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# Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD

**Solution:**

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Let's construct a diagram according to the given question.

In trapezium ABCD,

AB is parallel to CD and AB = 2 CD ---------- (1)

Diagonals AC and BD intersect at ‘O’

In ΔAOB and ΔCOD,

∠AOB = ∠COD (vertically opposite angles)

∠ABO = ∠CDO (alternate interior angles)

⇒ ΔAOB ~ ΔCOD (AA criterion)

⇒ Area of ΔAOB / Areaof ΔCOD = (AB)^{2 }/ (CD)^{2} [Theorem 6.6]

(2CD)^{2} / (CD)^{2} = 4CD^{2} / CD^{2} = 4 / 1 [From equation (1)]

Thus, Area of ΔAOB : Area of ΔCOD = 4:1

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**Video Solution:**

## Diagonals of a trapezium ABCD with AB || DC intersect each other at the point If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.4 Question 2

**Summary:**

Diagonals of a trapezium ABCD with AB || DC intersect each other at the point. If AB = 2 CD, the ratio of the areas of triangles AOB and COD is 4:1.

**☛ Related Questions:**

- In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area( ABC) / area(DBC) = AO/DO
- If the areas of two similar triangles are equal, prove that they are congruent.
- D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC
- Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

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