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Alex_Xolod [135]
2 years ago
14

What is the surface area of the square pyramid?

Mathematics
1 answer:
e-lub [12.9K]2 years ago
7 0

Answer:

SA = 16 units^2

Step-by-step explanation:

B - base

H - height

SA - surface area

B = 2

H = 3

SA = (B × B) + 4((1/2)×(B × H))

SA = (2×2) + 4((1/2)×(2×3))

SA = <u>16 units^2</u>

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<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

3) \triangle ABO, \triangle CDO are right triangles (a triangle with a right angle is a right triangle)

4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

5) \triangle ABO \cong \triangle CDO (LL)

<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

2) \overline{CD} \cong \overline{CD} (reflexive property)

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<u>Question 8</u>

1) \overline{CD} \perp \overline{AB}, point D bisects \overline{AB} (given)

2) \angle CDA, \angle CDB are right angles (perpendicular lines form right angles)

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5) \overline{CD} \cong \overline{CD} (reflexive property)

6)  \triangle ADC \cong \triangle BDC (LL)

7) \angle ACD \cong \angle BCD (CPCTC)

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