Calculate Areas of Rectangles, Triangles, Parallelograms and Circles We are Learning to…… Calculate Areas of Rectangles, Triangles, Parallelograms and Circles

What is the perimeter of this shape? To find the perimeter of a shape we add together the length of all the sides. What is the perimeter of this shape? Starting point 1 cm 3 Perimeter = 3 + 3 + 2 + 1 + 1 + 2 = 12 cm 2 3 Ask pupils if they know how many dimensions measurements of perimeter have. Establish that they only have one dimension, length, even though the measurement is used for two-dimensional shapes. Tell pupils that when finding the perimeter of a shape with many sides it is a good idea to mark on a starting point and then work from there adding up the lengths of all the sides. 1 1 2

Perimeter of a rectangle To calculate the perimeter of a rectangle we can use a formula. length, l width, w Using l for length and w for width, Explain the difference between the two forms of the formula. The first formula means double the length, double the width and add the two together. The second formula means add the length and the width and double the answer. Perimeter of a rectangle = l + w + l + w = 2l + 2w or = 2(l + w)

Perimeter Sometime we are not given the lengths of all the sides. We have to work them out from the information we are given. 9 cm 5 cm 12 cm 4 cm For example, what is the perimeter of this shape? The lengths of two of the sides are not given so we have to work them out before we can find the perimeter. Stress that to work out the perimeter we need to add together the lengths of every side. If we are not given some of the lengths, then we have to work them out before we can find the perimeter. a cm Let’s call the lengths a and b. b cm

Perimeter Sometime we are not given the lengths of all the sides. We have to work them out from the information we are given. 9 cm 12 – 5 a = = 7 cm 5 cm b = 9 – 4 = 5 cm 12 cm 4 cm Discuss how to work out the missing sides of this shape. The side marked a cm plus the 5 cm side must be equal to 12 cm, a is therefore 7 cm. The side marked b cm plus the 4 cm side must be equal to 9 cm, b is therefore 5 cm. a cm 7 cm P = 9 + 5 + 4 + 7 + 5 + 12 = 42 cm 5 cm b cm

Calculate the lengths of the missing sides to find the perimeter. 5 cm p = 2 cm p 2 cm q = r = 1.5 cm q r s = 6 cm t = 2 cm s 6 cm u = 10 cm Discuss how to find each missing length. 4 cm 4 cm P = 5 + 2 + 1.5 + 6 + 4 + 2 + 10 + 2 + 4 + 6 + 1.5 + 2 2 cm t 2 cm u = 46 cm

What is the perimeter of this shape? Remember, the dashes indicate the sides that are the same length. 5 cm 4 cm P = 5 + 4 + 4 + 5 + 4 + 4 = 26 cm

Area The area of a shape is a measure of how much surface the shape takes up. For example, which of these rugs covers a larger surface? Rug B Rug A Rug C Discuss how we can compare the area of the rugs by counting the squares that make up each pattern. Conclude that Rug B covers the largest surface.

Area of a rectangle Area is measured in square units. For example, we can use mm2, cm2, m2 or km2. The 2 tells us that there are two dimensions, length and width. We can find the area of a rectangle by multiplying the length and the width of the rectangle together. length, l width, w This formula should be revision from key stage 2 work. Area of a rectangle = length × width = lw

Area of a rectangle What is the area of this rectangle? 4 cm 8 cm The length and the width of the rectangle can be modified to make the arithmetic more challenging. Different units could also be used to stress that units must be the same before they are substituted into a formula. Area of a rectangle = lw = 8 cm × 4 cm = 32 cm2

Area of a triangle What proportion of this rectangle has been shaded? 4 cm 8 cm Drawing a line here might help. A line is drawn on the diagram to split the shape into two rectangles each with one half shaded. Pupils should conclude from this that one half of the whole rectangle is shaded. Establish that the area of the whole rectangle is equal to the base of the shaded triangle times the height of the shaded triangle. Conclude that the area of the shaded triangle is equal to half the base times the height. What is the area of this triangle? 1 2 Area of the triangle = × 8 × 4 = 4 × 4 = 16 cm2

Area of a triangle The area of any triangle can be found using the formula: Area of a triangle = × base × perpendicular height 1 2 base perpendicular height Ask pupils to learn this formula. Or using letter symbols, Area of a triangle = bh 1 2

Area of a triangle What is the area of this triangle? 6 cm 7 cm Area of a triangle = bh 1 2 Tell pupils that to work out the area of the triangle they must start by writing the formula. They can then substitute the correct values into the formula provided that they are in the same units. Stress that it is important to always write down the correct units at the end of the calculation. The numbers and units in the example may be modified to make the problem more challenging. = 1 2 × 7 × 6 = 21 cm2

Area of a parallelogram The area of any parallelogram can be found using the formula: Area of a parallelogram = base × perpendicular height base perpendicular height Ask pupils to learn this formula. Or using letter symbols, Area of a parallelogram = bh

Area of a parallelogram What is the area of this parallelogram? We can ignore this length 8 cm 7 cm 12 cm Area of a parallelogram = bh Tell pupils that to work out the area of the parallelogram they must start by writing the formula. They can then substitute the correct values into the formula provided that they are in the same units. Point out that the length of the diagonal can be ignored. Stress that it is important to always write down the correct units at the end of the calculation. The numbers and units in the example may be modified to make the problem more challenging. = 7 × 12 = 84 cm2

Area formulae of 2-D shapes You should know the following formulae: b h Area of a triangle = bh 1 2 b h Area of a parallelogram = bh Use this slide to summarize or review key formulae. a h b Area of a trapezoid = (a + b)h 1 2

The value of π For any circle the circumference is always just over three times bigger than the radius. The exact number is called π (pi). We use the symbol π because the number cannot be written exactly. π = 3.141592653589793238462643383279502884197169 39937510582097494459230781640628620899862803482 53421170679821480865132823066470938446095505822 31725359408128481117450284102701938521105559644 62294895493038196 (to 200 decimal places)! Explain that pi is just a number. We call it pi because it is not possible to write the number exactly. Even written to 200 decimal places, although extremely accurate, is an approximation.

Approximations for the value of π When we are doing calculations involving the value π we have to use an approximation for the value. For a rough approximation we can use 3. Better approximations are 3.14 or . 22 7 We can also use the π button on a calculator. Most questions will tell you what approximations to use. It is useful to approximate pi to a value of 3 when approximating the answers to calculations. When a calculation has lots of steps we write π as a symbol throughout and evaluate it at the end, if necessary.

The circumference of a circle For any circle, π = circumference diameter or, π = C d We can rearrange this to make an formula to find the circumference of a circle given its diameter. Pupils should be asked to learn these formulae. C = πd

The circumference of a circle Use π = 3.14 to find the circumference of this circle. C = πd 8 cm = 3.14 × 8 = 25.12 cm Tell pupils that when solving a problem like this they should always start by writing down the formula that they are using. This will minimize the risk of using the radius instead of the diameter, for example.

Finding the circumference given the radius The diameter of a circle is two times its radius, or d = 2r We can substitute this into the formula C = πd to give us a formula to find the circumference of a circle given its radius. C = 2πr

The circumference of a circle Use π = 3.14 to find the circumference of the following circles: 4 cm 9 m C = πd C = 2πr = 3.14 × 4 = 2 × 3.14 × 9 = 12.56 cm = 56.52 m 23 mm For each one, start by asking pupils what formula we have to use. Estimate each answer first using  = 3, or use this to check the answer. 58 cm C = πd C = 2πr = 3.14 × 23 = 2 × 3.14 × 58 = 72.22 mm = 364.24 cm

Finding the radius given the circumference Use π = 3.14 to find the radius of this circle. C = 2πr 12 cm How can we rearrange this to make r the subject of the formula? C 2π r = ? Link: A3 Formulae – changing the subject of a formula 12 2 × 3.14 = = 1.91 cm (to 2 d.p.)

Area of a circle This animation shows how the area of a circle can be approximated to the area of a parallelogram of base length r and height r. Watch the circle pieces rearrange into an approximate parallelogram an ask a volunteer to use the pen tool to label the length and the height in terms of r. Deduce from this that the area of a circle is r2.

Formula for the area of a circle We can find the area of a circle using the formula Area of a circle = π × r × r radius or Area of a circle = πr2

The circumference of a circle Use π = 3.14 to find the area of this circle. 4 cm A = πr2 = 3.14 × 4 × 4 = 50.24 cm2

Finding the area given the diameter The radius of a circle is half of its radius, or r = d 2 We can substitute this into the formula A = πr2 to give us a formula to find the area of a circle given its diameter. A = πd2 4

The area of a circle Use π = 3.14 to find the area of the following circles: 2 cm 10 m A = πr2 A = πr2 = 3.14 × 22 = 3.14 × 52 = 12.56 cm2 = 78.5 m2 23 mm Explain that rather than use the formula on the previous slide, it is usually easier to halve the diameter mentally to give the radius, before substituting it into the formula. The most common error is to neglect to half the diameter to find the radius and to substitute this value into the formula. Ensure that pupils do not make this mistake. 78 cm A = πr2 A = πr2 = 3.14 × 232 = 3.14 × 392 = 1661.06 mm2 = 4775.94 cm2

Circumference and Area of a Circle The Circumference of a circle can be calculated using the formulae: C = 2πr or C = πd The Area of a circle can be worked out by using the formula: A = πr² Where d is the diameter, r is the radius and π = 3.14 to 2 decimal places

Circumference problem The diameter of a bicycle wheel is 50 cm. How many complete rotations does it make over a distance of 1 km? Using C = πd and π = 3.14, The circumference of the wheel = 3.14 × 50 = 157 cm 1 km = 100 000 cm Explain that we can ignore any remainder when dividing 100000 by 157 because we are asked for the number of complete rotations. 50 cm The number of complete rotations = 100 000 ÷ 157 = 636

Compound Area If you are presented with a composite (non standard shape), don’t panic, you can still find its area easily Look to see how you can split the shape into shapes that you do know Try to split into rectangles, triangles, semi circles etc Find the area of each part and add them together

McGraw-Hill Page 11 #s 1 – 9 HW #14 To find the perimeter of any shape just add up the lengths of each of the sides To find the area of certain shapes, use the following formulae: Rectangle = Length x Width Triangle = ½ x Base x Height Parallelogram = Base x Vertical Height Trapezoid = ½ x Sum of the two Parallel sides x Height McGraw-Hill Page 11 #s 1 – 9 HW #14 BLM 1-4 #s 1,4,5,6a,7

BLM 1-4 #s 8 – 11