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to find the area we need to know about base and height.it is not changed in this question so it will be same.
therefore we need to find whose base is greater.
option a gives me that.
so A
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Bunuel
In a triangle ABC point D lies on segment AC. Is the area of triangle ABD greater than the area of triangle BDC?

(1) DC > AD
(2) BC > AB

Are You Up For the Challenge: 700 Level Questions

Solution


Step 1: Analyse Question Stem


    • ABC is a triangle and D lies on side AC.
    • We need to find if area of triangle ABD is greater than the area of triangle BDC.
    • Based on the above information let’s draw the below diagram:
      o In this diagram, BE is the perpendicular drawn from B to AC

    • Now, area of triangle \(ABD = \frac{1}{2}* base* height = \frac{1}{2}*AD*BE \)
    • Also, area of triangle \(BDC = \frac{1}{2}*base*height = \frac{1}{2}*CD*BE\)
    • we need to find if area of triangle ABD \(>\) area of triangle BDC
      o Or, \(\frac{1}{2}*AD*BE > \frac{1}{2}*CD*BE ⟹ AD > CD\)
Therefore, we need to find out if \(AD > CD.\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1: DC > AD
    • According to this statement: \(AD < CD\)
    •Thus, we can infer that area of triangle ABD \(<\) area of triangle BDC
Hence, statement 1 is sufficient and we can eliminate answer Options B, C and E.
Statement 2: BC > AB
    • From this statement, we can’t find anything about AD and CD.
    • So, let’s consider the
      o area of triangle \(ABD = \frac{1}{2}* base* height = \frac{1}{2}*AB*GD\)
      o and, area of triangle \(BDC = \frac{1}{2}*base*height = \frac{1}{2}*BC*DH\)
    • We know that \(BC > AB\), however we don’t which one is greater between DG and DH.
    • Thus, even after considering this case too, we can’t decidedly say whether area of triangle ABD is greater than area of triangle BDC or not.

Hence, statement 2 is not sufficient.
Thus, the correct answer is Option A.
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