) and arc \(QUR\) less the area of triangle \(OQR\). So the required area is \(\tfrac{1}{2}\pi(\sqrt{2})^2 + \tfrac{1}{4}\pi 2^2 - \tfrac{1}{2} \times 2 \times 2 = \pi + \pi - 2 = 2\pi - 2\).

The diagram shows that the two pieces will fit together to form a right-angled triangle which has base 8 and height 6. The length of the hypotenuse = \(\sqrt{6^2 + 8^2}\), that is 10, so the perimeter of the triangle is 24.

Triangles \(ABG\), \(GFE\) and \(EFA\) are similar, so \(AF : FE = EF : FG = GB : BA = 1 : 2\). Thus if \(AF = a\), then \(FE = 2a\), \(FG = 4a\), and the shaded area is \(8a^2\). By Pythagoras' Theorem, \(AE = \sqrt{5}a\), so \(AD = 2\sqrt{5}a\) and the area of the square is \(20a^2\). Thus the required fraction is 8/20, which is 2/5.

[Alternatively, as we have shown that \(FG = 4AF\), we can divide parallelogram \(AGCE\) into 10 congruent triangles, 8 of which make up rectangle \(EFGH\). So the area of the rectangle is 4/5 of the area of the parallelogram, which in turn is half the area of square \(ABCD\).]

Let the length of a short side of a rectangle be \(x\) and the length of a long side be \(y\). Then the whole square has side of length \((y + x)\), whilst the small square has side of length \((y - x)\). As the area of the whole square is four times the area of the small square, the length of the side of the whole square is twice the length of the side of the small square. Therefore \(y + x = 2(y - x)\) i.e. \(y = 3x\) so \(x : y = 1 : 3\).

As the diagram shows, the figure may be cut into two parts which fit together to form a square measuring 8 cm × 8 cm.