Question

Diagonals AC and BD of trapezium ABCD, in which AB\(||\)DC intersect each other at O. The triangle which is equal in area of \( \triangle \text{AOD} \) is :

• A. \( \triangle \text{AOB} \)
• B. \( \triangle \text{BOC} \)
• C. \( \triangle \text{DOC} \)
• D. \( \triangle \text{ADC} \)

Hint:

A key property for trapeziums: When the diagonals intersect, they form four triangles. The two triangles formed by a non-parallel side and segments of the diagonals are equal in area. In trapezium ABCD with AB \(||\) DC, and diagonals intersecting at O: Area(\(\triangle AOD\)) = Area(\(\triangle BOC\)). This is because \(\triangle ADC\) and \(\triangle BDC\) have the same base DC and are between the same parallels, so their areas are equal. Removing the common \(\triangle DOC\) from both leaves \(\triangle AOD\) and \(\triangle BOC\) with equal areas.

Answer 1

The Correct Option is B

Concept: This problem uses the property of triangles on the same base and between the same parallel lines, and how areas of triangles are related when diagonals of a trapezium intersect. Step 1: Property of triangles on the same base and between the same parallels Triangles on the same base (or equal bases) and between the same parallel lines are equal in area. In trapezium ABCD, AB \(||\) DC. Consider triangles \(\triangle ADC\) and \(\triangle BDC\). They share the same base DC and are between the same parallel lines AB and DC. Therefore, Area(\(\triangle ADC\)) = Area(\(\triangle BDC\)). Step 2: Decompose the areas We can write the areas as: Area(\(\triangle ADC\)) = Area(\(\triangle AOD\)) + Area(\(\triangle DOC\)) Area(\(\triangle BDC\)) = Area(\(\triangle BOC\)) + Area(\(\triangle DOC\)) Step 3: Equate the areas and simplify Since Area(\(\triangle ADC\)) = Area(\(\triangle BDC\)), we have: Area(\(\triangle AOD\)) + Area(\(\triangle DOC\)) = Area(\(\triangle BOC\)) + Area(\(\triangle DOC\)) Subtract Area(\(\triangle DOC\)) from both sides of the equation: Area(\(\triangle AOD\)) = Area(\(\triangle BOC\)) Step 4: Conclusion The triangle which is equal in area to \(\triangle AOD\) is \(\triangle BOC\). Visualization: Imagine the trapezium. The triangles \(\triangle ADB\) and \(\triangle ACB\) are also on the same base AB and between parallels AB and DC, so Area(\(\triangle ADB\)) = Area(\(\triangle ACB\)). Area(\(\triangle ADB\)) = Area(\(\triangle AOD\)) + Area(\(\triangle AOB\)) Area(\(\triangle ACB\)) = Area(\(\triangle BOC\)) + Area(\(\triangle AOB\)) Equating these: Area(\(\triangle AOD\)) + Area(\(\triangle AOB\)) = Area(\(\triangle BOC\)) + Area(\(\triangle AOB\)). Subtracting Area(\(\triangle AOB\)) from both sides gives Area(\(\triangle AOD\)) = Area(\(\triangle BOC\)).