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In the diagram below, triangle ABC is right-angled at B. D and E are

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In the diagram below, triangle ABC is right-angled at B. D and E are  [#permalink]

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New post 17 Jul 2017, 23:32
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

64% (01:56) correct 36% (02:18) wrong based on 59 sessions

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In the diagram below, triangle ABC is right-angled at B. D and E are  [#permalink]

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New post 18 Jul 2017, 01:21
Bunuel wrote:
In the diagram below, triangle ABC is right-angled at B. D and E are points on AC and AB, respectively. If BC = 6, what is the area of triangle BPC?
Image

(1) BE = BC and BD bisects ∠ABC.
(2) AB = 8.


Attachment:
2017-07-18_1031.png


It should be A.

Statement 1: Sufficient

(1) BE = BC and BD bisects ∠ABC.

If BE = BC, then BP splits the triangle BEC in two equal halves. We know sides of the triangle BEC so we can calculate its area.

For BEC,
\(Area = \frac{1}{2}*base*height = \frac{1}{2}*6*6 = 18.\)

For BPC
\(Area = \frac{1}{2} * 18 = 9.\)

Statement 2: Insufficient

(2) AB = 8.

With this information, we can calculate area of triangle ABC but with no information on points D, E or P, we have no way of calculating area of the triangle.
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Re: In the diagram below, triangle ABC is right-angled at B. D and E are  [#permalink]

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New post 19 Jul 2017, 00:04
lets take the easier statement2 first
stat2: not suff,,,we can calculate the area of triangle ABC..nothing else..

stat1: traingle BEC is isosceles and the angle bisector is the median as well,,, hence PE = PC.
median divides the traingle into two equal are... hence suff

ans A
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Re: In the diagram below, triangle ABC is right-angled at B. D and E are  [#permalink]

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New post 01 Mar 2019, 11:10
Another way of looking at 1) From BE=BC and angle bisector BD we know the 2 divided triangles have an equal side and angle. Since they share side BP they also have another equal side (SAS congruent, so each is half the area of EBC).
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Re: In the diagram below, triangle ABC is right-angled at B. D and E are   [#permalink] 01 Mar 2019, 11:10
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