Exercise practice 1: Area

3.1 Figure it Out: missing lengths and paths

3.1.1 FIO 1, Q1: identify the missing side lengths

Statement: two composite figures are made of rectangles with some lengths and all areas marked. Find the missing sides.

Figure (i): the four rectangles have areas 21, 28, 35 and 14 square inches, with marked lengths 7 in, 4 in, 3 in and 2 in, and one unknown marked ?.

Working from the 21 in² rectangle: its width is 7 in, so its height is 21 ÷ 7 = 3 in. The 28 in² rectangle stands on the same bottom level and rises 4 in higher, so its height is 3 + 4 = 7 in and its width is 28 ÷ 7 = 4 in. The 35 in² rectangle shares its right edge with the 28 in² one and sticks out 3 in further left, so its width is 4 + 3 = 7 in and its height is 35 ÷ 7 = 5 in. The 14 in² strip begins level with the top of the 35 in² rectangle and ends 2 in lower, so its height is 5 + 2 = 7 in and its width is 14 ÷ 7 = 2 in.

Answer: the missing side is 2 in.

Figure (ii): the top row is 4 m tall. The dotted 29 m² rectangle has width 29/4 m, and the hatched 11 m² rectangle has width 11/4 m. The thick outline encloses the dotted rectangle and the plain rectangle below it, so both of those rectangles share the width 29/4 m. The printed 50 m² label, with its arrow pointing into the plain lower rectangle, gives that rectangle's area; the dotted and hatched rectangles already carry their own labels. Hence the height of the lower rectangle is 50 ÷ (29/4) = 200/29 m, about 6.9 m.

Answer: the top sides are 29/4 m and 11/4 m, and the side marked ? is 200/29 m, about 6.9 m.

3.1.2 FIO 1, Q2: the path around the park

(i) The path is the part of the outer rectangle ABCD not covered by the park EFGH. To find its area, measure the length and width of ABCD and the length and width of EFGH. Choose values: park 8 m by 5 m, path 1 m wide all round, so outer rectangle 10 m by 7 m. Area of path = 70 − 40 = 30 m². Formula: area of path = AB × AD − EF × FG.

(ii) The widths of the path alone are not enough: a 1 m path around a small park covers less than a 1 m path around a large park. Also measure the park's length l and width b. Break the path into four rectangles: the left and right strips each have area width × b, and the top and bottom strips each have area width × (l + left + right). Adding these gives the formula.

(iii) Moving the outer rectangle while keeping the park inside changes neither rectangle's dimensions, so neither area changes. The area of the path stays the same; only the individual widths around the four sides change.

3.1.3 FIO 1, Q3: the crosspath

The plot is 14 m by 12 m and has one strip running across and one running down. Besides the plot's sides, measure the width of each strip. If the across strip is w₁ wide and the down strip w₂ wide, the two strip areas are 14 × w₁ and 12 × w₂. Their overlap is a w₁ × w₂ rectangle counted twice, so subtract it once: area of crosspath = 14w₁ + 12w₂ − w₁w₂.

Example: w₁ = 2 m and w₂ = 3 m gives 28 + 36 − 6 = 58 m².

3.1.4 FIO 1, Q4: the spiral tube

The small bent tube on the left has two arms of length 5 and width 1 meeting at a right angle. Counted separately, the arms give 5 + 5 = 10 unit squares, but the corner square is counted twice, so the tube's area is 10 − 1 = 9 square units. A straight tube of the same width and area must therefore have length 9.

For the spiral, read the straight runs from the figure: 20, 20, 20, 15, 15, 10, 10, 5 and 5 units, with a 1 by 1 overlap at each of 8 corners. The arms add to 120, so the area is 120 − 8 = 112 square units.

3.1.5 FIO 1, Q5: doubling the side of the square

Let the square have side s. The diagonal splits it into two equal triangles of area s²/2, and region 3 is one of them. The other half is split at the midpoint of the diagonal into regions 1 and 2, two triangles with equal bases and the same height, so each has area s²/4.

When every length is doubled, every area is multiplied by 4. The increase for region 1 is 4 × s²/4 − s²/4 = 3s²/4, the same for region 2, and for region 3 the increase is 4 × s²/2 − s²/2 = 3s²/2. Doubling the side always multiplies an area by 4, so the increase is always 3 times the original area.

3.1.6 FIO 1, Q6: the square with a hole

Cut the square along two perpendicular lines that run from side to side and cross inside the square. Slide the four pieces, without turning them, to diagonally opposite positions so that the cut edges form the outside. The outer boundary becomes a square whose side equals the length of one cut, and the old corners point inwards around a square hole.

No area is created: hole area = (new square area) − (original square area). For a 4 cm square with cuts of 5 cm, the new square is 25 cm² and the original is 16 cm², so the hole is 9 cm².

3.2 Figure it Out: triangles and equal areas

3.2.1 FIO 2, Q1: areas of three triangles

(i) The base is BC = 4 cm and the perpendicular height from A is 3 cm: area = ½ × 4 × 3 = 6 cm². (ii) The base is EF = 5 cm and the perpendicular from D is 3.2 cm, so area = ½ × 5 × 3.2 = 8 cm². (iii) The right triangle has legs 3 cm and 4 cm; each leg is a height for the other, so area = ½ × 3 × 4 = 6 cm².

3.2.2 FIO 2, Q2: the altitude BY

The area from base BC = 6 units and height AX = 4 units is ½ × 6 × 4 = 12 square units. Using AC = 8 units as base, the same area is ½ × 8 × BY = 4 BY. So 4 BY = 12 and BY = 3 units.

3.2.3 FIO 2, Q3: area of an isosceles triangle

In the isosceles triangle SUB, the perpendicular SE from the apex bisects the base UB, so UE = EB. Triangles SUE and SEB have equal bases and the same height SE, hence equal areas. Since area of SEB is 24, area of SUE is also 24, and the whole triangle has area 48 square units.

3.2.4 FIO 2, Q4: rectangle to triangle

Take a rectangle ABCD with base l and height w. Extend AD beyond D by an equal length so that the new height is 2w, and join the new top point to B. The triangle with base l and height 2w has area ½ × l × 2w = lw, the rectangle's area. Cutting the rectangle along a line from the midpoint of DC to B and moving the corner triangle shows the dissection; a 6 × 4 rectangle becomes a triangle with base 6 and height 8, area 24.

3.2.5 FIO 2, Q5: triangle to rectangle

Take a triangle with base b and height h. Draw a rectangle on the same base whose other side is h/2. Its area is b × h/2 = ½bh, equal to the triangle's area. Or halve the base and keep the full height. A triangle with base 6 and height 4 (area 12) becomes a 6 by 2 rectangle.

3.2.6 FIO 2, Q6: the three identical squares

Let each square have side s. The red region is a triangle with base s and height 2s, so its area is s². If the red area is 49, then s² = 49.

The blue region is a triangle cut off by the diagonal DH crossing AB at P. Since AD = BH = s and the angles at A and B are right angles, triangles PAD and PBH are congruent, so AP = PB = s/2. The blue area is ½ × s × s/2 = s²/4 = 49/4 = 12.25 square units.

If red plus blue is 180, then s² + s²/4 = 180, that is 5s²/4 = 180, so s² = 144. Each square has area 144 square units.

3.2.7 FIO 2, Q7: the midpoint triangle

Join NY. Since N is the midpoint of XZ, triangles XYN and ZYN have equal bases XN = NZ and the same height from Y, so area XYN = ½ area XYZ. In triangle XYN, M is the midpoint of XY, so triangles XMN and YMN have equal bases and the same height from N, giving area XMN = ½ area XYN. Therefore area XMN = ¼ area XYZ.

3.2.8 FIO 2, Q8: Gopal's shortest path

Reflect the water tank in the near bank of the river to a point T'. Join the house H to T' by a straight line; the point P where this line meets the bank is where Gopal should collect water. The shortest path is H to P to T. Any other bank point Q gives QT = QT' because Q lies on the mirror line, so the walk through Q equals the walk H to Q to T', which is shortest when it is the straight line H T'.

3.3 Figure it Out: polygons and quadrilaterals

3.3.1 FIO 3, Q1: quadrilateral split by a diagonal

The diagonal AC = 22 cm splits ABCD into triangles ABC and ACD. Their heights onto AC are BM = 3 cm and DN = 3 cm, on opposite sides of AC. Area of ABCD = ½ × 22 × 3 + ½ × 22 × 3 = 33 + 33 = 66 cm².

3.3.2 FIO 3, Q2: area of the shaded region

The rectangle ABCD is 18 cm long and 10 cm wide, so its area is 180 cm². Two right-triangle corners are unshaded: triangle AEF has legs 10 cm and 6 cm, area 30 cm²; triangle EBC has legs 8 cm and 10 cm, area 40 cm². Shaded area = 180 − 30 − 40 = 110 cm².

3.3.3 FIO 3, Q3: measurements for a regular hexagon

Joining the centre of a regular hexagon to its six vertices makes six equilateral triangles of side equal to the hexagon's side s. So one length, the side s, fixes the whole shape. To use the triangle formula, measure the distance a from the centre to the middle of a side; each triangle then has area ½ × s × a and the hexagon has area 3sa. The corner-to-corner distance 2s is another measurement that determines the figure.

3.3.4 FIO 3, Q4: fraction of the rectangle in the blue region

The blue region is two triangles that meet at one interior point. One stands on the top edge of the rectangle and the other on the bottom edge, so both have base equal to the rectangle's width w. Their heights measured from the interior point add up to the rectangle's height h. The total blue area is therefore ½ × w × (part of h) + ½ × w × (rest of h) = ½ × w × h, exactly half the rectangle's area. Fraction = ½, wherever the meeting point sits.

3.3.5 FIO 3, Q5: a quadrilateral with half the area

Mark the midpoints of the four sides of the given quadrilateral ABCD and join them in order to get quadrilateral PQRS. To see that it has half the area, draw diagonal AC. In triangle ABC, the segment joining midpoints P and Q is parallel to AC and the corner triangle PBQ has one quarter of the area of ABC; in triangle ACD the corner triangle RDS has one quarter of ACD's area. So those two corners are a quarter of ABCD. Using diagonal BD in the same way shows that the other two corner triangles are another quarter, so the four cut-off corners make half the quadrilateral. What remains, PQRS, is the other half.