Question

9. Diagonals AC and BD of a trapezium ABCD with AB II DC intersect each other to
Prove that ar (AOD) = ar (BOC).
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Answer 1

Step-by-step explanation:

Given:

• ABCD is trapezium with AB || DC.
• Diagonals of trapezium are AC & BD.

To Prove:

• ar(AOD) = ar(BOC)

Solution: Refer to the attachment. We have

• AB || DC
• Diagonals AC & BD intersect at point O.

Now in ∆ADB and ∆BCA

• These two triangles lie on the same base AB and are between same parallels AB & DC therefore areas of these triangles will be equal.

So, we have now

• ar (∆ADB) = ar (∆BCA)

As we can see there is a common triangle i.e ∆AOB in the center of these two ∆s.

∴ Subtracting ar (∆AOB) both the sides.

➟ ar (∆ADB) – ar (∆AOB) = ar (∆BCA) – ar (∆AOB)

➟ ar (∆AOD) = ar (∆BOC)

Hence, proved.

• A trapezium is a type of quadrilateral with 4 vertices and 4 edges.

• Area is given by 1/2 × Sum of parallel sides × Distance between them.

• Perimeter - Sum of all sides.

Answer 2

Triangles with same base and between the same parallel are equal in area.


△ADC & △BDC

DC is common base


Both lie in Between AB & DC

⇢ Area of △ADC
⇢ Area of △BDC

Area of ADC - area of △DOC = area of △BDC - area of △DOC


area of (AOD) = area of (BCO)


• Hence proved