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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 quadrilateral ABCD be the original shape of the field.

The proposal may be implemented as follows:

Join diagonal BD and draw a line parallel to BD through point A.

Let it meet the extended side CD of
ABCD at point E.

Join BE and AD. Let them intersect each other at O.

Then, portion \(\triangle{AOB}\) can be cut from the original field so that the new shape of the field will be \(\triangle{BCE}\) as shown below.

We have to prove that the area of \(\triangle{AOB}\) (portion that was cut so as to construct Health Centre) is equal to the area of \(\triangle{DEO}\)(portion added to the field so as to make the area of the new field so formed equal to the area of the original field).

It can be observed that \(\triangle{DEB}\) and \(\triangle{DAB}\) lie on the same base BD and are between the same
parallels BD and AE.

Area (\(\triangle{DEB}\)) = Area (\(\triangle{DAB}\))

Now subtracting area (\(\triangle DOB \) ) On both side

Area (\(\triangle{DEB}\)) - Area (\(\triangle{DOB}\)) = (\(\triangle{DAB}\)) - Area (\(\triangle{DOB}\))

\(\therefore \) Area (\(\triangle{DEO}\)) = Area (\(\triangle{AOB}\))

- In Figure, E is any point on median AD of a \(\triangle{ABC}\). Show that ar (ABE) = ar (ACE)
- In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = \(\frac{1}{4} \) ar (ABC).
- Show that the diagonals of a parallelogram divide it into four triangles of equal area.
- In Figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).
- D, E and F are respectively the mid-points of the sides BC, CA and AB of a \(\triangle{ABC}\). Show thati) BDEF is a parallelogram.ii) ar(DEF) = \(\frac{1}{4} \) ar(ABC) iii) ar(BDEF) = \(\frac{1}{2} \) ar(ABC)
- In Figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:i) ar (DOC) = ar (AOB)ii) ar (DCB) = ar (ACB)iii) DA || CB or ABCD is a parallelogram.[Hint: From D and B, draw perpendiculars to AC.]
- D and E are points on sides AB and AC respectively of \(\triangle{ABC}\) such that ar (DBC) = ar (EBC). Prove that DE || BC.
- XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF)
- The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Figure). Show that ar (ABCD) = ar (PBQR).[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
- Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
- In Figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show thati) ar (ACB) = ar (ACF)ii) ar (AEDF) = ar (ABCDE).
- 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.]
- In Figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
- Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
- In Figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

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