Answer:
PQ = √84 = 2√21 in ≈ 9.165 in
Explanation:
The base edges AB = 10 in, AC = 17 in, and BC = 21 in
Q ∈ BC and AQ is the altitude of the base.
Let BQ = x in ⇒ CQ = (21 - x) in
ΔAQB a right triangle at Q
∴ AQ² = AB² - BQ² = 10² - x² ⇒(1)
ΔAQC a right triangle at Q
∴ AQ² = AC² - CQ² = 17² - (21-x)² ⇒(2)
Equating (1) and (2)
∴ 10² - x² = 17² - (21-x)²
∴ 10² - x² = 17² - (21² - 42x + x²)
∴ 10² - x² = 17² - 21² + 42x - x²
∴ 10² - 17² + 21² = 42x
∴ 42x = 252
∴ x = 252/42 = 6
Substitute at (1)
∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²
∴ AQ = 8 in
∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base
∴ PA⊥AQ
∴ ΔPAQ is a right triangle at A,
PA = 2sqrt5 in and AQ = 8 in
∴ PQ (hypotenuse) = √(PA² + AQ²)
∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in