BTGmoderatorDC wrote:In triangle ABC, angle ABC is 105 degrees. Is the area of triangle ABC less than 5?
(1) Segment AB = 2√2
(2) Angle BCA is 30 degrees.
OA C
Source: Veritas Prep
None of the statements alone is sufficient to answer the question, so we combine them.
See the image with construction.
Drop a perpendicular from B to AC; it meets at point D. /_BDC = /_BDA = 90º; /_CBD = 60º; /_BAD = 45º
We have to get the area of ∆ABC = 1/2 * BD * AC = BD(AD + DC)/2
1. Given ∆ABD, a 45-45-90 triangle and AB = 2√2, we have BD = AD = (2√2)/√2 = 2
2. Given ∆BDC, a 30-60-90 triangle and BD = 2, we have CD = 2√3
Thus, area of ∆ABC = BD(AD + DC)/2 = 2(2 + 2√3)/2 = 2 + 2√3 = A unique value
Though the value of 2 + 2√3 can be calculated, it is not required to be computed. There can be three scenarios.
1. 2 + 2√3 = 5. In this case, the answer is No.
2. 2 + 2√3 > 5. In this case, the answer is No.
3. 2 + 2√3 < 5. In this case, the answer is Yes.
Note that since 2 + 2√3 is a unique value, only one of the three scenarios would occur and we would get a unique answer. Sufficient.
The correct answer:
C
Hope this helps!
-Jay
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