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Pythagorean Theorem and Space Figures

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Pythagorean Theorem and Space Figures ... = 10 The slant height of the pyramid is the perpendicular bisector of MK, so PSK is a right . (SK)2 + (PS)2 = (PK) ... – PowerPoint PPT presentation

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Title: Pythagorean Theorem and Space Figures


1
Pythagorean Theorem and Space Figures
Lesson 9.8
2
Rectangular Solid
  • Face
  • Edge AB is one of 12 edges
  • Diagonal HB is one of 4 diagonals

ABFE is one rectangular face out of the 6 faces
H
G
E
F
O
C
A
B
3
Regular Square Pyramid
  • Square base Bottom of the pyramid.
  • Vertex
  • Altitude
  • Slant height

Point where the edges of the triangles meet.
Distance from vertex to the base. It is
perpendicular to the center of the base.
Height of the triangles, perpendicular to the
base of the triangle.
4
Look at the right angles inside and out.
5
Look for the right angles here.
6
Find HB
Keep your answer in reduced radical form.
?ABD, 32 72 (BD)2 v58 BD
?HDB, 52 (v58)2 (HB)2 25 58 (HB) 2
v83 HB
7
  1. JK ¼ of JKMO ¼ (40) 10
  2. The slant height of the pyramid is the
    perpendicular bisector of MK, so PSK is a right
    ?.
  3. (SK)2 (PS)2 (PK)2
  4. 52 (PS)2 132
  5. PS 12

C. The altitude of a regular pyramid is
perpendicular to the base at its center. Thus,
RS ½ (JK) 5, and PRS is a right ?. (RS)2
(PR)2 (PS)2 52 (PR)2 122
PR v119
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