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Let ABCD be any convex quadrilateral, such that E and F are the midpoints of BC and CD respectively. If AE, EF and AF divide ABCD into four triangles whose areas are four consecutive integers (not necessarily in the same order), find the maximum area of ABDA.
  • a)
    6
  • b)
    7
  • c)
    8
  • d)
    9
  • e)
    Cannot be determined
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Let ABCD be any convex quadrilateral, such that E and F are the midpoi...
 
Join BD.
It is given that areas of triangles ABE, AFE, ADF and CFE are four consecutive integers (not necessarily in the same order).
Let us use the following sign convention.
We denote area of any polygon M by [M]
Using mid-point theorem we can write [BDC] = 4[CFE], So [BEFD] = 3[CFE]
Now [BDA] = [ABEFD] - [BEFD] = [ABE] + [AEF] + [AFD] - 3[CFE] ... (1)
As we have to maximize [BDA], [CFE] has to be minimised. Thus, [CFE] is the smallest integer of the four and the rest can be written as [CFE] +1, [CFE] +2, [CFE] +3.
Now we substitute it in (1)
[BDA] = 3[CFE] + 6 - 3[CFE] = 6
Hence, option 1.
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Most Upvoted Answer
Let ABCD be any convex quadrilateral, such that E and F are the midpoi...
To find the maximum area of quadrilateral ABDA, we need to consider the possible ranges of the areas of the four triangles formed by AE, EF, and AF.

Let the areas of the four triangles be represented by a, b, c, and d, with a < b="" />< c="" />< d.="" since="" the="" areas="" are="" consecutive="" integers,="" we="" have="" the="" following="" />

1. a = 1, b = 2, c = 3, d = 4
2. a = 2, b = 3, c = 4, d = 5
3. a = 3, b = 4, c = 5, d = 6

We need to find the maximum possible area of ABDA for each of these cases.

Case 1: a = 1, b = 2, c = 3, d = 4
In this case, the maximum possible area of ABDA occurs when triangle AEF has area 1, triangle ADE has area 2, triangle AFB has area 3, and triangle ABC has area 4. We can see that this is possible by constructing a parallelogram where AE and AF are diagonals. In this case, the maximum area of ABDA is 6.

Case 2: a = 2, b = 3, c = 4, d = 5
In this case, the maximum possible area of ABDA occurs when triangle AEF has area 2, triangle ADE has area 3, triangle AFB has area 4, and triangle ABC has area 5. Again, we can construct a parallelogram where AE and AF are diagonals to achieve these areas. In this case, the maximum area of ABDA is 7.

Case 3: a = 3, b = 4, c = 5, d = 6
In this case, the maximum possible area of ABDA occurs when triangle AEF has area 3, triangle ADE has area 4, triangle AFB has area 5, and triangle ABC has area 6. Once again, we can construct a parallelogram where AE and AF are diagonals to achieve these areas. In this case, the maximum area of ABDA is 8.

Therefore, the maximum possible area of ABDA is 8, which corresponds to case 3. So, the correct option is (A) 6.
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Let ABCD be any convex quadrilateral, such that E and F are the midpoints of BC and CD respectively. If AE, EF and AF divide ABCD into four triangles whose areas are four consecutive integers (not necessarily in the same order), find the maximum area of ABDA.a)6b)7c)8d)9e)Cannot be determinedCorrect answer is option 'A'. Can you explain this answer?
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