Download 8√3 - Dr Frost Maths

Dr Frost
Year 10 – End of Year Revision
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Non-Right Angled Triangles
Using formula for area of nonright 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
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
? and tangent
between chord
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
? is half
circumference
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
=
? cm2
2.3
434mm3
=
? cm3
0.434
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
Congruence and Similarity
a) ED / 8 = 6 / 4
?
So 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
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.
x
y
16
10
8
5?
24?
15
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.
x=
=
4  √(16 – (4 x 3 x -2))
6
?
4  √40
= 1.72 or -0.38
6
Factorisation and Simplification
= (2x + y)(x + y) = 2x + y
(x + y)(x - 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
a)
b)
c)
d)
p9
q3
2u
3wy3
?
?
?
?
?
i) 1
ii) 8
?
iii) = (8/27)2/3 = (2/3)2 = 4/9
i.e. ‘Flip (if negative power), then Root
?
(if fractional power), then Power’.
Probability
a) 2/7 x 1/6 = 2/42 = 1/21
?
b) 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
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 = 9n?2 – 6n + 4
= 9n2 – 6n + 6 – 2
= 3(n2 – 3n + 2) – 2
Therefore it must be a
term in the sequence,
more specifically, the
(n2 – 3n + 2)th term.
Trial and Improvement
x = 1.5
x = 1.8
x = 1.9
x = 1.85
: 1.53 + (10 x 1.5) = 18.375
: 1.83 + (10 x 1.8) = 23.832
: 1.93 + (10 x 1.9) = 25.859
: 1.853 + (10 x 1.85) = 24.832
Too small
Still too small
Too big
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.
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
?