The area of a triangle
Area of triangle ABC = 1/2ab sin C
or, 1/2ac sin B
or,1/2bc sin A
We can use this formula when we are given two sides and the included angle.
Question 1
Find the area of triangle ABC.
The Answer
Did you get 12.4 cm2?
Well done - the area of the triangle is 1/2 x 5.2 x 7.1 x sin 42 = 12.4cm2
Remember to show all your working. It is easy to make a mistake when using your calculator, and you won't get any marks for a wrong answer.
The angle between the tangent and the radius is 90°
A tangent to a circle is a line which just touches the circle.
Remember A tangent is always at right-angles to the radius where it touches the circle.
Tangents from a point outside the circle are equal in length
Two tangents to a circle from a point are equal. Have a look at the line of symmetry. The two tangents fit together.
The angle between the tangent and the chord at the point of contact is equal to the angle in the alternate segment
This is sometimes known as the alternate segment theorem.
Remember
This is the circle property which is the most difficult to spot. Look out for a triangle with one of its vertices resting on the point of contact of the tangent.
The angle between a tangent and a chord is equal to the angle made by that chord in the alternate segment.
Question 1
What is the size of:
- angle x?
- angle y?
The Answer
- Did you get x = 60°? Well done!
- Did you get y = 80°? Well done! You remembered that the angles in a triangle add up to 180°.
- Make sure that you learn these circle properties. If you are asked to find the angles in a circle, you will then be able to see which of them apply to the question.
- Do not be afraid to find the sizes of other angles first. It is not always possible to find the required angle immediately!
Similar areas and volumes
We already know that if two shapes are similar, their corresponding sides are in the same ratio, and their corresponding angles are equal.
Look at the two cubes below:
The cubes are similar and the ratio of their lengths is a:b or .
Question 1
What is the ratio of:
- the areas of their faces?
- their volumes?
The Answer
-
Cube 'a' has a face area of a2 Cube 'b' has a face area of b2 The ratio of their areas is a2 :b2 or
-
Cube 'a' has a volume of a3 Cube 'b' has a volume of b3 The ratio of their volumes is a3 :b3 or
For any pair of similar shapes, the following is true:
Ratio of lengths = a:b or
Ratio of areas = a2:b2 or
Ratio of volumes = a3:b3 or
Question 2
These two shapes are similar. What is the length of x?
The Answer
The ratio of the areas is 25:36 (a2:b2)
The ratio of the lengths is a:b Therefore we find the square roots of 25 and 36. Ratio of lengths = 5:6
5:6 = 2:x
So
(multiply both sides by 2) = x
x = 2.4 cm
I'm sure that you found that question straightforward. Now you try one!
Question 3
Two similar pyramids have volumes of 64 cm3 and 343cm3. What is the ratio of their surface areas?
The Answer
Did you get 16:49? Well done!
Teacher's Note
If not, try to fill in the blanks below:
Ratio of volumes = 64:? To find the ratio of the lengths, we find the cube roots of 64 and ? Therefore, the ratio of the lengths is ?:7. To find the ratio of the areas, we square ? and 7.
The ratio of the areas is 16:49
Problems in three dimensions
It is easier to work in two dimensions than three! Try to split problems of this type into a series of two-dimensional questions. Confused? Look at the following example.
The pyramid VABCD has a square base of length 4cm. The height of the pyramid is 3cm.
Question 1
Calculate the length of VA
The Answer
We need to use the right-angled triangle VAO. We know that VO is 3cm, but we do not have any more information about the triangle. However, AO 1/2 AC, and it is possible to calculate the length of AC.
Using Pythagoras' theorem,
AC2 = 42 + 42
AC2 = 32
Therefore, AC =
We now know that AO =
Applying Pythagoras' theorem to the triangle AVO, we get
AV2 =
AV2 =
AV2 = 17
AV = = 4.12 cm (3sf).
Question 2
Calculate the angle between VA and the base ABCD
The Answer
The angle between VA and the base ABCD is the angle VÂO.
Using trigonometry, sin VÂO =
VÂO = 46.7° (1d.p.)
Easy? Try this one yourself!
The triangular prism PQRSTU has a length of 10cm. PTQ and SUR are equilateral triangles with sides of length 4cm.
Question 3
Find the perpendicular height of triangle SUR
The Answer
The perpendicular height of triangle SUR is or 3.46 cm (3sf).
Question 4
the length of QU
The Answer
Applying Pythagoras' theorem to the triangle QUR, we find that QU = or 10.8 cm (3sf).
Question 5
Hence find the angle between QU and the base PQRS.
The Answer
The angle between QU and the base is therefore 18.8° (1dp).
Remember
Keep your answers in surd form or store them in the memory of your calculator. Using rounded values later on in a question will lead to inaccurate answers.
You might also be expected to apply the sine or cosine rules to a three-dimensional problem. Look at the following example.
Question 6
Sally and Kate stand some distance away from a building (B) which is 10m high. Sally is on a bearing of 030° from the building, and from where she is standing the angle of elevation of the top of the building is 35°. Kate is on a bearing of 090° from the building, and from where she is standing the angle of elevation of the top of the building is 50°. This information is shown on the diagram.
What is the distance between Sally and Kate?
The Answer
Using trigonometry on the right-angled triangles OBS and OBK, we find that OS = 14.3m (3sf) and OK = 8.39m (3sf).
Applying the cosine rule to triangle KOS, we get:
KS2 = 8.392 + 14.32 - ( 2 x 8.39 x 14.3 x cos60°)
KS2 = 154.534
KS = 12.4m (3sf)
Vector 'arithmetic'
Equal vectors
If two vectors have the same magnitude and direction, then they are equal.
Adding vectors
Vector followed by vector represents a movement from P to R.
Subtracting vectors
Vector , followed by a backwards movement along , is equivalent to a movement from X to Z.
Multiplication by a scalar
e.g.
Question 1
If x = , y = and z = , find:
- -y
- x - y
- 2x + 3z
The Answer
-
. Did you remember to change the signs?
To travel from X to Z, it is possible to move along vector , followed by . It is also possible to go directly along .
is therefore known as the resultant of and
Geometric problems
Question 1
Write as single vectors:
-
f + g
-
a + b
-
e - b - a
The Answer
-
e
-
-c Did you remember the minus sign?
-
-d
Remember
Two vectors are equal if they have the same magnitude and direction, regardless of where they are on the page.
Question 2
Triangles ABC and XYZ are equilateral. X is the midpoint of AB, Y is the midpoint of BC and Z is the midpoint of AC. = a , = b and = c. Express each of the following in terms of a, b and c.
