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Shape and Space Triangles The aim of this unit is to teach pupils to:

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1 Shape and Space Triangles The aim of this unit is to teach pupils to:
Identify and use the geometric properties of triangles, quadrilaterals and other polygons to solve problems; explain and justify inferences and deductions using mathematical reasoning Understand congruence and similarity Identify and use the properties of circles Material in this unit is linked to the Key Stage 3 Framework supplement of examples pp Triangles

2 Area of a right-angled triangle
What proportion of this rectangle has been shaded? 4 cm 8 cm What is the shape of the shaded part? Discuss the fact that a right-angled triangle always has half the area of the rectangle made by its height and its width. It is not necessary to know the length of the longest side opposite the right angle (the hypotenuse) to find the area. What is the area of this right-angled triangle? 1 2 Area of the triangle = × 8 × 4 = 4 × 4 = 16 cm2

3 Area of a right-angled triangle
We can use a formula to find the area of a right-angled triangle: height, h base, b Area of a triangle = 1 2 × base × height Introduce the formula for the area of a right-angled triangle. = 1 2 bh

4 Area of a right-angled triangle
Calculate the area of this right-angled triangle. To work out the area of this triangle we only need the length of the base and the height. 8 cm 6 cm 10 cm We can ignore the third length opposite the right angle. Talk through this example. Make sure that pupils are able to identify which length is the base and which length is the height. The lengths of the sides may be modified to make the arithmetic more difficult. Area = 1 2 × base × height = × 8 × 6 1 2 = 24 cm2

5 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

6 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

7 The area of a triangle h b
Any side of the triangle can be taken as the base, as long as the height is perpendicular to it: b h b h

8 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

9 Area of a triangle Use this activity to deduce that the area of any triangle can be found by multiplying the length of the base by the perpendicular height. Start with a right-angled triangle. Modify the triangle while keeping the length of the base and the height constant. Set the pen tool to draw straight lines and use it to draw a rectangle around the triangle. Ask pupils to use this to find the area of the triangle. Repeat for a variety of triangles.

10 The area of a triangle Modify the triangle to change its height and its width. If required use the pen tool to draw a rectangle around the triangle to justify its formula. The square grid can be turned off once the derivation of the formula has been established. As an extension activity discuss how we could use Pythagoras’ Theorem to calculate the perimeter of the triangle.


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