Pythagorean Theorem Pythagorean Theorem and Space Figuresand Space Figures

Lesson 9.8

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

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.

Look at the right angles Look at the right angles inside and out.inside and out.

Look for the right angles here.Look for the right angles here.

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

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