# Year 10 – End of Year Revision - PowerPoint PPT Presentation

1 / 26

ζ. Dr Frost. Year 10 – End of Year Revision. Make sure you’re viewing these slides in Presentation Mode (press F5) so that you can click the green question marks to reveal answers. Non-Right Angled Triangles. ?. Using formula for area of non-right angled triangle:

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Year 10 – End of Year Revision

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

#### Presentation Transcript

ζ

Dr Frost

Year 10 – End of Year Revision

Make sure you’re viewing these slides in Presentation Mode(press F5) so that you can click the green question marks to reveal answers.

Non-Right Angled Triangles

?

Using formula for area of non-right angled triangle:

A = ½ × 2x × x ×sin(30) = ½ x2

So 2A = x2

So x = √(2A)

Non-Right Angled Triangles

Angle CAB = 67°

Using sine rule:

(CB / sin67) = (8.7 / sin64)

So CB = 8.91015cm

Area = ½ x 8.7 x 8.91015 x sin 49

= 29.3cm2

?

Trigonometry

?

Length BD = √(162 – 122) = √112

sin(40) = √112 / CD

CD = √112 / sin(40) = 16.46cm

Circle Theorems

?

55°. ABO is a right angle because line AC is a tangent to the circle.

55°. Reason 1: By the alternate segment theorem, angle between chord and tangent is same as angle in alternate segment. Reason 2: Since BE is the diameter, BDE is 90°. And angles in triangle BDE add up to 180°.

Circle Theorems

a)

54

b)

27

Because angle at circumference is half angle at centre.

?

?

Circle Theorems

?

Angles of quadrilateral ADOB add up to 360, but angles ABO and ABO are 90 (as AD and AB are tangents of the circle) so DOB = 130.

BCD = 65 as angle at circumference is half angle at centre.

(Alternatively, if you added a line from D to B, you could have used the alternate segment theorem)

Angles

Exterior angle of hexagon is 360/6 = 60, so interior angle is 180 – 60 = 120.

Exterior angle of octagon is 360/8 = 45, so interior angle is 180 – 45 = 135.

Therefore

x = 360 – 120 – 135 = 105.

?

Angles

?

Since Tile B is regular, it’s an equilateral triangle with angles 60. Interior angle of Tile A must be (360 – 6)/2 = 150 by using angles around this point.

Therefore exterior angle is 30.

Then if Tile A has n sides, 360/n = 30, so n = 12.

Converting between units

15m2=150,000 cm2

23km3=23,000,000,000 m3

230mm2=2.3 cm2

434mm3= 0.434 cm3

?

?

?

?

Congruence and Similarity

a)

If congruent, we need to prove either all the sides are the same, or a mixture of sides and angles (i.e. SSS, SAA, SSA or AAA)

Since ABC is equilateral, AB = AC. Both triangles share the same side AD.

Therefore triangles ABD and ACD are congruent.

b)

BD = DC (since congruent triangles)

AB = BC (since equilateral triangle)

BC = BD + DC = 2BD

Therefore BD = ½ AB

?

?

Congruence and Similarity

?

ED / 8 = 6 / 4So ED = 12cm

The ratio of lengths AB to CE is 2:3. So the ratio of length AC : length CD is 2:3.If the total length is 25cm, then AC = (2/5) x 25 = 10cm

?

Proportionality

?

M = kL3

160 = k x 23

So k = 160 / 8 = 20

When L = 3, M = 20 x 33 = 540

Given that y is proportional to x, find the missing values.

?

?

Simultaneous Equations

?

x = 3, y = -2

Area and Perimeter of Shapes

Perimeter:

x – 1 + 3x + 3x + 1 = 56

So 7x = 56, therefore x = 8

Area is therefore:

½ x 24 x 7 = 84.

?

Enlargement

Method 1: Draw enlarged triangle and find its area.

Points are (1,1), (3,1) and (2.5, 2.5). Therefore area is

½ x 2 x 1.5 = 1.5.

Method 2: Area of original triangle is ½ x 4 x 3 = 6.

If length is enlarged by scale factor of ½, then area enlarged by scale factor of (½)2 = ¼ . So area is 6 x ¼ = 1.5.

?

Surds

a)

5√2 / 2

b)

4 + 2√3 + 2√3 + 3 – (4 – 2√3 – 2√3 + 3)

Notice how I’ve used brackets for the second expanded expression, so that I negate the terms properly...

= 4 + 4√3 + 3 – 4 + 4√ 3 – 3

= 8√3

Alternatively, we could have noticed we have the difference of two squares:

(2 + √3 + 2 – √3)(2 + √3 – 2 + √3)

= 4(2√3) = 8√3

?

?

?

= 4√2 – 3

So a = 4 and b = -3

Surds

?

= 6 + 2√8 + 3√2 + √16

= 6 + 2√4√2 + 3√2 + 4

= 6 + 4√2 + 3√2 + 4

= 10 + 7√2

Coordinate Geometry

?

On line AD, when x = 0, y = 6 (giving point D). When y = 0, x = 3 (giving point A).

Find the perpendicular line to y = -2x + 6 at the point A. We get y = 0.5x – 1.5.

When x = 0, y = -1.5 (giving point P).

Therefore length PD

= 1.5 + 6 = 7.5

Factorisation and Simplification

= (x+7)(x-7)

?

= 3x4y3/2

?

= (2t + 1)(t + 2)

?

In the above factorisation, we have the product of two numbers which must both be at least 1.

?

?

4 √(16 – (4 x 3 x -2))

6

x =

4 √40

6

= = 1.72 or -0.38

Factorisation and Simplification

?

(2x + y)(x + y)

(x + y)(x - y)

2x + y

x – y

=

=

?

5(2x+1)2 = (5x-1)(4x+5)

5(4x2 + 4x + 1) = 20x2 + 21x – 5

20x2 + 20x + 5 = 20x2 + 21x – 5

20x + 5 = 21x – 5

10 = x

?

Multiply everything by x first:

x2 + 3 = 7x

So x2 – 7x + 3 = 0

This won’t factorise, so use quadratic formula. But make sure you leave your answer in surd form and not decimal form, otherwise your answer won’t be be ‘exact’!

Factorisation and Simplification

?

Simpliy (m-2)5

Answer: m-10 or 1 / m10

?

p9

q3

2u

3wy3

?

?

?

?

1

8

= (8/27)2/3 = (2/3)2 = 4/9i.e. ‘Flip (if negative power), then Root (if fractional power), then Power’.

?

?

Probability

2/7 x 1/6 = 2/42 = 1/21

Possibilities for tiles are 1-2, 1-3 and 2-3. Probabilities for each are 2/7 x 3/6 = 6/42, 2/7 x 2/6 = 4/42 and 3/7 x 2/6 = 6/42. So total probability is 16/42 = 8/21.

?

?

Sequences

?

x2 + 2xy + y2

3n – 2

Remember that a term is in the sequence if there is some integer k such that 3k – 2 equals our term.Square of a term in the sequence:(3n – 2)2 = 9n2 – 6n + 4 = 9n2 – 6n + 6 – 2 = 3(n2 – 3n + 2) – 2Therefore it must be a term in the sequence, more specifically, the (n2 – 3n + 2)th term.

?

?

Trial and Improvement

?

x = 1.5: 1.53 + (10 x 1.5) = 18.375Too small

x = 1.8: 1.83 + (10 x 1.8) = 23.832Still too small

x = 1.9: 1.93 + (10 x 1.9) = 25.859Too big

x = 1.85: 1.853 + (10 x 1.85) = 24.832Too small

Therefore solution correct to 1dp must be 1.9.

It’s important that once you’ve identified the two closest solutions either side (1.8 and 1.9) you try the midpoint (1.85). Otherwise you won’t know which of the two is the correct solution.

Inequalities

Answer: -1, 0, 1, 2, 3

Solve the following:

2x > x - 6

x > - 6

?

-x + 1 ≤ 6

x ≥ -5

?

1 ≤ 2x + 3 < 9

-1 ≤ x < 3

?

?