1. Pythagorean Theorem and Space Figures Lesson 9.8

2. Rectangular Solid  Face  EdgeAB is one of 12 edges  DiagonalHB is one of 4 diagonals E HG A C B O F ABFE is one rectangular face out of the 6 faces

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. ΔABD, 3 2 + 7 2 = (BD) 2 √58 = BD ΔHDB, 5 2 + (√58) 2 = (HB) 2 25 + 58 = (HB) 2 √83 = HB

7. A.JK = ¼ of JKMO = ¼ (40) = 10 B.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. 5 2 + (PS) 2 = 13 2 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 5 2 + (PR) 2 = 12 2 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 5 2 + (PR) 2 = 12 2 PR = √119