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aivan3 [116]
3 years ago
7

Do you know how do this because I need help?

Mathematics
1 answer:
Kaylis [27]3 years ago
5 0

I am guessing.

I think it is the answer you have in, I am not in that grade level yet so, I may be incorrect. Take care!

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The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC
DIA [1.3K]

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Step-by-step explanation:

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Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

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ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

∴  10² - x² = 17² - (21-x)²

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∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

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∴ ΔPAQ is a right triangle at A,

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∴ PQ (hypotenuse) = √(PA² + AQ²)

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3 years ago
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