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As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t

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As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink] As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t   03 Feb 2020, 01:30
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[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 \(AC\)?

Attachment:
2.3ps.png
2.3ps.png [ 11.56 KiB | Viewed 3437 times ]


A. \(3 \)

B. \(4\)

C. \(5\)

D. \(6\)

E. \(7\)
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E
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As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink] As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t   Updated on: 27 Feb 2020, 08:50
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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.

Originally posted by DUtkarsh on 03 Feb 2020, 10:46.
Last edited by DUtkarsh on 27 Feb 2020, 08:50, edited 1 time in total.
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink] As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t   05 Feb 2020, 03:06
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=>

Attachment:
2.3ps(a).png
2.3ps(a).png [ 11.59 KiB | Viewed 3220 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
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink] As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t   14 Feb 2020, 23:12
Where are the assumption provided that FC=EC=BC–BE=BC–BD, is it a rule .
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As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink] As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t   Updated on: 15 Feb 2020, 09:18
Hello Moderators,

Please confirm if my assumption and understanding to the solution provided by Math Revolution are correct.

Concept - Length of the two tangents, only two possible, drawn from the same point are equal.

Points from which the tangents are drawn are A,B,C

Tangents from A : - AD and AF ; AD = AF = 3

Tangents from B :- BD and BE ; BD = BE = 5

Tangents from C :- CE and CF; CE = CF = 4 (BC = 9 ; BE = 5; EC = FC = (9-5 = 4)

Total length of AC = AF+FC = 3+4 = 7


Thanks

Originally posted by adizephyr on 15 Feb 2020, 08:54.
Last edited by adizephyr on 15 Feb 2020, 09:18, edited 1 time in total.
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink] As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t   15 Feb 2020, 09:14
DUtkarsh wrote:
Considering the radius of Incircle as r. Know that DI, EI and FI are perpendiculars on their respective sides.
In triangles BDI and BEI,
BI2 + DI2 = BI2 = BE2 + EI2
BI2 + DI2 = BE2 + EI2
BI2 = BE2 (Since, DI = EI = r)
Hence, BI = 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.


DUtkarsh

can u please elaborate above highlighted portion....
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink] As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t   27 Feb 2020, 08:54
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Rohit2015 wrote:
DUtkarsh wrote:
Considering the radius of Incircle as r. Know that DI, EI and FI are perpendiculars on their respective sides.
In triangles BDI and BEI,
BI2 + DI2 = BI2 = BE2 + EI2
BI2 + DI2 = BE2 + EI2
BI2 = BE2 (Since, DI = EI = r)
Hence, BI = 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.


DUtkarsh

can u please elaborate above highlighted portion....


Rohit2015 - Thanks for pointing out the typo, it'd be BD and not BI in the first half of the equation. I hope it is clear now, and I have also updated the aforementioned post.
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink] As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t   04 Mar 2020, 10:35
MathRevolution wrote:
=>

Attachment:
2.3ps(a).png


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


Hello MathRevolution,
could you please explain why AD=AF and FC=BC?
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink] As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t   14 May 2023, 00:10
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
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