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Tuesday, 29 April 2014
FRACTIONS - Basic Operations
${2\over 7}+{3\over 7}={2+3\over 7}={5\over 7}$
SURFACE AREA AND VOLUME
A PRISM is a solid whose cross sectional area is the same all along its length, and all its sides are flat.
In a RIGHT PRISM the edges of the sides are perpendicular to the base. If the sides are not perpendicular to the base it is an OBLIQUE PRISM.
A COMPOSITE SOLID is one which is made up of two or more solids (e.e. a cube and a hemisphere stuck together).
The APEX of a pyramid or cone is the very top point (vertex).
A RIGHT PYRAMID is one which has its apex aligned directly above the center of the base. The surface area of a pyramid comprises the area of the sides plus the base (unless stated otherwise).
The surface area of a closed (right) cylinder is $SA=2\pi r^2+2\pi rh\quad\quad\quad \mbox{CLOSED CYLINDER}$ where $h=$height, $r=$radius of the base.
A RIGHT CONE is one which has its apex aligned directly above the center of the base. The surface area of a right cone is
$SA=\pi r l+\pi r^2\quad\quad\quad \mbox{RIGHT CONE}$ where $l=$
slant height, $r=$radius of the base.
A SPHERE is a surface whose shape is the outside of a ball.\\ The surface area of a sphere of radius $r$ is $SA=4\pi r^2\quad\quad\quad \mbox{SPHERE}$
The surface area of composite solids involving a sphere and other solids excludes common areas (unless stated otherwise)\\ The volume of a right prism is $V=Ah \quad\quad\quad \mbox{RIGHT PRISM}$ where $h=$height of the prism, and $A=$cross-sectional area.
The volume of a cylinder is $V= \pi r^2 h\quad\quad\quad \mbox{CYLINDER}$ (really same as $V=Ah$ where $A=\pi r^2$ here.)
The volume of a sphere of radius $r$ is $V={4\pi r^3 \over 3 }\quad\quad\quad \mbox{SPHERE}$
The volume of a pyramid is $V={1\over 3} Ah\quad\quad\quad \mbox{PYRAMID}$ where $h=$perpendicular height, $A=$area of base.
The volume of a cone is $V={1\over 3} \pi r^2 h\quad\quad\quad \mbox{CONE}$ (really same as ${1\over 3}Ah$ where $A=\pi r^2$ here.)
Volume of composite solids is the sum of its volume parts. Conversion between volume and capacity: $1 m^2=1 kL, 1 cm^2=1 mL,1 L =1000 mL$
AREAS OF SIMILAR FIGURES ARE RELATED:
If the matching sides of two similar figures are in the ratio $m:n$ then the areas are in the ratio $m^2:n^2$.
So if you double the sides of a figure its area becomes $4$ (since $2^2=4$) times bigger.
VOLUMES OF SIMILAR FIGURES ARE RELATED:
If the matching sides of two similar figures are in the ratio $m:n$ then (1) the ratio of their surface areas is $m^2:n^2$.
(2) the ratio of their volumes is $m^2:n^2$. So if you double the sides of a figure its volume becomes $8$ (since $2^3=8$) times bigger.
PYTHAGORAS' THEOREM applies to right angled triangles: $a^2+b^2=c^2$ where the longer side is $c$. This is useful when finding some surface areas and volumes for pyramids where right angled-triangles can occur. }
In a RIGHT PRISM the edges of the sides are perpendicular to the base. If the sides are not perpendicular to the base it is an OBLIQUE PRISM.
A COMPOSITE SOLID is one which is made up of two or more solids (e.e. a cube and a hemisphere stuck together).
The APEX of a pyramid or cone is the very top point (vertex).
A RIGHT PYRAMID is one which has its apex aligned directly above the center of the base. The surface area of a pyramid comprises the area of the sides plus the base (unless stated otherwise).
The surface area of a closed (right) cylinder is $SA=2\pi r^2+2\pi rh\quad\quad\quad \mbox{CLOSED CYLINDER}$ where $h=$height, $r=$radius of the base.
A RIGHT CONE is one which has its apex aligned directly above the center of the base. The surface area of a right cone is
$SA=\pi r l+\pi r^2\quad\quad\quad \mbox{RIGHT CONE}$ where $l=$
slant height, $r=$radius of the base.
A SPHERE is a surface whose shape is the outside of a ball.\\ The surface area of a sphere of radius $r$ is $SA=4\pi r^2\quad\quad\quad \mbox{SPHERE}$
The surface area of composite solids involving a sphere and other solids excludes common areas (unless stated otherwise)\\ The volume of a right prism is $V=Ah \quad\quad\quad \mbox{RIGHT PRISM}$ where $h=$height of the prism, and $A=$cross-sectional area.
The volume of a cylinder is $V= \pi r^2 h\quad\quad\quad \mbox{CYLINDER}$ (really same as $V=Ah$ where $A=\pi r^2$ here.)
The volume of a sphere of radius $r$ is $V={4\pi r^3 \over 3 }\quad\quad\quad \mbox{SPHERE}$
The volume of a pyramid is $V={1\over 3} Ah\quad\quad\quad \mbox{PYRAMID}$ where $h=$perpendicular height, $A=$area of base.
The volume of a cone is $V={1\over 3} \pi r^2 h\quad\quad\quad \mbox{CONE}$ (really same as ${1\over 3}Ah$ where $A=\pi r^2$ here.)
Volume of composite solids is the sum of its volume parts. Conversion between volume and capacity: $1 m^2=1 kL, 1 cm^2=1 mL,1 L =1000 mL$
AREAS OF SIMILAR FIGURES ARE RELATED:
If the matching sides of two similar figures are in the ratio $m:n$ then the areas are in the ratio $m^2:n^2$.
So if you double the sides of a figure its area becomes $4$ (since $2^2=4$) times bigger.
VOLUMES OF SIMILAR FIGURES ARE RELATED:
If the matching sides of two similar figures are in the ratio $m:n$ then (1) the ratio of their surface areas is $m^2:n^2$.
(2) the ratio of their volumes is $m^2:n^2$. So if you double the sides of a figure its volume becomes $8$ (since $2^3=8$) times bigger.
PYTHAGORAS' THEOREM applies to right angled triangles: $a^2+b^2=c^2$ where the longer side is $c$. This is useful when finding some surface areas and volumes for pyramids where right angled-triangles can occur. }
Labels:
Measurement,
Prism,
Pythagoras,
Similar Figures,
Volume & Surface Area
COORDINATE GEOMETRY
Consider the two points $$A(x_1, y_1) \quad\mbox{and}\quad B(x_2, y_2)$$
The distance between the points $A$ and $B$ is $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\quad\quad\quad \mbox{DISTANCE}$$
The coordinates of the midpoint of the interval joining the points $A$ and $B$ are $$\left( {x_1+x_2 \over 2 }, {y_1+y_2 \over 2}\right)\quad\quad\quad \mbox{MIDPOINT}$$
The slope or gradient of the line through the points $A$ and $B$ is $$m={y_2-y_1 \over x_2-x_1 }={ \mbox{rise}\over \mbox{run}}\quad\quad\quad \mbox{GRADIENT}$$
The GRADIENT-INTERCEPT form of the equation of a line is $$y=mx+b$$ where $m$ is the gradient or slope of the line, and $b$ is the $y$-intercept.
The POINT-GRADIENT form of a line is $$y-y_1=m(x-x_1)$$ where $m$ is the gradient of the line, and $(x_1,y_1)$ is ANY point on the line.
The GENERAL FORM OF A LINE is $$ax+by+c=0$$ where $a,b,c$ are constants.The slopes or gradients of parallel lines are equal, that is, $m_1=m_2$.
The slopes of perpendicular lines are negative reciprocals of each other. This can be written as $$m_2=-{1\over m_1}\quad\quad\mbox{or}\quad\quad m_1\times m_2=-1$$
The $x$-intercept is found by putting $y=0$ and solving for $x$ in the equation of the line. The $y$-intercept is found by putting $x=0$ and solving for $y$ in the equation of the line.
VERTICAL LINES have the form, $$x=k$$ where $k$ is a constant (number).
HORIZONTAL LINES have the form, $$y=k$$ where $k$ is a constant (number). }
The distance between the points $A$ and $B$ is $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\quad\quad\quad \mbox{DISTANCE}$$
The coordinates of the midpoint of the interval joining the points $A$ and $B$ are $$\left( {x_1+x_2 \over 2 }, {y_1+y_2 \over 2}\right)\quad\quad\quad \mbox{MIDPOINT}$$
The slope or gradient of the line through the points $A$ and $B$ is $$m={y_2-y_1 \over x_2-x_1 }={ \mbox{rise}\over \mbox{run}}\quad\quad\quad \mbox{GRADIENT}$$
The GRADIENT-INTERCEPT form of the equation of a line is $$y=mx+b$$ where $m$ is the gradient or slope of the line, and $b$ is the $y$-intercept.
The POINT-GRADIENT form of a line is $$y-y_1=m(x-x_1)$$ where $m$ is the gradient of the line, and $(x_1,y_1)$ is ANY point on the line.
The GENERAL FORM OF A LINE is $$ax+by+c=0$$ where $a,b,c$ are constants.The slopes or gradients of parallel lines are equal, that is, $m_1=m_2$.
The slopes of perpendicular lines are negative reciprocals of each other. This can be written as $$m_2=-{1\over m_1}\quad\quad\mbox{or}\quad\quad m_1\times m_2=-1$$
The $x$-intercept is found by putting $y=0$ and solving for $x$ in the equation of the line. The $y$-intercept is found by putting $x=0$ and solving for $y$ in the equation of the line.
VERTICAL LINES have the form, $$x=k$$ where $k$ is a constant (number).
HORIZONTAL LINES have the form, $$y=k$$ where $k$ is a constant (number). }
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