![Page 1: Pythagorean Theorem and Space Figures Lesson 9.8](https://reader036.vdocuments.us/reader036/viewer/2022072015/56649ecf5503460f94bdd730/html5/thumbnails/1.jpg)
Pythagorean Theorem Pythagorean Theorem and Space Figuresand Space Figures
Lesson 9.8
![Page 2: Pythagorean Theorem and Space Figures Lesson 9.8](https://reader036.vdocuments.us/reader036/viewer/2022072015/56649ecf5503460f94bdd730/html5/thumbnails/2.jpg)
Rectangular SolidRectangular Solid
FaceFace
EdgeEdge AB is one of 12 edgesAB is one of 12 edges
DiagonalDiagonal HB is one of 4 diagonalsHB is one of 4 diagonals
E
H G
A
C
B
O
F
ABFE is one rectangular face out of the 6 faces
![Page 3: Pythagorean Theorem and Space Figures Lesson 9.8](https://reader036.vdocuments.us/reader036/viewer/2022072015/56649ecf5503460f94bdd730/html5/thumbnails/3.jpg)
Regular Square Regular Square PyramidPyramid
Square baseSquare base Bottom of the pyramid.Bottom of the pyramid.
VertexVertex
AltitudeAltitude
Slant heightSlant 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.
![Page 4: Pythagorean Theorem and Space Figures Lesson 9.8](https://reader036.vdocuments.us/reader036/viewer/2022072015/56649ecf5503460f94bdd730/html5/thumbnails/4.jpg)
Look at the right angles Look at the right angles inside and out.inside and out.
![Page 5: Pythagorean Theorem and Space Figures Lesson 9.8](https://reader036.vdocuments.us/reader036/viewer/2022072015/56649ecf5503460f94bdd730/html5/thumbnails/5.jpg)
Look for the right angles here.Look for the right angles here.
![Page 6: Pythagorean Theorem and Space Figures Lesson 9.8](https://reader036.vdocuments.us/reader036/viewer/2022072015/56649ecf5503460f94bdd730/html5/thumbnails/6.jpg)
Find HBFind HB
Keep your answer in reduced radical form.
ΔABD, 32 + 72 = (BD)2
√58 = BD
ΔHDB, 52 + (√58)2 = (HB)2
25 + 58 = (HB) 2
√83 = HB
![Page 7: Pythagorean Theorem and Space Figures Lesson 9.8](https://reader036.vdocuments.us/reader036/viewer/2022072015/56649ecf5503460f94bdd730/html5/thumbnails/7.jpg)
A.JK = ¼ of JKMO = ¼ (40) = 10B.The slant height of the
pyramid is the perpendicular bisector of MK, so PSK is a right Δ.A. (SK)2 + (PS)2 = (PK)2
B. 52 + (PS)2 = 132
C. 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 Δ. (RS)2 + (PR)2 = (PS)2
52 + (PR)2 = 122
PR = √119
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 Δ. (RS)2 + (PR)2 = (PS)2
52 + (PR)2 = 122
PR = √119