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505-555 Level|   Geometry|      
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A projectile is fired at a target at point C, which lies 100 feet from 19 Aug 2019, 11:00
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A
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15% (low)

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90% (02:29) correct 10% (03:02) wrong based on 39 sessions

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A projectile is fired at a target at point C, which lies 100 feet from the base of the platform from which the projectile was fired at point A. If AB = 90 feet, by what distance, in feet, did the projectile fall short of its target?

A. \(100 - 40 \sqrt{5}\)

B. \(100 - 40 \sqrt{3}\)

C. 20

D. 25

5. 40
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A
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A projectile is fired at a target at point C, which lies 100 feet from 19 Aug 2019, 11:51
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Using Phythagoras therom

Imo - A) 100-40√5

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A projectile is fired at a target at point C, which lies 100 feet from 19 Aug 2019, 22:07
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According to Pythagoras theorem,
90^2=10^2-X^2
X^2=8000
X=√8000=40√5
So, BC=100-40√5
Option A

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A projectile is fired at a target at point C, which lies 100 feet from 20 Aug 2019, 11:30
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SajjadAhmad
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A projectile is fired at a target at point C, which lies 100 feet from the base of the platform from which the projectile was fired at point A. If AB = 90 feet, by what distance, in feet, did the projectile fall short of its target?

A. \(100 - 40 \sqrt{5}\)

B. \(100 - 40 \sqrt{3}\)

C. 20

D. 25

5. 40

100 - (\sqrt{90^2-10^2}
= 100 -\sqrt{8000}
= 100-40\sqrt{5}
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A projectile is fired at a target at point C, which lies 100 feet from 25 Aug 2019, 18:33
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SajjadAhmad
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BHBH.jpg

A projectile is fired at a target at point C, which lies 100 feet from the base of the platform from which the projectile was fired at point A. If AB = 90 feet, by what distance, in feet, did the projectile fall short of its target?

A. \(100 - 40 \sqrt{5}\)

B. \(100 - 40 \sqrt{3}\)

C. 20

D. 25

5. 40

We can let x = the distance from point B to the base of the platform, so by the Pythagorean theorem, we have:

x^2 + 10^2 = 90^2

x^2 + 100 = 8100

x^2 = 8000

x = √8000 = √1600 x √5 = 40√5

Since 100 is the distance from point C to the base of the platform, the projectile fell short by 100 - 40√5 feet.

Answer: A
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