Exercise practice 2: Area

4.1 Figure it Out: parallelograms

4.1.1 FIO 4, Q1: seven parallelograms on a grid

All seven parallelograms stand on the same base of 5 grid units between the same pair of parallel lines, so all have the same height of 3 grid units. Since area = base × height, all seven have the same area of 15 square units, however far they lean.

The perimeters are not equal. The perimeter is twice the sum of the base and a slanted side, and the slanted side gets longer as the parallelogram leans further. The one that leans least, closest to a rectangle, has the least perimeter; the one that leans most has the greatest perimeter.

4.1.2 FIO 4, Q2: four parallelogram areas

(i) Base 7 cm and height 4 cm give 28 cm². (ii) Base 5 cm and height 3 cm give 15 cm². The diagonal drawn in the figure is extra information. (iii) The height 4.8 cm is drawn to the slanted 5 cm side, so use area = 5 × 4.8 = 24 cm². (iv) The height 4.4 cm belongs to the slanted 2 cm side, so area = 2 × 4.4 = 8.8 cm².

The general rule is that each side must be paired with the height measured perpendicular to that side.

4.1.3 FIO 4, Q3: find QN

The area of the parallelogram is the same whichever side is used as the base. Using SR = 12 cm and its height QM = 6 cm gives area = 12 × 6 = 72 cm². Using PS = 7.6 cm and its perpendicular height QN gives 7.6 × QN = 72, so QN = 72 ÷ 7.6 = 180/19 cm, about 9.47 cm.

4.1.4 FIO 4, Q4: rectangle or parallelogram

Stand both shapes on the 5 cm side. The rectangle's height is its 4 cm side, so its area is 20 cm². The parallelogram's 4 cm side is slanted, so its true height is shorter than 4 cm, and its area is less than 20 cm². The rectangle has the greater area. Equal side lengths do not guarantee equal areas unless the parallelogram is a rectangle.

4.1.5 FIO 4, Q5: a rectangle twice a triangle

Choose a side of the triangle as the base. Draw the rectangle that has that base and extends to the line through the opposite vertex parallel to the base; its other side is the triangle's height. The rectangle's area is base × height, and the triangle's area is half of that, so the rectangle is twice the triangle. Making a second copy of the triangle and joining the copies into a parallelogram gives another route to the same rectangle.

4.1.6 FIO 4, Q6: a rectangle equal to a triangle

Take the triangle with base b and height h and draw a rectangle on the same base whose other side is h/2. Its area is b × h/2 = ½bh, the same as the triangle. A dissection proof: join the midpoints of the two sides from the top vertex, cut along that midline and along the perpendicular from the vertex to it, then turn the two small top triangles about the midpoints; they complete a rectangle of base b and height h/2.

4.1.7 FIO 4, Q7: isosceles triangle to rectangle

In an isosceles triangle, the altitude from the apex is also a median, so it splits the triangle into two congruent right triangles. Cut along this altitude, turn one half over, and join the two halves along their equal slanted sides. The two right triangles form a rectangle of height h and width b/2, whose area is h × b/2 = ½bh, the triangle's area.

4.1.8 FIO 4, Q8: rectangle to isosceles triangle

Let the rectangle be l long and b wide. Cut along a diagonal, turn one half over, and join it to the other half along the side of length b so that the two sides of length l form one straight line. The result is a triangle with base 2l and height b; its two slanted sides are both copies of the diagonal, so the triangle is isosceles. Its area is ½ × 2l × b = lb, the rectangle's area.

4.1.9 FIO 4, Q9: equilateral triangle or square

Let the side be s. The square's area is s². The equilateral triangle has height shorter than s, so its area is less than ½ × s × s, well under the square. Two equilateral triangles together have area 2 × (√3/4)s² = (√3/2)s² ≈ 0.866s², which is still less than s². The square has the greater area in both comparisons.

4.2 Figure it Out: rhombus, trapezium and real-life areas

4.2.1 FIO 5, Q1: rhombus from its diagonals

Area of a rhombus = ½ × product of diagonals = ½ × 20 × 15 = 150 cm².

4.2.2 FIO 5, Q2: rectangle to rhombus

Let the rectangle have base l and height b, and let M be the midpoint of the base. Cut from M to the two top corners, producing a middle triangle and two congruent right triangles. Slide the corner triangles up so their right-angle vertices meet at a point T above the midpoint of the top side. The four sides of the new figure are equal, so it is a rhombus with diagonals l and 2b. Its area is ½ × l × 2b = lb, the rectangle's area.

4.2.3 FIO 5, Q3: four trapezium areas

(i) Parallel sides 10 ft and 7 ft, distance 16 ft: area = ½ × 16 × (10 + 7) = 136 sq ft. (ii) Parallel sides 24 m and 36 m, distance 14 m: area = ½ × 14 × (24 + 36) = 420 sq m. (iii) Parallel sides 14 in and 6 in, distance 10 in: area = ½ × 10 × (14 + 6) = 100 sq in. (iv) Parallel sides 12 ft and 18 ft, distance 8 ft: area = ½ × 8 × (12 + 18) = 120 sq ft.

4.2.4 FIO 5, Q4: isosceles trapezium to rectangle

Drop a perpendicular from one end of the shorter parallel side and cut off the right triangle it forms. Because the trapezium is isosceles, this triangle is congruent to the piece missing at the other end; turn it over and fit its slanted side against the other slanted side. The pieces form a rectangle with the trapezium's height and length equal to the average of the parallel sides, so the area is unchanged.

4.2.5 FIO 5, Q5: trapezium to rectangle by vertices

Let I and J be the midpoints of the two non-parallel sides AD and BC. Draw through I and J the lines perpendicular to the base DC. They meet DC at G and F and the line AB extended at H and E. EFGH is a rectangle. The triangles AHI and DGI are congruent (AI = ID, vertically opposite angles, right angles), and BEJ and CFJ are congruent in the same way. The rectangle gains the outside triangles and loses the inside ones, which are equal, so the areas match.

4.2.6 FIO 5, Q6: construct a trapezium of area 144 cm²

Start with a rectangle of area 144 cm², say 12 cm by 12 cm. Use height h = 12 cm and parallel sides with average 12, for example 10 cm and 14 cm. Draw the 14 cm base, a parallel line 12 cm above it, mark the 10 cm side on that line, and join the ends. Check: ½ × 12 × (10 + 14) = 6 × 24 = 144 cm².

4.2.7 FIO 5, Q7: hexagon split into three pieces

Joining the centre of a regular hexagon to its vertices makes six congruent equilateral triangles. The single equilateral triangle in the figure is one of them; the rhombus is two of them joined along a side; the trapezium is the remaining three. The ratio of the areas is 3 : 1 : 2 for trapezium : triangle : rhombus, and 3 + 1 + 2 = 6 unit triangles make the whole hexagon.

4.2.8 FIO 5, Q8: the trapezium and triangle of equal area

B is the point where ZA extended meets WX extended. Triangles ZYA and BXA are congruent: YA = XA because A is the midpoint of XY, the angles at A are vertically opposite, and the angles at Y and X are alternate angles between the parallel lines. So the two triangles have equal areas. The trapezium ZYXW equals quadrilateral ZAXW plus triangle ZYA, and triangle ZWB equals the same quadrilateral plus triangle BXA. Since the added triangles have equal areas, the trapezium and the triangle have equal areas.