Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

It can be observed that ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD.

∴ Area (ΔDAC) = Area (ΔDBC)

⇒ Area (ΔDAC) - Area (ΔDOC) = Area (ΔDBC) - Area (ΔDOC)

⇒ Area (ΔAOD) = Area (ΔBOC)

In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (ACB) = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)

**Answer :**

(i) ΔACB and ΔACF lie on the same base AC and are between

The same parallels AC and BF.

∴ Area (ΔACB) = Area (ΔACF)

(ii) It can be observed that

Area (ΔACB) = Area (ΔACF)

⇒ Area (ΔACB) + Area (ACDE) = Area (ACF) + Area (ACDE)

⇒ Area (ABCDE) = Area (AEDF)

A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

**Answer**

Let ABCD be the plot of the land of the shape of a quadrilateral.

Construction,

Diagonal BD is joined. AE is drawn parallel BD. BE is joined which intersected AD at O. △BCE is the shape of the original field and △AOB is the area for constructing health centre. Also, △DEO land joined to the plot.

To prove:

ar(△DEO) = ar(△AOB)

Proof:

△DEB and △DAB lie on the same base BD and between the same parallel lines BD and AE.

ar(△DEB) = ar(△DAB)

⇒ ar(△DEB) - ar△DOB) = ar(△DAB) - ar(△DOB)

⇒ ar(△DEO) = ar(△AOB)

ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY). [Hint : Join CX.]

**Answer**

Given,

ABCD is a trapezium with AB || DC.

XY || AC

Construction,

CX is joined.

To Prove,

ar(ADX) = ar(ACY)

Proof:

ar(△ADX) = ar(△AXC) --- (i) (On the same base AX and between the same parallels AB and CD)

also,

ar(△ AXC)=ar(△ ACY) --- (ii) (On the same base AC and between the same parallels XY and AC.)

From (i) and (ii),

ar(△ADX)=ar(△ACY)