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03 Dec 2020, 12:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
88% (01:34) correct
12%
(02:01)
wrong
based on 34
sessions
History
Date
Time
Result
Not Attempted Yet
Find the length of DB in the figure above.
(A) 8
(B) 10
(C) 12
(D) 15
(E) 20
Attachment:
2020-12-03_22-56-13.png [ 84.62 KiB | Viewed 4176 times ]
03 Dec 2020, 12:09
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2 right triangles. X should be 12 IMO
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ananya3
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Schools: Fuqua '24 Haas '24 ISB '23 Johnson '24 Kelley '24 Kenan-Flagler '24 McDonough '24 NUS Ross '24 Tepper '24 Tuck '24 Rotman '24 (A)
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03 Dec 2020, 19:46
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IMO C
ACB and ACD are right angled triangles right angles at C
from the Pythagorean triplets (8,15,17) => CD = 8
for ACB the Pythagorean triplet (3,4,5) is multiplied by 5
=> we have AB = 25, AC = 15
=> missing side = 20
x = 20-8 = 12
OR
=> for ACD, CD =\sqrt{(\(17^2\)-\(15^2\))}
=> CD = 8
=> for ACB, CB = \sqrt{\(25^2\)-\(15^2\)}
=> CB = 20
x = 20-8 = 12
ACB and ACD are right angled triangles right angles at C
from the Pythagorean triplets (8,15,17) => CD = 8
for ACB the Pythagorean triplet (3,4,5) is multiplied by 5
=> we have AB = 25, AC = 15
=> missing side = 20
x = 20-8 = 12
OR
=> for ACD, CD =\sqrt{(\(17^2\)-\(15^2\))}
=> CD = 8
=> for ACB, CB = \sqrt{\(25^2\)-\(15^2\)}
=> CB = 20
x = 20-8 = 12
