Given a right triangle:
sin(A)=cos(B)$$sin(A) = cos(B)$$
cos(A)=sin(B)$$cos(A) = sin(B)$$
tan(A)=tan(B)1$$tan(A) = \frac{1}{tan(B)}$$
30° - 60° - 90° Trigonometric Ratios:
sin(30°)=21$$sin(30°) = \frac{1}{2}$$
cos(30°)=23$$cos(30°) = \frac{\sqrt{3}}{2}$$
tan(30°)=33$$tan(30°) = \frac{\sqrt{3}}{3}$$
45° - 45° - 90° Trigonometric Ratios:
sin(45°)=22$$sin(45°) = \frac{\sqrt{2}}{2}$$
cos(45°)=22$$cos(45°) = \frac{\sqrt{2}}{2}$$
tan(45°)=1$$tan(45°) = 1$$
Cone
Volume: V=31πr2h$$V = \frac{1}{3}πr^2h$$
Surface Area: S.A.=πrl+πr2$$S.A. = πrl + πr^2$$, where l$$l$$ is the slant height.
Rectangular Prism
Volume: V=lwh$$V = lwh$$
Surface Area: S.A.=2lw+2wh+2lh$$S.A. = 2lw + 2wh + 2lh$$
Regular Pyramid
Volume: V=31lwh$$V = \frac{1}{3}lwh$$
Cylinder
Volume: V=πr2h$$V = πr^2h$$
Surface Area: S.A.=2πr2+2πrh$$S.A. = 2πr^2 + 2πrh$$
Cavalieri's Principle: If two solids have congruent altitudes and every plane parallel to the bases results in cross-sections of equal area, then the solids have equal volumes.
Population Density: PopulationDensity=AreaPopulation$$Population Density = \frac{Population}{Area}$$
Rectangle → Cylinder
Triangle → Cone
Semi-Circle → Sphere
Secant and Tangent
2 Tangents
2 Secants
Measure of angle formed =21$$= \frac{1}{2}$$(larger arc – smaller arc)
Tangent and Secant: A2=B(B+C)$$A^2 = B(B + C)$$
2 Secants: A(A+B)=C(C+D)$$A(A + B) = C(C + D)$$
2 Chords: (A)(B)=(C)(D)$$(A)(B) = (C)(D)$$
2 Tangents Formulas
Area of a Sector: A=(360θ°)πr2$$A = (\frac{θ}{360}°)πr^2$$, where θ is the central angle in degrees.
Arc Length: ArcLength=2πr(360θ°)$$Arc Length= 2πr(\frac{θ}{360}°)$$, where θ is the central angle in degrees.
Equation of circle with center at origin: x2+y2=r2$$x^2 + y^2 = r^2$$
Equation of a circle with center (h,k)$$(h, k)$$: (x−h)2+(y−k)2=r2$$(x − h)^2 + (y − k)^2 = r^2$$
Example: Given (x+5)2+(y−7)2=81$$(x + 5)^2 + (y − 7)^2 = 81$$, Center: (−5,7)$$(-5, 7)$$; Radius= 81=9$$\sqrt{81} = 9$$
Inscribed Angle: ∡BCD=2∡BAC$$\measuredangle BCD = 2\measuredangle BAC$$
Angle Inscribed in a Semi-Circle: ∡ABD=90°$$\measuredangle ABD = 90°$$
Inscribed Quadrilateral Theorem: Opposite angles of an inscribed quadrilateral are supplementary.
Cylinder
Horizontal Plane: Circle
Vertical Plane: Rectangle
Cone
Horizontal Plane: Circle
Vertical Plane: Triangle
Rectangular Prism
Horizontal Plane: Rectangle
Vertical Plane: Rectangle
Triangular Prism
Horizontal Plane: Triangle
Vertical Plane: Rectangle
Sphere
Horizontal Plane: Circle
Vertical Plane: Circle
Triangle Midsegment Theorem: If AB=DB$$AB = DB$$ and AE=EC$$AE = EC$$, then DE∣∣BC$$DE || BC$$ and DE=21BC$$DE = \frac{1}{2} BC$$.
Triangle Inequality Theorem:
a+b>c$$a + b > c$$
a+c>b$$a + c > b$$
b+c>a$$b + c > a$$
Geometry Brain Dump – Second Semester
Given a right triangle:
sin(A)=cos(B)
cos(A)=sin(B)
tan(A)=tan(B)1
30° - 60° - 90° Trigonometric Ratios:
sin(30°)=21
cos(30°)=23
tan(30°)=33
45° - 45° - 90° Trigonometric Ratios:
sin(45°)=22
cos(45°)=22
tan(45°)=1
Cone
Volume: V=31πr2h
Surface Area: S.A.=πrl+πr2, where l is the slant height.
Rectangular Prism
Volume: V=lwh
Surface Area: S.A.=2lw+2wh+2lh
Regular Pyramid
Volume: V=31lwh
Cylinder
Volume: V=πr2h
Surface Area: S.A.=2πr2+2πrh
Cavalieri's Principle: If two solids have congruent altitudes and every plane parallel to the bases results in cross-sections of equal area, then the solids have equal volumes.
Population Density: PopulationDensity=AreaPopulation
Rectangle → Cylinder
Triangle → Cone
Semi-Circle → Sphere
Secant and Tangent
2 Tangents
2 Secants
Measure of angle formed =21(larger arc – smaller arc)
Tangent and Secant: A2=B(B+C)
2 Secants: A(A+B)=C(C+D)
2 Chords: (A)(B)=(C)(D)
2 Tangents Formulas
Area of a Sector: A=(360θ°)πr2, where θ is the central angle in degrees.
Arc Length: ArcLength=2πr(360θ°), where θ is the central angle in degrees.
Equation of circle with center at origin: x2+y2=r2
Equation of a circle with center (h,k): (x−h)2+(y−k)2=r2
Example: Given (x+5)2+(y−7)2=81, Center: (−5,7); Radius= 81=9
Inscribed Angle: ∡BCD=2∡BAC
Angle Inscribed in a Semi-Circle: ∡ABD=90°
Inscribed Quadrilateral Theorem: Opposite angles of an inscribed quadrilateral are supplementary.
Cylinder
Horizontal Plane: Circle
Vertical Plane: Rectangle
Cone
Horizontal Plane: Circle
Vertical Plane: Triangle
Rectangular Prism
Horizontal Plane: Rectangle
Vertical Plane: Rectangle
Triangular Prism
Horizontal Plane: Triangle
Vertical Plane: Rectangle
Sphere
Horizontal Plane: Circle
Vertical Plane: Circle
Triangle Midsegment Theorem: If AB=DB and AE=EC, then DE∣∣BC and DE=21BC.
Triangle Inequality Theorem:
a+b>c
a+c>b
b+c>a