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03 Feb 2020, 01:30
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E
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:
71% (02:16) correct
29%
(02:31)
wrong
based on 129
sessions
History
Date
Time
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Not Attempted Yet
[GMAT math practice question]
As the figure shows \(AD = 3, BD = 5\) and \(BC = 9\). Moreover, point \(I\) is the incenter of triangle \(ABC\). What is the length of \(AC\)?
2.3ps.png [ 11.56 KiB | Viewed 3810 times ]
A. \(3 \)
B. \(4\)
C. \(5\)
D. \(6\)
E. \(7\)
As the figure shows \(AD = 3, BD = 5\) and \(BC = 9\). Moreover, point \(I\) is the incenter of triangle \(ABC\). What is the length of \(AC\)?
Attachment:
2.3ps.png [ 11.56 KiB | Viewed 3810 times ]
A. \(3 \)
B. \(4\)
C. \(5\)
D. \(6\)
E. \(7\)
DUtkarsh
Current Student
Joined: 25 May 2013
Last visit: 06 Feb 2022
Posts: 22
Given Kudos: 78
Location: India
Concentration: Finance, Strategy
Schools: ISB'22 (D) Marshall '24 (S) Kenan-Flagler '24 (S) Foster '24 (WL) Owen '24 (A) Imperial (A) NUS (A) WashU Olin '24 (WL) Jones (WD)
GMAT 1: 710 Q49 V38
GPA: 2.8
WE:Account Management (Consulting)
Products:
Schools: ISB'22 (D) Marshall '24 (S) Kenan-Flagler '24 (S) Foster '24 (WL) Owen '24 (A) Imperial (A) NUS (A) WashU Olin '24 (WL) Jones (WD)
GMAT 1: 710 Q49 V38
Posts: 22
Considering the radius of Incircle as r. Know that DI, EI and FI are perpendiculars on their respective sides.
In triangles BDI and BEI,
BD^2 + DI^2 = BI^2 = BE^2 + EI^2
BD^2 + DI^2 = BE^2 + EI^2
BD^2 = BE^2 (Since, DI = EI = r)
Hence, BD = BE = 5
Therefore, EC = 9 – 5 = 4
Similarly, using triangles ADI and AFI
AD = AF = 3
And, using triangles CEI and CFI
EC = CF = 4
Solution = AF + CF = 3 + 4 = 7 i.e. Option E.
In triangles BDI and BEI,
BD^2 + DI^2 = BI^2 = BE^2 + EI^2
BD^2 + DI^2 = BE^2 + EI^2
BD^2 = BE^2 (Since, DI = EI = r)
Hence, BD = BE = 5
Therefore, EC = 9 – 5 = 4
Similarly, using triangles ADI and AFI
AD = AF = 3
And, using triangles CEI and CFI
EC = CF = 4
Solution = AF + CF = 3 + 4 = 7 i.e. Option E.
05 Feb 2020, 03:06
Kudos
Bookmarks
=>
2.3ps(a).png [ 11.59 KiB | Viewed 3593 times ]
Since \(AD = AF,\) we have \(AF = 3.\)
Since \(BC = 9\) and \(BD = 5,\) we have \(FC = EC = BC – BE = BC – BD = 9 – 5 = 4.\)
Then we have \(AC = AF + FC = 3 + 4 = 7.\)
Therefore, E is the answer.
Answer: E
Attachment:
2.3ps(a).png [ 11.59 KiB | Viewed 3593 times ]
Since \(AD = AF,\) we have \(AF = 3.\)
Since \(BC = 9\) and \(BD = 5,\) we have \(FC = EC = BC – BE = BC – BD = 9 – 5 = 4.\)
Then we have \(AC = AF + FC = 3 + 4 = 7.\)
Therefore, E is the answer.
Answer: E
