# 4.1 Compare quadrilaterals for similarities and differences  (Page 2/2)

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The height is often a line drawn inside the triangle. This is the case in four of the six triangles above. But if the triangle is right-angled, the height can be one of the sides . This can be seen in the fourth triangle. In the sixth triangle you can see that the height line needs to be drawn outside the triangle.

Summary:

In summary, if you want to use the area formula you need to have a base and a height that make a pair, and you must have (or be able to calculate) their lengths. In some of the following problems, you will have to calculate the area of a triangle on the way to an answer.

Here is a reminder of the Theorem of Pythagoras; it applies only to right-angled triangles, but you will encounter many of those from now on.

I

In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

If you are a bit vague about applying the theorem, go back to the work you did on it before and refresh your memory.

• Using the formula, calculate the area of Δ ABC where A = 90°, BC = 10 cm and AC = 8 cm. A reasonably accurate sketch will be helpful. This is a two-step problem: first use Pythagoras and then the area formula.
• When calculating the area of quadrilaterals, the same principle applies as with triangles: when we refer to height it is always with reference to a specific base .
• We can use the formula for a triangle’s area to develop some formulae for our six quadrilaterals.
• A square consists of two identical triangles, as in the sketch. Let us call the length of the square’s side s . Then the area ( A ) of the square is:

A = 2 × area of 1 triangle = 2 (½ × base × height) = 2 × ½ × s × s = s 2 = side squared.

You probably knew this already!

• It works the same for the rectangle: The rectangle is b broad and l long, and its area ( A ) is:

A = first triangle + second triangle

= (½ × base × height) + (½ × base × height)

= (½ × b × $\ell$ ) + (½ × $\ell$ × b) = ½ b $\ell$ + ½ b $\ell$ = b $\ell$

= breadth times length.

You probably knew this already!

• The parallelogram is a little harder, but the sketch should help you understand it. If we divide it into two triangles, then we could give them the same size base (the long side of the parallelogram in each case). If we call this line the base of the parallelogram, we can use the letter b . You will see that the heights ( h ) of the two triangles are also drawn (remember a height must be perpendicular to a base).
• Can you convince yourself (maybe by measuring) that the two heights are identical? And what about the two bases? The area is: A = triangle + triangle = ½ bh + ½ bh = bh = base times height.
• A challenge for you: Do the same for the rhombus. (Answer: A = bh , like the parallelogram). Let’s see what we can do to find a formula for the trapezium. It is different from the parallelogram, as its two parallel sides are NOT the same length.
• Let us call them P s 1 and P s 2 . Again, the two heights are identical.
• Then from the two triangles in the sketch we can write down the area:

A = triangle 1 + triangle 2 = ½ × P s 1 × h + ½ × P s 2 × h

= ½ h ( P s 1 + P s 2 ) = half height times sum of parallel sides.

(Did you notice the factorising?)

• Finally, we come to the kite, which has one long diagonal (which is the symmetry line) and one short diagonal, which we can call sl (symmetry line) and sd (short diagonal).
• The kite can be divided into two identical triangles along the symmetry line. Because a kite has perpendicular diagonals, we know that we can apply the formula for the area of a triangle easily.
• This means that the height of the triangles is exactly half of the short diagonal. h = ½ × sd . Look out in the algebra below where we change h to ½ sd ! Both sorts of kite work the same way, and give the same formula.
• Refer to the sketches.

Area = 2 identical triangles

= 2( ½ × sl × h ) = 2 × ½ × sl × ½ × sd

= sl × ½ × sd = ½ × sl × sd

= half long diagonal times short diagonal.

In the following exercise the questions start easy but become harder – you have to remember Pythagoras’ theorem when you work with right angles.

Calculate the areas of the following quadrilaterals:

1 A square with side length 13 cm

2 A square with a diagonal of 13 cm (first use Pythagoras)

3 A rectangle with length 5 cm and width 6,5 cm

4 A rectangle with length 12 cm and diagonal 13 cm (Pythagoras)

5 A parallelogram with height 4 cm and base length 9 cm

6 A parallelogram with height 2,3 cm and base length 7,2 cm

7 A rhombus with sides 5 cm and height 3,5 cm

8 A rhombus with diagonals 11 cm and 12 cm

(What fact do you know about the diagonals of a rhombus?)

9 A trapezium with the two parallel sides 18 cm and 23 cm that are 7,5 cm apart

10 A kite with diagonals 25 cm and 17 cm

## Assessment

 LO 3 Space and Shape (Geometry)The learner will be able to describe and represent cha­racteristics and relationships between two-dimensional shapes and three–dimensional objects in a variety of orientations and positions. We know this when the learner: 3.2 in contexts that include those that may be used to build awareness of social, cultural and environmental issues, describes the interrelationships of the properties of geometric figures and solids with justification, including: 3.2.2 transformations. 3.3 uses geometry of straight lines and triangles to solve problems and to justify relationships in geometric figures; 3.4 draws and/or constructs geometric figures and makes models of solids in order to investigate and compare their properties and model situations in the environment.

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