# Unit 4: Measurement and Geometry ## Angle Theorems 1. ```Transversal Parallel Line Theorems``` (TPT) a. Alternate Angles are Equal ```(Z-Pattern)``` b. Corresponding Angles Equal ```(F-Pattern)``` c. Interior Angles add up to 180 ```(C-Pattern)``` - 2. ```Supplementary Angle Triangle``` (SAT) - When two angles add up to 180 degrees - 3. ```Opposite Angle Theorem (OAT)``` (OAT) - Two lines intersect, two angles form opposite. They have equal measures - 4. ```Complementary Angle Theorem``` (CAT) - The sum of two angles that add up to 90 degrees - 5. ```Angle Sum of a Triangle Theorem``` (ASTT) - The sum of the three interior angles of any triangle is 180 degrees - 6. ```Exterior Angle Theorem``` (EAT) - The measure of an exterior angle is equal to the sum of the measures of the opposite interior angles - 7. ``` Isosceles Triangle Theorem``` (ITT) - The base angles in any isosceles triangle are equal - 8. ```Sum of The Interior Angle of a Polygon``` - The sum of the interioir angles of any polygon is ```180(n-2)``` or ```180n - 360```, where ```n``` is the number of sides of the polygon - 9. ```Exterior Angles of a Convex Polygon``` - The sum of the exterior angle of any convex polygon is always ```360 degrees``` - ## Properties of Quadrilaterals - Determine the shape using the properties of it |Figure|Properties| |:-----|:---------| |Scalene Triangle|no sides equal|Length of line segment| |Isosceles Triangle| two sides equal|Length of line segment| |Equilateral Triangle|All sides equal|Length of line segment| |Right Angle Triangle|Two sides are perpendicular to each other| |Parallelogram|Opposite sides are parallel and have equal length. Additionally, the diagonals bisect each other| |Rectangle|Adjacent sides are perpendicular to each other. Furthermore, the diagonals bisect each other and are equal in length| |Square|All sides are equal in length. The adjacent sides and diagonals are perpendicular. The adjacent sides are equal in length, so as the diagonals| |Rhombus|Opposite sides are parallel and all sides are equal to each other, the diagonals are perpendicular| |Trapezoid|There is one pair of opposite sides and they are parallel and unequal in length| |Kite|The diagonals are perpendicular| ## 2D Geometry Equations |Shape|Formula|Picture| |:----|:------|:------| |Rectangle|```Area```: $`lw`$
```Perimeter```: $`2(l+w)`$|| |Triangle|```Area```: $`\frac{bh}{2}`$
```Perimeter```: $`a+b+c`$|| |Circle|```Area```: $`πr^2`$
```Circumference```: $`2πr`$ or $`πd`$|| |Trapezoid|```Area```: $` \frac{(a+b)h}{2}`$
```Perimeter```: $`a+b+c+d`$|| ## 3D Geometry Equations |3D Object|Formula|Picture| |:----|:------|:------| |Rectangular Prism|```Volume```: $`lwh`$
```SA```: $`2(lw+lh+wh)`$|| |Square Based Pyramid|```Volume```: $`\frac{1}{3} b^2 h`$
```SA```: $`2bs+b^2`$|| |Sphere|```Volume```: $`\frac{4}{3} πr^3`$
```SA```: $`4πr^2`$|| |Cone|```Volume```: $` \frac{1}{3} πr^2 h`$
```SA```: $`πrs+πr^2`$|| |Cylinder|```Volume```: $`πr^2h`$
```SA```: $`2πr^2+2πh`$|| |Triangular Prism|```Volume```: $`ah+bh+ch+bl`$
```SA```: $` \frac{1}{2} blh`$|| ## Optimization (For Maximimizing Area/Volume, or Minimizing Perimeter/Surface Area) ### 2D Objects |Shape|Maximum Area|Minimum Perimeter| |:----|:-----------|:----------------| |4-sided rectangle|A rectangle must be a square to maximaze the area for a given perimeter. The length is equal to the width
$`A = lw`$
$`A_{max} = (w)(w)`$
$`A_{max} = w^2`$|A rectangle must be a square to minimaze the perimeter for a given area. The length is equal to the width.
$`P = 2(l+w)`$
$`P_{min} = 2(w + w)`$
$`P_{min} = 2(2w)`$
$`P_{min} = 4w`$| |3-sided rectangle|$`l = 2w`$
$`A = lw`$
$`A_{max} = 2w(w)`$
$`A_{max} = 2w^2`$|$`l = 2w`$
$`P = l+2w`$
$`P_{min} = 2w+2w`$
$`P_{min} = 4w`$| ### 3D Objects |3D Object|Maximum Volumne|Minimum Surface Area| |:--------|:--------------|:-------------------| |Cylinder(closed-top)|The cylinder must be similar to a cube where $`h = 2r`$
$`V = πr^2h`$
$`V_{max} = πr^2(2r)`$
$`V_{max} = 2πr^3`$|The cylinder must be similar to a cube where $`h = 2r`$
$`SA = 2πr^2+2πrh`$
$`SA_{min} = 2πr^2+2πr(2r)`$
$`SA_{min} = 2πr^2+4πr^2`$
$`SA_{min} = 6πr^2`$| |Rectangular Prism(closed-top)|The prism must be a cube,
where $`l = w = h`$
$`V = lwh`$
$`V_{max} = (w)(w)(w)`$
$`V_{max} = w^3`$|The prism must be a cube,
where $`l = w = h`$
$`SA = 2lh+2lw+2wh`$
$`SA_{min} = 2w^2+2w^2+2w^2`$
$`SA_{min} = 6w^2`$| |Cylinder(open-top)|$`h = r`$
$`V = πr^2h`$
$`V_{max} = πr^2(r)`$
$`V_{max} = πr^3`$|$`h = r`$
$`SA = πr^2+2πrh`$
$`SA_{min} = πr^2+2πr(r)`$
$`SA_{min} = πr^2+2πr^2`$
$`SA_{min} = 3πr^2`$| |Square-Based Rectangular Prism(open-top)|$`h = \frac{w}{2}`$
$`V = lwh`$
$`V_{max} = (w)(w)(\frac{w}{2})`$
$`V_{max} = \frac{w^3}{2}`$|$`h = \frac{w}{2}`$
$`SA = w^2+4wh`$
$`SA_{min} = w^2+4w(\frac{w}{2})`$
$`SA_{min} = w^2+2w^2`$
$`SA_{min} = 3w^2`$| ## Labelling - Given any polygons, labelling the vertices must always: 1. use ```CAPITAL LETTERS``` 2. they have to be labeled in ```clockwise``` or ```counter-clockwise``` directions - For a triangle, the side lengths are labeled in ```LOWERCASE LETTERS``` associated to the opposite side of the vertex - ## Median - Each median divides the triangle into 2 smaller triangles of equal area - The centroid is exactly the way of each median from the vertex, or the way from the midpoint of the opposite side, or ```2:1``` ratio - The three medians divide the triangle into ```6``` smaller triangles of equal area and ```3 pairs``` of congruent triangles - ## Terms: - ```Altitude``` The height of a triangle, a line segment through a vertex and perpendicular to the opposite side - ```Orthocenter```: where all 3 altitudes of the triangle intersect - - ```Midpoint```: A point on a line where the length of either side of the point are equal - ```Median```: A line segment joining the vertex to the midpoint of the opposite side - ```Midsegment```: A line joining 2 midpoints of the 2 sides of a triangle - ```Centroid```: The intersection of the 3 medians of a triangle - ## Proportionality theorem: - The midsegment of a triangle is ```half``` the length of the opposite side and ```parallel``` to the opposite side - Three midsegment of a triangle divide ```4 congruent``` triangles with the same area - The Ratio of the outer triangle to the triangle created by the 3 midsegments is ```4 to 1``` - ## Tips - Make sure to know your optimization formualas - Read the word problems carefully, determine which formual to use - Never **ASSUME**, be sure to **CALCULATE** as most of the time the drawings are **NOT ACCURATE** - To find ```missing area```, take what you have, subtract what you don't want - Don't be afraid to draw lines to help you solve the problem