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ABCD is a trapezium and AB is parallel to CD. AB has a length, which is two times the length of CD. If the diagonals of the trapezium intersect at O, then the ratio of areas of ΔAOB and ΔCOD triangle is 
  • a)
    1: 4
  • b)
    4: 1
  • c)
    1: 2
  • d)
    2: 1 
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
ABCD is a trapezium and AB is parallel to CD. AB has a length, which i...
The triangles AOB and COD are similar. Hence the ratio of areas is equal to the ratio of square of the sides. Since AB is twice of CD, so ratio of areas of ΔAOB and ΔCOD is 4: 1. 
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Most Upvoted Answer
ABCD is a trapezium and AB is parallel to CD. AB has a length, which i...
To find the ratio of the areas of triangles AOB and COD, we need to find the ratio of their bases and heights.

Let the length of CD be x. Then the length of AB is 2x.

Since AB is parallel to CD, triangles AOB and COD are similar. This means that the ratio of their corresponding sides is equal.

Let the height of triangle AOB be h1 and the height of triangle COD be h2. Then we have:

h1/h2 = AB/CD = 2x/x = 2

So the ratio of the heights of the triangles is 2:1.

The ratio of the areas of the triangles is equal to the ratio of their bases multiplied by the ratio of their heights. Since the ratio of the heights is 2:1, the ratio of the areas is also 2:1.
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ABCD is a trapezium and AB is parallel to CD. AB has a length, which is two times the length of CD. If the diagonals of the trapezium intersect at O, then the ratio of areas of ΔAOB and ΔCOD triangle isa)1: 4b)4: 1c)1: 2d)2: 1Correct answer is option 'B'. Can you explain this answer?
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