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Subject of a FormulaSmile 1500
a
A = i (a + b)h is the formula used to calculate the area of a trapezium where:
A = area of a trapezium
a = length of one of the parallel sides
b = length of other parallel side
h = height
A is the subject of this formula A = (a + b)h
A = i (a + b)h can be rearranged to make b the subject of the formula b = ^ - a ^ h
jere are two methods to show how to make b the subject of the formula:
Using rearrangement
multiply both sides by 2
divide both sides by h
A =
2A = (a + b)h
= a + b
subtract a from both sides ^ - a =
b = 2A84-a
Using flag diagrams
A =
b = 2A -a h
Choose either method to check that: the formula is
the formula is
a
h
2A _ b
2A" a + b
I when a is the subject.
| when h is the subject.
Substitute the values a = 4, b = 6 and h = 7 into the formula where A is the subject, to find a value for A.
Substitute the values for a, h and A into the formula where b is the subject, to check that b = 6.
Substitute the values for b, h and A into the formula where a is the subject, to check that a = 4.
Substitute the values for a, b and A into the formula where h is the subject, to check that h = 7.
/ / L \T = 2n"\j \ g I is the formula for calculating the time of a swinging pendulum where: T = timeL = length of the string
g = acceleration of gravity.
T is the subject of this formula: T = 2ir\j (-)\w /
T = 2n~J {* can be rearranged to make L the subject of the formula L = g (-M" \ 9 ' \2717
Using rearrangement
divide both sides by 2n
square both sides
multiply both sides by g
I ="\/( L 2rc Mg.
fj) \2nlI) 2 = L 2nl g
-\ = L
271
Using flag diagrams
Choose either method to check the formula is g= L '
| when g is the subject.
Substitute values for T, L and g to check that the rearrangements are correct.
Rearrange each formula to make the letter in brackets the subject of the formula.
1. A = 3b (b) 2. v = Ik (I) 3. v = u + at (t)
4. m = (x + y) (x)
7. d = V(11.5h) (h)
10. v2 = u2 +2as (u)
5. F = nw (v)
8. A = 3(p + 5) (p)
11. a = 6-12 (r)
12
6. F = H3V {r)
9. v2 = u2 +2as (s)
,2. s = &±*& (v)
13. Make c the subject of the formula T = 4^
Use your formula to find c when T = 3T = 2.4
14. The formula for the perimeter P of a rectangle is P = 2L + 2W where L = length and W = width. Make L the subject of this formula and find the length of a rectangle whose perimeter is 36cm and whose width is 7.3cm.
15. The formula for the surface area of a cylinder is S = 2rcrh + 2nr2 where r = radius and h = height. Make h the subject of the formula and find the height of a cylinder with surface area of 84cm2 and radius 2cm.
©RBKC SMILE 1995.
Smile 1501
Changing the SubjectIn the equation x = ab + 2b 'xl is the subject of the equation, because the equation is in the form x
1 and 'x* only occurs once.
The equation can be rearranged to make 'b' the subject, 'b1 appears twice in the equation. To make 'b1 occur once a factor of b has to be taken.
Take a factor of b
Divide both sides by (a + 2)a + 2
x = ab + 2b
x = b(a + 2)
= b
or b =a + 2
Rearrange the equation to check that with 'a1 the subject, the equation is a =
Make the letter in brackets the subject of each of these equations.
1. Z = 3p + pq (p) 2. s = 2ac + 4ab (a) 3. h = d2 -3hR (h)
In the equation x = 3v + 4 'x' is the subject. It can be rearranged to make y the subject. 2-y
Multiply both sides by (2 - y)
Expand the brackets
Add xy to both sides
Subtract 4 from both sides
Take a factor of y
Divide both sides by (3 + x)
or
x = 3Yt4 2-y
x(2-y) = 3y + 4
2x-xy = 3y + 4
2x = 3y + xy + 4
2x - 4 = 3y + xy
2x-4 = y(3 + x)
2*-^4 = y 3 + x y
y - 2X^4
y 3 + x
Make the letter in brackets the subject of each of these equations.
y =3 (x)
(s)
5. p = lmv2 u2 (m) 6.
9. v = uf u-f (f)
7 Z =x + 2
(x)
©RBKC SMILE 1995.
AreasSmile 1504'
This speed-time graph shows the journey of a cyclist.
time (h)
She travels at a speed of 10 kilometres per hour (10km/h or 10kmrr1 ) for 23A hours then slows down for 1 /2 hour to a speed of 7 kmrr1 which she maintains for another 1 3A hours.
Over the first hour she travelled at a speed of 10kmrr1 .
10kmh-1 x1h = 10km.
She covered a distance of 10km in the first hour|
The shaded rectangle represents the 10km the cyclist travelled in the first hour.
The shaded square represents 1kmlr1 x 1h, a distance of 1km.
The area under a speed-time graph represents distance travelled.
To find the total distance travelled by the cyclist calculate the total area under the graph.
4 5
time (h)
Rectangle A = 10kmrr1 x 2.75h = 27.5km
Trapezium B =
mRectangle C =
(10 + 7)kmh-1 x 0.5h2
4.25km
7kmrr1 x1.75h 12.25km
The total distance travelled by the cyclist
= (27.5 + 4.25 + 12.25)
= 44km
1. Copy and complete the information shown by these graphs.
a) This graph shows flow of water, in litres per minute (/min'1 ) overtime, in minutes (min).
10
3 6 Time (min)
The shaded area represents
5/min-1 x 3min = litres
b) This graph shows egg yield, in dozens per day over time, in days.
18
2 9^*c *^
O) O)til 0 5 10
Time (days)
The shaded area represents
9 dozen per day x 10 days =
c) This graph shows velocity in metres per second (ms-1 ) overtime in seconds (s).
^20I £10
0 5 10 Time (s)
The shaded area represents
Write similar information for these graphs,
d) e) f)
8 16 Time(s)
0 1 2
Mass (kg)
0 4 Area (cm2)
2. This graph represents the rate at which water flows from a reservoir in a 12 hour period.
40
c~ 30 .s=SH. WE£ »z 20
m 3
SS 10
6am 9am 12
Time
3pm 6pm
a) What volume of water is represented by the shaded area?Find the approximate area under the graph to calculate the volume of water used during the 12 hours.
b) Use the graph to decide if more water is used from 6am - 9am than 3pm - 6pm. Explain your answer.
c) If the water flows into the reservoir at a steady rate of 28 thousand litres per hour during the 12 hour period, will the water be lower or higher at 6pm than at 6am? Explain your answer.
3. This graph shows a person's pulse rate over a ten minute period.
Notice that the vertical scale starts at 60.
100
Pulse rate per mi u> -^i co o0 0 0 2 /
X
\
11s\^ F
\ '
\\
«31234567891
Time (min)
a) What does the shaded square represent?
b) Estimate the average pulse rate for the first 3 1 /2 minutes.
c) What was the normal resting pulse rate?
Estimate the average pulse rate for the 10 minutes shown.
©RBKC SMILE 1995
Smile 1511
1. Match each of these inequalities with one of the graphs below. The unshaded region shows the inequality,
i) x>2 ii) y^6 Hi) x + y^.3 vi) 2x + 3y^ 12
v) y + 2x^50 vi) xy ^ 144 vii) y^2x viii) x^2y
a)
e)
-1;: -i
18
12
-12
12
-2
111
150
25
> 0
-25
-50
-75
Hi
11'̂t1i
Hiiiiii^11
1
$$$m
ill
b)
C)
g)
-9
>
h)
llllllll
llll
-
Jm
llll11 m
ininiti
>3
2
1
t
mtill
|||
illllll::':::::::':"::::::
"y
kmmil
%jjj.iSiSSSS
III
SswSi
AM
SwSjS:
fill
Jill
' X
ip:-:-;-:-:*lf!; :;-:
IB
Turn over.
2. Which of these graphs is shaded correctly to show all the following inequalities?
x ^ 0y ^ o2x + y > 4
x + 3y > 9x + y < 6
a) Vit!illlimm*;
lill• ::: ::::>:W:Sij:ffffffffgilK;
llllil
"HI^i
h»iiiiii::::: :::::: :::::::::::::::i :|:|:|:|:|:i:i:|:|:|jxy
-mmNil
2&J1S&:8i£:;:!?!
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:x-x-x-:'i'>>i';-x
Hi•:•:-:•:•:•:•:•:•:•:•>:*;
::::::::: :: ::: :::::: ;:x::
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b)
c) W888*
Pl| iiiilmmmSSSsSjSi
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lill
•IiPiiillli•̂iifci illli;:::::X;X;:xX:::X| |::::::::>:::|:i:|:|:i:::ij
mmm
W:-SS:x::::::ik
iiiiili
iiisiii
feP
:::llll!iiii
lill
tiplii •*WmKxSjiig:!:¥::: ::::: :::¥ij
Illli
iSSSSSS:
iii!ilim 1 ^
•:-:-X-:-:-:-:-:-:-:-:-:H;X;:;X;X;X;:::;:-N
II 111
II III
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iffisHSS
ISx^iwS-iiHIIP
iiiiii
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d) &£»H| «w•if miHIWiiSSwiiiji:SSiKfW*::
Illli
111
liil
P«i III:::::::::x :: :::::: : :: ::: : i
iili;i;l
^ip;•*«
Illli
llllil
%t̂i'\oi- !•:•!• :-i: i''''::-''" :
Illli
illillli
*;8ii;$iiiiiB
e)
You may like to check your answers using the MicroSMILE program REGIONS.
^if;^
ifcfcriipfiSffiSiSi:¥:W:::!*: : M-:*^:-:-:-:-:?.;$:«?*
:::-:W:':::vj»:
IIIIII
illli
V^111:
•111! Illlillllil
>v ills ilps.;X;X;:vXv:;'?!«
ll^llllil
lii "v.'ftw Sjt:*:*:*:??!*
liili
© 1992 RBKC SMILE
Smile 1517
Trig ProblemsThe sine and cosine rules are true for any triangle.
Sine Rule
sinA = sinB = sinC a b c
Cosine Rule
a2 = b2 +c* -2bccosA
b2 = a2 + c2 -2accosB c2 = a2 +b2 -2abcosC
These rules can be used to solve problems In triangles. The rule used will depend upon what information is known and what is to be found.
3.2cm-
To find the length of side x.The length of the two other sides and the angle between them are known.Use the cosine rule.
x2 = (4.8)2 + (3.2)2 - 2 (4.8 x 3.2) cos 56 C
= 16.102
x = V16.102
x = 4.01cm
To find ZA.
Length x is now known.Use the sine rule.
sinA = 4.01
sinA
sin110 6.8
= 4.01sin110 6.8
= 0.5541
= 33.65°
1. Find side b to the nearest metre. 2. Find side x to the nearest metre.
3. Find side b to the nearest metre. 4. Find x.
5. The bearing from A to B is 180°. The bearing from A to C is 215°. The distance from A to B is 677km. The distance from B to C is 1500km. The distance from C to A is 2000km.
• The bearing from B to C is between 180° and 270°. What is this bearing to the nearest degree?
6. You do not have to use the sine or cosine rule for this question.
tanSO = _h_ AC B *————30m———-
• Find a similar equation connecting h, AC and 16°.
• Solve the simultaneous equations and find h and AC.
©RBKC SMILE 1995.
You will need a pack of playing Smile 1520
GameThis is o gome for two or more.
Deal 6 cards to each player.
Sort your cards into pairs and find the differences.
Add up your score.
example:
The first to reach 100 wins!
- 17Here's another way to play ..... first to 100 loses!!
Smile 1522
Eight Cubes
You will need 1522A (8 cubes)
Make a yellow cube. Make a blue cube.
Smile 1523
A Red CubeYou will need 1523A (27 cubes)
Make a red cube.
Smile 1524
You will need multilink cubes or centicubes
4 Cube Solids
With 3 cubes it is only possible to make 2 different solids:
There are 8 different solids using 4 cubes.
Find them.
Some people think there are only 7. What do you think?
4 Cube SolidYou will need multilink cubes or centicubes.
With 3 cubes it is only possible to make 2 different solids.
There are 8 different solids using 4 cubes.• Find the 8 different solids using 4 cubes.• Some people think there are only 7. What do you think?
Smile 1524
© RBKC SMILE Mathematics 2005
4 Cube SolidYou will need multilink cubes or centicubes.
With 3 cubes it is only possible to make 2 different solids.
There are 8 different solids using 4 cubes.• Find the 8 different solids using 4 cubes.• Some people think there are only 7. What do you think?
Smile 1524
RBKC SMILE Mathematics 2005
Smile Worksheet 1525
Economical Weaving
Colour this pattern so that the same colour never crosses itself.
Try to use as few colours as possible.
Can you do it using only 4 colours?
turn over
© RBKC SMILE 2001
Fraction Wall 2y///////Ar////jy/^^^
11 1This is a fraction wall using 2 » 4 and 8
Smile 1
1 Whole Strip
1 2
1 *
s s
1 *1 1F ff
21*
It ff
1*1 1ff ff
y looking at the fraction wall you should notice that
1=2 = 4 248
You can also use the fraction wall to add fractions.
14
38
2 + 1 4 4
8
Copy and complete these fractions:
8
D i = :2 8
4) 1 + 1 = !
2) i = :4 8
5) i + i = :
3)8
6) 2. + 1 =
I.I$ 10)
I\\ 15)
\
Z
2 +4
1 +2
1 -
5 -8
b
1 =8
1 +4
2 =8
2 =8
b
*8
1 =8
*
8
**
4
8) 12
*
8
13) 24
16) 12
8 b b 4 b
+ 2 = 1 9) 1 + 5 = I88 488
11) 1 - 1 = *8 8
-1=1 14) z - a = !88 888
- 1 = *8
©1992 RBKC SMILE
ProportionSmile 1533
If two quantities p and q, are directly in proportion to each other, the notation p °c q is used. The quantities can be linked by an equation of the form p = kq, where '/c1 is the constant of proportionality.
Statement
'Velocity, v, is directly I proportional to time, f.'
'Stopping distance, d, of a car is proportional to the square of its speed, s.'
The pressure, p, is inversely proportional to the volume, v.'
Notation
Voc f
C^oc S2
DOC -i V
Formula
v=kt
d=ks2
'•$
Graph
V
d
P
V t
y5
v^v
1. Express these statements
(i) in notation, (ii) as formulae, sketch the graph.
a) The increase in length, d, of a rod is directly proportional to the increase in temperature, f.
b) The circumference, c, of a circle is directly proportional to its radius, r.
c) The mechanical energy of motion, e, of a car is proportional to the square of its velocity, v.
d) The volume, v, of a sphere is proportional to the cube of its radius, r.
e) The distance to the horizon, d, is proportional to the square root of the observer's height , n, above the surface of the sea.
If there is more information connecting the two quantities the constant of proportionality (k) can be calculated.
A quantity y, is inversely proportional to the square of a quantity x, and when x= 5, y=4.
•y is inversely proportional to x2 ' y
Substitute x = 5, y=4
The constant of proportionality is
The relationship between x and y is
A sketch of the graph of this relationship.
y = y x2
4 =25
k =100
„ _100
2. y is proportional to x and wheny = 12, x=2.
• Find the expression connecting x and y and find y when x = 5.
• Sketch a graph of this relationship.
3. y is inversely proportional to the square of x and when x = 4, y = 3.
• Find the expression connecting x and y and find y when x = 8.
• Sketch a graph of this relationship.
y is proportional to x3, so y^x3 and y = ax3.
• Find the constant of proportionality, a, if x= 2 when y= 0.1
• Complete the table of values:
0.1 12.5
Sketch a graph of this relationship.
Some corresponding values of x and y are shown in the table:
X
y15
5125
10500
202000
Which of the following could be true?a) y oc x2b) y = 5xc) y = 5x*d) y = 30x-25
6. Some corresponding values of Fand Ware shown in the table:
FW
10030
20015
30010
5006
• Which of the following is a possible relationship between Fand IV?
a) W is directly proportional to F
b) Fis inversely proportional to W
c) W = 3/ioF
d) FW = 3000
7. Given that y <* complete the table of values:
20 17240
14 10693-6
• Sketch a graph of this relationship.
8. Find the equation for which the following are corresponding pairs:
(20, 35), (24, 42), (48, 84), (50, 87-5).
• Sketch a graph of this relationship.
Here are some familiar formulae containing a constant of proportionality.
Volume of a sphere
Time of a swinging pendulum
Formula
,.*v
r-2^ Lg
v is proportional to the cube of r.
T is proportional to the square root of L
Constant of proportionality
1"
2rtVs
©RBKC SMILE 1995.
Smile 1537
Simultaneous equations and inequalities
grapn can be used to solve:
the simultaneous equations
f 2x + y =12 I x-y = 3
and the
simultaneous inequalities
(2x + y <12 I x-y > 3
5 \=:1:Q::.. 15
-15 t
The solution to the equations is x = 5, y = 2
The solution set for the inequalities is the unshaded region R.Choose a point in the unshaded region to check that its coordinates satisfy both equations.
•
You can either draw graphs on paper or use the MicroSMILE program REGIONS to do the following questions.
BJBEB The program REGIONS shades the unwanted region only a short way from the line.
1. a) Solve the simultaneous equations f y = x + 5 1 x + 2y = 1
b) Shade out the unwanted regions to show the solution set R to the simultaneous inequalities f y > x + 5
1 x + 2y < 1
2. Solve the simultaneous equations:
a) y-4x = 1 b) 2x-y = 2x + y = 4
3. Show the solution set for each pair of simultaneous inequalities
a) fy-x>0 b) fy-x<0 c) fy<x + 1 d) fx-y>2 [ y + x > -2 I -2y - x < 3 1 y > 1 - 4x I -2x + y > -2.
©RBKC SMILE 1995.
Solving Simultaneous^quationsTwo equations connecting x andy are called simultaneous equations. It is sometimes possible to solve them.
Here are three methods trjat can be used to find solutions to this pair of simultaneous equations
2x+y = ~\2 (1) x-y = 3 (2)
Add equation (1) to equation (2)
substitute x = 5 in equation (1)
-j> = 12+3
3x = 15 x = 5
10+.y = 12 y = 2
2x+y = "\2 (1) x-y = 3 (2)
Rearrange equation (2) to make x the subject of the equation, substitute x =y + 3 into equation (1)
substitute y = 2 into equation (3)
x =y +3 (3)
2{y +3)+y =122y +Q+y =12
3.y +6 =123y = 6y = 2x = 2 + 3x = 5
Each method leads to the unique solution: x = 5 and>> = 2.
( x — v == 12 (1)
x-y = 3 (2)
Graphs representing the two equations are plotted.
Smile 1538o
The solution is found by the point where the graphs intersect (5, 2). So x = 5, y = 2
Use any method to solve the following pairs of simultaneous equations. Try to use each method at least once. = 10 2. f x-v = 2 3.(2x + 3v=7 4. t2x+y= J\ 5.Jjc+y=101.fjc + 2y=10 2.\x-y =
\x-y = 4 \y=-x + y =-3x 2x + 5y = 9
'••{ >. f 2jc + 3.y I3;c-=
©RBKCSMILE1995.
Smile 1540
Is There a Solution ?
This shows graphs of the simultaneous equations f 2x + y = 12x-y = 3
The unique solution to the simultaneous equations is the point of intersection (5, 2) so x = 5, y = 2.
This shows graphs of the simultaneous equations / x + 2y = 32x + 4y=10
The graphs are parallel, they will never intersect. There is no solution to these simultaneous equations.
ORBKC SMILE 1905.
Smile 1541
Cones
^^^^M^^iii^^f^^f^MfM^^^^i iilliiililiHiii
of circular base = jiri + 7tr2
1. Find the curved surface area of this cone.
11cm
2. The volume of this cone is 36cm3 .
Find its height.Find its slant height.Find its curved surface area.
3. Find the total surface area of this cone.
7.5cm
4. The volume of this cone is 400cm3.
Find a) the area of the base.b) the radius of the base.c) the total surface area.
5. The formula for the total surface area of this cone is A = rcr2 + Ttrl.
A is the subject of the formula.
Rearrange this formula to make I the subject.
6. Find the area of materialneeded to make this lampshade.
©RBKC SMILE 1995,
Smile 1543
Composite FunctionsThe function 'subtract 2 and multiply by 3' can be written x 3(x - 2).
Write these functions in the form x (i) 'multiply by 3 and add 2'(ii) 'divide by 3 and add 2'(iii) 'add 2 and multiply by 3'(iv) 'add 2 and divide by 3'(v) 'square and subtract 7'(vi) 'subtract 7 and square'
Is the function x ——* 3x2 the same as the function x ——>(3x)2? Explain your answer.
s is the function 'square' s(x) = x2
d is the function 'add 2' d(x) = x + 2
The composite function ds(x) means 'do function s, then do function d to your answer'.
s(x) = d(x) =
square^- add 2
ds(x) = x2 + 2
3. Copy and complete the flag diagram to find the composite function sd(x).
d(x) = x + 2 s(x) = x2
sd(x) =
4. The functions f and g are defined as f(x) = 4x + 1 and g(x) = 2x -1
(i) Copy and complete the flag diagram to find fg(0).
g(x) = 2x-1 f(x) = 4x +1
Find (ii) fg(-2)
Complete fg(x)
gfM
iii) gf(0) (iv) gf(-2)
5. The functions f and g are defined as f(x) = 2x -1 and g(x) = 3x - 2.
Find (i) fg(2)
(iii) gf(0)
(v) fg(V2)
Complete fg(x) =
gf(x) =
«) gf(2)
(iv) fg(0)
vi) gf('/2)
6. f(x) = Vx and g(x) = x + 2.
Complete fg(x) =
gf (x) =
f(x) = x + 1 and gf(x) = x. Complete g(x) =^mmmsm
8. f(x) = 2x + 1 and fg(x) = x. Complete g(x) =
©RBKC SMILE 1995
Mystic Rose Smile Worksheet 1555
How many lines are used to create this Mystic Rose?
it might help to draw simpler ones.
Which patterns do not have a hole at the centre?
Can you explain why?
©RBKC SMILE 2001
CD
I
09
6261-89
8688
26
U.
0170201-6
oCD
CD
o00
15
00
CD CO CD CM
23 24 25
96 96 co
(M CM
CM CO
CM
ooCM
CM
CO CO
CO
CD CO
CO
CD
CD IO
98
99O) CO
co co
co
CO O)
Spirals Smile Worksheet 1557
turn over
© RBKC SMILE 2001
Smile 1559
The trapeziumhas been enlarged by scale factors,2, 12, 2, 22, 3, 82
to give a set of similar trapezia.
D Use a rotagram to check the corresponding angles of each trapezium are equal.
D Copy and complete this table.a) b) c) d) e)
Scale factor
1 2
i*2
2*3
3*
original . corresponding length • new length
1:2 = 2:1
1:12 = 2:3
1:2
1:22 = 2:5
1:3
1:35 = H:H
original area (cm^)
4
4
4
4
4
4
new area (cnrr]
1
9
16
B
•B
original . new area * area
4:1
4:9
4:16= 1:4
• :•
fl:B = fl:B
B:B
What do you notice about the ratios in columns b) and e)?
This triangle has an area of 8cm 2 A
4cm
D Enlarge it by scale factors 2, li 2, 2\, 3, 82. You may wish to use MicroSMILE TRANSFORM.
D Complete a table of results.
a) b) c)
4cm
d) e)
Scale factor
1 2
original . corresponding length • new length
4:2 = 2:1
original area (cm2 )
8
new area (cm*)
2
original . new area • area
8:2 = 4:1
What do you notice about the ratios in column b) and e)?
This shape has an area of 4cm2 .
D What would its area be if it was enlarged by scale factor a) 2
b) 2-c)
This shape has an area of 6cm2 .
D What would its area be if it-was enlarged by scale factor a) 2
b) 5c) 3^
D Copy and complete this summary:
When a shape is enlarged by scale factor n • the corresponding angles are
• the ratio of the sides is 1: B
• the ratio of the areas is 1:
©RBKC SMILE 1995.
Smile 1560
Similarity ProblemsWhen a shape is enlarged by a scale factor n, the original shape and the new shape are similar.
• The ratio of each original side to the corresponding new side is 1: n.
• The ratio of the original area to the new area is1:n2 .
1. A circle with radius 2cm is enlarged by scale factor 2.
• Calculate the diameter, circumference and area of each circle.
• Do your results agree with the summary in the box?
Drawings are not to scale.
2. The area of this postcard is 60cm2 .
6cm
A similar version of the same postcard has a height of 3cm. 3cm
What is the area of the small version?
3. A photograph is 20cm long and 15cm wide.
20cm
15cm
The length of a similar, smaller photograph is 5cm.
5cm
a. What is the width of the smaller photograph? b. What is the area of the smaller photograph?
4. One square metre of paper weighs 90g.
• How many kilograms would a 10 metre by 10 metre square of paper weigh?
5. 18 square floor tiles each 30cm long are needed to cover a floor.
30cm
30cm
How many tiles would be needed if square tiles of 7.5cm long are used?
6. This is the map of a deciduous forest.
Scale
1:50000
• Calculate the area of the forest in km2 .
©RBKC SMILE 1995
Smile 1561
Combining Transformations^ Draw a grid with x-axis from -7 to 14 and y-axis from -7 to 10.
Plot the points (5,1), (7,1), (7, 2), (6, 2), (6, 4), (5, 4), (5, 1) and join them in order. Shade the 'U shape and label it L0 .
^ Draw the resulting '!_' shapes for the transformations in the table below. The first transformed 'L' shape L1 has been completed for you.
i^iiBS iiiiliiiiiiLO
L,
L0
L3
LO
L5
L0
L,
L0
L9
translate |]
translate /4\r/
reflect in x-axis
reflect in y-axis
reflect in y = x
rotate 180° about (0,0) anticlockwise
rotate 90° about (0, 0) anticlockwise
rotate 180° about (0, 0)
reflect in x = 3
reflect in x=-2
fiBiliiii illiPlliililllL,
L2
L,
L4
L5
L6
L7
L8
L,
L10
Describe the single transformation to
a) L2 onto L0b) L4 onto L0c) L6 onto L0d) L8 onto L0e) L10 onto L0
map:
Use MicroSMILE Transform to check your work.
©RBKC SMILE 1995.
Smile 1562
Ifi! in;
Draw a grid with x-axis from -8 to 8 and y-axis from -8 to 8. Draw the lines y = x and y = -x
Plot the points (2,1), (7.1) (7,4) and join up to form a triangle. Label it A.
Use the information in the table below to draw 7 more triangles.
Starting Shape
A
B
C
D
E
F
G
Transformation
reflect in y = x
reflect in y-axis
reflect in y = -x
reflect in x-axis
reflect in y = x
reflect in y-axis
reflect in y = -x
Label of new shape
B
C
D
E
F
G
H
Describe the single transformation to map:a) A onto Eb) B onto Gc) D onto Hd) E onto B
Use MicroSMILE Transform to check your work.
©RBKC SMILE 1995.
Smile Worksheet 1565
SymmetryThe dotted lines are lines of symmetry. Use reflection to complete the pictures.
7
\
\\
9 x 2009?
© RBKC SMILE 2001
You will need a calculator Smile 1566
Finding Square RootsThe Square Root of 9 is 3, because 3x3 = 9.
It can be written \ 9 = 3.
^Some square roots are harder to find:-
*$
Guesses for
34
3.5 3.4
3x3= 9 4 x 4= 16 3.5 x 3.5= 12.25 3:4 x 3.4 = 11.56
Too high Try 3.4
Too low TryH'
1. Find Y12 as accurately as you can./— i— i—2. Find another square root. eg. Y 1000, \ 32, Y 24
Velocity from Distance-Time GraphsII The rate at which distance is travelled is called speed.
III Velocity is a measure of speed with the direction of motion specified.
11 If one direction is regarded as positive velocity, a speed in the opposite direction has negative velocity.
Smile 1568
«
This graph describes the journey of a cyclist from A to B.
The graph can be used to find her average velocity from A to B.
This is found from the gradient of the chord AB.
Average Velocity = gradient of chord
= increase in distance increase in time
= 300-0 60-0
= 5m/s10 20 30 40 50 60
Time (seconds)
The graph can be used to estimate the cyclist's maximum velocity between A and B. This is found from the gradient of the tangent to the curve (drawn by eye) at the steepest point, C.
At point C, t = 30,
maximum velocity = gradient of tangent
= 300-0 30-16
= 300 14
= 21.4m/s
500
400
300
% 200- 5
100
U
&
10 20 30 40 50 60
Time (seconds)
The cyclist stops at 60 seconds. The gradient of the tangent to the curve at B is 0.
1. Sketch a possible distance-time graph of a car journey assuming that the car travels at constant velocity except when it is held up by traffic lights.• What is the gradient of the tangent to the curve when the car stops at traffic lights?
2. Draw an accurate distance-time graph of this 3-stage journey:- a steady velocity of 15m/s for 3 minutes- a 1 /2 minute stop- a steady velocity of 10m/s for 4 minutes.
• Use the graph to find the average velocity for the journey.• Check your answers by calculation.
Turn over
3. This table shows the distance of a car from home at 15 minute intervals.
Time
Distance (km)
10.00
0
10.15
10
10.30
20
10.45
33
11.00
58
11.15
80
11.30
98
11.45
116
12.00
120
• Draw a distance-time graph.
a) Calculate the average velocity for the whole journey.
b) Calculate the average velocity between 10.30 and 11.30.
c) Draw a chord on the graph to estimate the average speed from when the car was 30km from the start to when it was 90km from the start.
d) Draw a tangent at the steepest point of the curve and use it to estimate the maximum velocity.
4. This table shows the time a train passed kilometre posts on a journey.
Distance (km)
Time (hours)
0
0
10
0.1720
0.26
30
0.32
40
0.42
45
0.53
50
0.75
60
0.86
70
0.91
80
1.0
Draw a distance-time graph and use it to estimate:
a) the average velocity for the first half hour of the journey
b) the velocity when the train was 40km from the start
c) the maximum velocity.
5. Trace this graph and draw suitable tangents to estimate:
a) the velocity at 4 seconds
b) the time when the velocity is 5m/s. 25
CD Onj "GO
b
1234
Time (seconds)
©RBKC SMILE 1995.
Smile-1569 0
^^
Velocity = cnar|ge in distance y change in time
Acceleration = change in velocity change in time
Finding acceleration from a velocity-time graph
This velocity-time graph represents a two minute cycle ride.
The cyclist's velocity increases ily for the first 50 seconds, 0 to 10 m/s.
She then travels at a constant velocity for 40 seconds.
Then she slows down until she stops.
20 40 60 80 100 120 time(s)
The gradient of a velocity-time graph is a measure of acceleration.
To find the acceleration of the cyclist between .. .
0 and 50 seconds . . . The gradient of the graph for the range 0 <; t <; 50= 10-0
50-0
I = °'2W The velocity is increasing by 0.2 m/s every second.
The acceleration is 0.2 metres per second per second, or 0.2 m/s2 or 0.2 ms'2 .
The gradient of the graph for the range 50 <, t <, 90
= 10-1090-50
= 0The cyclist has 0 acceleration, she is travelling at constant velocity.
90' and 120 seconds. The gradient of the graph for the range 90 s t ^ 120
50 and 90 seconds
120-90
30
= -0-33 to 2 dps.
The velocity is decreasing by 0-33 m/s2 . The deceleration is 0-33 m/s2 . The acceleration is -0-33 m/s2 .
This velocity-time graph represents the journey of an object with velocity vm/s, at time t seconds given by the equation.
v= 16t-4t2
time (s)
The graph is a curve. To estimate acceleration a tangent to the curve is drawn by eye and the gradient calculated.
At t = 1 second
At t = 2 seconds
At t = 2-5 seconds
the acceleration is estimated from the gradient of the tangent AB.
Acceleration 12-8 1 -0-5
8m/s2
the gradient of the tangent to the curve CD is zero.
The acceleration is zero and the velocity is constant.
the acceleration is estimated from the gradient of the tangent EF.
Acceleration 17-14 2-2-75
-4m/s2
1. This acceleration-time graph represents a journey with constant acceleration.
time
Which of these velocity-time graphs could represent the same journey?
i) ii) iii) iv)
1
time time time time
2. A cyclist's journey after t seconds has a velocity of vm/s where v = Vst2 .
a) Sketch a graph of her velocity against time for the first 5 seconds.
fc b) Draw tangents at suitable points to find the acceleration at: i) t = 2 ii) t = 4.
c) Find the approximate time when her acceleration was 1 m/s2 .
Finding distance from velocity-time graphs and velocity from acceleration-time graphs.
This velocity-time graph shows a journey of constant velocity of 20m/s for 10 seconds. 20'
In 10 seconds the distance covered is 200m, this is represented by the area under the graph.
>> 10--
5time(s)
_L10
The area under a velocity-time graph is a measure of distance.
This velocity-time graph represents a two minute cycle ride.To find the distance travelled by the cyclist find the area under the graph.
Triangle A
Rectangle B
Triangle C
Total
1 /2 (50x10) 250
40x10 400
1 /2(30x10) 150
10"
20 40
= 80060 80 100 120 time(s)
The area is 800 units, the cyclist travelled 800 metres.
This velocity-time graph represents the journey of an object with velocity vm/s, at time t seconds given by the equation.
v= 16t-4t2
As it is a curve it is more difficult to calculate the distance travelled.Methods that can be used are:o estimationo counting squareso finding the areas of strips which will approximate closely to trapezia.
Area of trapezium 1 = 1 /2(0 + 7) x 0-5
Area of trapezium 2 = 1 /2(7 + 12) x 0-5
Area of trapezium 3 = 1 /2(12 + 15} x 0-5
Area of trapezium 4 = 1 /2(15 + 16) x 0-5
Area of trapezium 5 = 1 /2(16 + 15) x 0-5
Area of trapezium 6 = 1 /2(15 + 12) x 0-5
Area of trapezium 7 = 1 /2(12 + 7) x 0-5
Area of trapezium 8 = 1 A>(7 + 0) x 0-5
Total area = 42
The distance travelled is approximately 42m.
This acceleration-time graph shows a journey of constant acceleration of 3m/s2 for 10 seconds.
In 10 seconds the velocity has increased by 30m/s this is represented by the area under the graph.
3 .
£ 2.§> o>
5time(s)
10
The area under a acceleration-time graph is a measure of velocity.
The velocity-time graphs below each show an increase in velocity of 30m/s during a 10 second interval.
246time(s)
246time(s)
They all show acceleration of 3m/s2 .
Any one of them could represent the same journey as the velocity-time graph above.
1. Sketch at least two possible velocity-time graphs which correspond to these two acceleration-time graphs.
6 "
03J5 2 .
10 15 20 time(s)
2'
-4
10 15 20 time(s)
2. This velocity-time graph represents a journey.
1/T 30
0-20
O
% 10/ \\0 5 10 15 20
time(s)
Which of these statements are true?
a) The initial velocity is 10cm/s.
b) The total distance covered is 225cm.
c) The acceleration for 0 < t < 5 is 2cm/s2 .
d) The acceleration for 10 < t < 20 is 1 cm/s2 .
3. In this velocity-time graph
20
| 15 ^>> 10 'o o 0) c:
0 1234 5 time(s)
a) Describe the acceleration during the first 2 seconds of the journey.
b) What is happening from t = 2 to t = 3?
c) What is the acceleration and velocity at time t = 4?
d) Calculate the distance travelled between t = 3 and t = 5.
4. The acceleration, acm/s2 , of an object is represented by the equation a = -t 2 + 4t + 6, where t is time in seconds.
a) Sketch a graph of the acceleration of the object from t = 0 to t = 4.
b) Use the trapezium method to estimate the velocity of the object at t = 1 , t = 2, t = 3, and t = 4.
c) Use these values of velocity to sketch a velocity-time graph of the object for the first 4 seconds.
d) Estimate the total distance travelled by the object.
© RBKC SMILE 1995.
Smile Worksheet 1570
Pounds and PenceHere are four ways of showing £6.50
——————1 ri1 i I
i £6.50 ir
650p 6.5
Actual money ... amount written ... amount written ... amount shown in pounds in pence on a calculator
1. Cut out the pieces below and group them into equal sets of money. _•-£?T T r
0.7
r
"
"
T
£7.00
T
70p
700p~l
0.07~l
£0.70
£0.07
r£50
7p
7.
7000p
r
r£70
2, Cut out the pieces below and group them into equal sets of money.r ______T
£63
r
r
T
6.03
63p
[®f T
0.63"1
603p
£10
~I
£6.03
r
If£0.63
6-3
630p
6300p
r
£6.30
r
63-© RBKC SMILE 2001
t?3 J° %OTZ
.3P%I
ooi3 p
25%
of £
125
% o
f £4
75%
of £
250
% o
f £10
60p
£2.0
0
£1.0
0lO
p
£6.0
050p
75p
25p
£25.
00£3.00
These cards and those from 1572A should be cut out and put in envelope 1572.
r '25p
75p
r
3P
£25.00
£1.50
£4.00
40p
50p
20p
60p
£1.00 £5.00
r
£3.00 £6.00
lp
£10.00
£8.00
£2.00
lOp
£15.00
L _ _
These cards and those from 1572B should be cut out and put in envelope 1572. <?
\
l%of£3
75% of £4
r75% of £1
l%of£l
10%of£l
100% of £15 l%of£7
100% of £6 10% of £2
50% of £20 10% of £4
25% of £2 25% of £4
50% of £4 25%of£l '
25% of £100 10% of £6 50% of £10
100% of £4 50% of £16 75% of £2
Smile 1589
You will need a calculator
Square Roots InvestigationChoose a number. Keep on finding sqi with a calculator.
e.g. 5.6
2.3664319
1.583211
1.2402907
Investigate what happens when you start with different numbers.
Try x
x etc.
Write about any patterns you see in your answers.
©RBKC SMILE 1998.
Smile 1591
You will need a set of dominoes
Domino Sums
1. Make as many Domino Sums as you can with a complete set of dominoes. You may only use each domino once.
2. How many dominoes did you have left over?
3. Try this game again. See if you can finish with fewer dominoes this time.
Two Cuts InvestigationYou will need colour pencils.
A square
Smile Worksheet 1592
with two cuts make:
2 quadrilaterals and 2 triangles
1 hexagon and2 right-angled triangles
1. Colour the triangles in green.
2. Colour the quadrilaterals in blue.
3. Colour the hexagon in red.
4. Using two cuts on a square, investigate what shapes you can make. Use different colours to highlight your shapes.
A mathematical dictionary will be useful.
©RBKC SMILE 2001
