Geometry Brain Dump – Second Semester

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Flashcards for Geometry Brain Dump - Second Semester

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25 Terms

1
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Right Triangles and Trigonometry Relationships

𝑠𝑖𝑛 (𝐴) =𝑐𝑜𝑠 (𝐵) , 𝑐𝑜𝑠 (𝐴) =𝑠𝑖𝑛 (𝐵) , 𝑡𝑎𝑛 (𝐴) = 1 𝑡𝑎𝑛 (𝐵)

2
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Cavalieri's Principle for Three-Dimensional Solids

If every plane parallel to the two bases of two solids results in cross sections of equal area and the two solids have congruent altitudes, then the solids have equal volumes.

3
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Population Density Formula

D=Population/Area

4
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Shape resulting from 360° Rotation of a Rectangle

Cylinder

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Shape resulting from 360° Rotation of a Triangle

Cone

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Shape resulting from 360° Rotation of a Semi-Circle

Sphere

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Angle formed by 2 Secants/Tangents

Measure of angle formed = 1/2(larger arc – smaller arc)

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Tangent and Secant Formula

𝐴²= 𝐵(𝐵 + 𝐶)

<p>𝐴²= 𝐵(𝐵 + 𝐶)</p>
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2 Secants Formula

𝐴(𝐴 + 𝐵) = 𝐶(𝐶 + 𝐷)

<p>𝐴(𝐴 + 𝐵) = 𝐶(𝐶 + 𝐷)</p>
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2 Chords Formula

(𝐴)(𝐵) = (𝐶)(𝐷)

<p>(𝐴)(𝐵) = (𝐶)(𝐷)</p>
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2 Tangents

A=B

<p>A=B</p>
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Area of a Sector

𝐴 = (θ/360)𝜋𝑟², where θ represents the central angle in degrees

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Arc Length

Arc Length= 2𝜋𝑟(θ/360), where θ represents the central angle in degrees

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Equation of circle with center at origin

𝑥² + 𝑦² = 𝑟²

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Equation of a circle with center (ℎ, 𝑘)

(𝑥 − ℎ)² + (𝑦 − 𝑘)² = 𝑟²

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Inscribed Angle

∡𝐵𝐶𝐷 = 2∡𝐵𝐴𝐶

<p>∡𝐵𝐶𝐷 = 2∡𝐵𝐴𝐶</p>
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Angle Inscribed in a Semi-Circle

∡𝐴𝐵𝐷 = 90°

<p>∡𝐴𝐵𝐷 = 90°</p>
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Inscribed Quadrilateral Theorem

The opposite angles of an inscribed quadrilateral to a circle are supplementary.

<p>The opposite angles of an inscribed quadrilateral to a circle are supplementary.</p>
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Cross Section of Cylinder (Horizontal, Vertical Plane)

Circle, Rectangle

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Cross Section of Cone (Horizontal, Vertical Plane)

Circle, Triangle

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Cross Section of Rectangular Prism (Horizontal, Vertical Plane)

Rectangle, Rectangle

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Cross Section of Triangular Prism (Horizontal, Vertical Plane)

Triangle, Rectangle

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Cross Section of Sphere (Horizontal, Vertical Plane)

Circle, Circle

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Triangle Midsegment Theorem

If 𝐴𝐵 = 𝐷𝐵 and 𝐴𝐸 = 𝐸𝐶, then 𝐷𝐸 ∥ 𝐵𝐶 and 𝐷𝐸 = 1/2𝐵𝐶.

<p>If 𝐴𝐵 = 𝐷𝐵 and 𝐴𝐸 = 𝐸𝐶, then 𝐷𝐸 ∥ 𝐵𝐶 and 𝐷𝐸 = 1/2𝐵𝐶.</p>
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Triangle Inequality Theorem

𝑎 + 𝑏 > 𝑐 , 𝑎 + 𝑐 > 𝑏 , 𝑏 + 𝑐 > 𝑎

<p>𝑎 + 𝑏 &gt; 𝑐 , 𝑎 + 𝑐 &gt; 𝑏  , 𝑏 + 𝑐 &gt; 𝑎</p>