year 10 – end of year revision

26
ζ Year 10 – End of Year Revision Dr Frost Make sure you’re viewing these slides in Presentation Mode (press F5) so that you can click the green question marks to reveal answers.

Upload: rae-roberson

Post on 02-Jan-2016

30 views

Category:

Documents


0 download

DESCRIPTION

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: - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Year  10  –  End of Year Revision

ζYear 10 – End of Year RevisionDr Frost

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

Page 2: 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)

?

Page 3: 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

?

Page 4: Year  10  –  End of Year Revision

Trigonometry

Length BD = √(162 – 122) = √112

sin(40) = √112 / CDCD = √112 / sin(40) = 16.46cm

?

Page 5: Year  10  –  End of Year Revision

Circle Theorems

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

b) 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°.

?

Page 6: Year  10  –  End of Year Revision

Circle Theorems

a)54

b)27Because angle at circumference is half angle at centre.

?

?

Page 7: 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)

?

Page 8: 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.

Thereforex = 360 – 120 – 135 = 105.

?

Page 9: 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.

?

Page 10: 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

?

?

?

?

Page 11: 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 = 2BDTherefore BD = ½ AB

?

?

Page 12: Year  10  –  End of Year Revision

Congruence and Similarity

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

b) 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

?

?

Page 13: 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

?

x 16 8 24y 10 5 15?

?

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

Page 14: Year  10  –  End of Year Revision

Simultaneous Equations

x = 3, y = -2?

Page 15: Year  10  –  End of Year Revision

Area and Perimeter of Shapes

Perimeter:x – 1 + 3x + 3x + 1 = 56So 7x = 56, therefore x = 8

Area is therefore:½ x 24 x 7 = 84.

?

Page 16: 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.

?

Page 17: Year  10  –  End of Year Revision

Surdsa)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 – 3So a = 4 and b = -3

?

?

?

Page 18: 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

?

Page 19: 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

?

Page 20: 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.

??

?

?

x = 4 √(16 – (4 x 3 x -2))

6

4 √406 = = 1.72 or -0.38

?

Page 21: Year  10  –  End of Year Revision

Factorisation and Simplification

(2x + y)(x + y)(x + y)(x - y)

2x + yx – y

= =

5(2x+1)2 = (5x-1)(4x+5)5(4x2 + 4x + 1) = 20x2 + 21x – 520x2 + 20x + 5 = 20x2 + 21x – 520x + 5 = 21x – 510 = x

?

?

Multiply everything by x first:x2 + 3 = 7xSo x2 – 7x + 3 = 0This 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’!

?

Page 22: Year  10  –  End of Year Revision

a) p9

b) q3

c) 2ud) 3wy3

Simpliy (m-2)5 Answer: m-10 or 1 / m10?

????

i) 1ii) 8iii) = (8/27)2/3 = (2/3)2 = 4/9

i.e. ‘Flip (if negative power), then Root (if fractional power), then Power’.

??

?

Factorisation and Simplification

Page 23: Year  10  –  End of Year Revision

Probability

a) 2/7 x 1/6 = 2/42 = 1/21b) 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.

?

?

Page 24: Year  10  –  End of Year Revision

Sequences

a) x2 + 2xy + y2

b) 3n – 2

c) 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.

?

?

?

Page 25: Year  10  –  End of Year Revision

Trial and Improvement

x = 1.5 : 1.53 + (10 x 1.5) = 18.375 Too smallx = 1.8 : 1.83 + (10 x 1.8) = 23.832 Still too smallx = 1.9 : 1.93 + (10 x 1.9) = 25.859 Too bigx = 1.85 : 1.853 + (10 x 1.85) = 24.832 Too 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.

?

Page 26: Year  10  –  End of Year Revision

Inequalities

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

2x > x - 6

-x + 1 ≤ 6

x > - 6

x ≥ -5

?

?

1 ≤ 2x + 3 < 9 -1 ≤ x < 3? ?

Solve the following: