1. NOTES 9.8
Pythagorean Theorem and Space Figures

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 ΔABD, 32 + 72 = (BD)2 √58 = BD ΔHDB, 52 + (√58)2 = (HB)2
Keep your answer in reduced radical form.
ΔABD,
= (BD)2
√58 = BD
ΔHDB,
52 + (√58)2 = (HB)2
= (HB) 2
√83 = HB

7. JK = ¼ of JKMO = ¼ (40) = 10
The slant height of the pyramid is the perpendicular bisector of MK, so PSK is a right Δ.
(SK)2 + (PS)2 = (PK)2
52 + (PS)2 = 132
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