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25 Mar 2019, 06:49
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
75% (02:10) correct
25%
(02:10)
wrong
based on 108
sessions
History
Date
Time
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Not Attempted Yet
If the ratio of AB to DC is 3:1 and the ratio of AD to DC is 5:4 then the perimeter of the shaded region is
A) 18
B) 24
C) 30
D) 32
E) 36
Attachment:
AQyWxBA.png [ 25.13 KiB | Viewed 2974 times ]
Chethan92
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Concentration: Operations, General Management
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25 Mar 2019, 09:32
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AB:DC = 3:1
AB = 12. Then DC = 4
Also, AD:DC = 5:4. AD = 5
AC = 5+4 = 9
Hypotenuse, BD = \(\sqrt{144+25}\) = 13
Hypotenuse, BC = \(\sqrt{81+144}\) = 15
Perimeter of BDC = 15+4+13 = 32
D is the answer.
AB = 12. Then DC = 4
Also, AD:DC = 5:4. AD = 5
AC = 5+4 = 9
Hypotenuse, BD = \(\sqrt{144+25}\) = 13
Hypotenuse, BC = \(\sqrt{81+144}\) = 15
Perimeter of BDC = 15+4+13 = 32
D is the answer.
21 Aug 2020, 21:24
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Given that ratio of AB to DC is 3:1
Therefore \(\frac{12}{DC} = \frac{3}{1}\)
DC = 4
Now AD is to DC = 5:4
\(\frac{AD}{4} = \frac{5}{4}\)
AD = 5
AD + DC = 5 + 4 = 9
In Right \(\triangle\) ABD, by Pythagoras's Theorem
\(BD^2 = AB^2 + AD^2\)
\( BD^2 = 12^2 + 5^2\)
\(BD^2\) = 169
BD = 13
Similarly, In Right \(\triangle\) ABC, by Pythagoras's Theorem
\(BC^2 = AB^2 + AC^2\)
\( BC^2 = 12^2 + 9^2\)
\(BC^2\) = 225
BC = 15
Perimeter of \(\triangle\) BDC = BD + BC + DC = 13 + 15 + 4 = 32
Option D
Arun Kumar
Therefore \(\frac{12}{DC} = \frac{3}{1}\)
DC = 4
Now AD is to DC = 5:4
\(\frac{AD}{4} = \frac{5}{4}\)
AD = 5
AD + DC = 5 + 4 = 9
In Right \(\triangle\) ABD, by Pythagoras's Theorem
\(BD^2 = AB^2 + AD^2\)
\( BD^2 = 12^2 + 5^2\)
\(BD^2\) = 169
BD = 13
Similarly, In Right \(\triangle\) ABC, by Pythagoras's Theorem
\(BC^2 = AB^2 + AC^2\)
\( BC^2 = 12^2 + 9^2\)
\(BC^2\) = 225
BC = 15
Perimeter of \(\triangle\) BDC = BD + BC + DC = 13 + 15 + 4 = 32
Option D
Arun Kumar
