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05 Jun 2019, 02:40
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C
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Difficulty:
15%
(low)
Question Stats:
85% (02:03) correct
15%
(01:36)
wrong
based on 61
sessions
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Points A, B, C, and D are on the perimeter of a triangular garden. The length of AB is 3 feet, the length of BC is 5 feet, and the length of AD is 4 feet. How many feet greater than the distance from point D to point B is the distance from point D to point C?.
A. \(\sqrt{41}-5\)
B. \(2\)
C. \(4\sqrt{5}-5\)
D. \(5\)
E. \(4\sqrt{5}-\sqrt{7}\)
Attachment:
Triangle.JPG [ 4.93 KiB | Viewed 2679 times ]
05 Jun 2019, 02:54
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Option C)
From Question We need find the value of (CD-BD)
Since, ABD is right triangle and given AB = 3, AD = 4, Hence with right triangle property -> BD = 5.
Also, ADC is right triangle and given AC = 8, AD = 4, Similarily with right traingke property -> CD = √80 = 4√5
CD-BD = 4√5−5
From Question We need find the value of (CD-BD)
Since, ABD is right triangle and given AB = 3, AD = 4, Hence with right triangle property -> BD = 5.
Also, ADC is right triangle and given AC = 8, AD = 4, Similarily with right traingke property -> CD = √80 = 4√5
CD-BD = 4√5−5
Archit3110
Major Poster
05 Jun 2019, 03:28
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DC= 64+16 ; √80 ; 4√5
and DB= 5
IMO C: \(4\sqrt{5}-5\)
and DB= 5
IMO C: \(4\sqrt{5}-5\)
Bunuel
