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Year 10 – End of Year Revision PowerPoint PPT Presentation


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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:

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Year 10 – End of Year Revision

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Year 10 end of year revision

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.


Year 10 end of year revision

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)


Year 10 end of year revision

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

?


Year 10 end of year revision

Trigonometry

?

Length BD = √(162 – 122) = √112

sin(40) = √112 / CD

CD = √112 / sin(40) = 16.46cm


Year 10 end of year revision

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°.


Year 10 end of year revision

Circle Theorems

a)

54

b)

27

Because angle at circumference is half angle at centre.

?

?


Year 10 end of year revision

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)


Year 10 end of year revision

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.

?


Year 10 end of year revision

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.


Year 10 end of year revision

Converting between units

15m2=150,000 cm2

23km3=23,000,000,000 m3

230mm2=2.3 cm2

434mm3= 0.434 cm3

?

?

?

?


Year 10 end of year revision

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.

Angle ADB = ADC.

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

?

?


Year 10 end of year revision

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

?


Year 10 end of year revision

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.

?

?


Year 10 end of year revision

Simultaneous Equations

?

x = 3, y = -2


Year 10 end of year revision

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.

?


Year 10 end of year revision

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.

?


Year 10 end of year revision

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


Year 10 end of year revision

Surds

?

= 6 + 2√8 + 3√2 + √16

= 6 + 2√4√2 + 3√2 + 4

= 6 + 4√2 + 3√2 + 4

= 10 + 7√2


Year 10 end of year revision

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


Year 10 end of year revision

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


Year 10 end of year revision

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’!


Year 10 end of year revision

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’.

?

?


Year 10 end of year revision

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.

?

?


Year 10 end of year revision

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.

?

?


Year 10 end of year revision

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.


Year 10 end of year revision

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

?

?


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