Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Solutions
Question - 11 : - In the given figure, ABCDEis a pentagon. A line through B parallel to AC meets DC produced at F. Showthat
(i) ar (ACB) = ar (ACF)
(ii)ar (AEDF) = ar (ABCDE
Answer - 11 : -
(i) ΔACB and ΔACF lie onthe same base AC and are between
The same parallels AC andBF.
∴Area (ΔACB) = Area (ΔACF)
(ii) It can be observedthat
Area (ΔACB) = Area (ΔACF)
⇒Area (ΔACB) + Area (ACDE) = Area (ACF) + Area (ACDE)
⇒ Area (ABCDE) = Area (AEDF)
Question - 12 : - Avillager Itwaari has a plot of land of the shape of a quadrilateral. The GramPanchayat of the village decided to take over some portion of his plot from oneof the corners to construct a Health Centre. Itwaari agrees to the aboveproposal with the condition that he should be given equal amount of land inlieu of his land adjoining his plot so as to form a triangular plot. Explainhow this proposal will be implemented.
Answer - 12 : - 
Let quadrilateral ABCD bethe original shape of the field.
The proposal may beimplemented as follows.
Join diagonal BD and drawa line parallel to BD through point A. Let it meet
the extended side CD ofABCD at point E. Join BE and AD. Let them intersect each other at O. Then,portion ΔAOB can be cut from the original field so that the new shape of thefield will be ΔBCE. (See figure)
We have to prove that thearea of ΔAOB (portion that was cut so as to construct Health Centre) is equalto the area of ΔDEO (portion added to the field so as to make the area of thenew field so formed equal to the area of the original field)
It can be observed thatΔDEB and ΔDAB lie on the same base BD and are between the same parallels BD andAE.
∴Area (ΔDEB) = Area (ΔDAB)
⇒Area (ΔDEB) − Area (ΔDOB) = Area (ΔDAB) − Area (ΔDOB)
⇒ Area (ΔDEO) = Area (ΔAOB)
ABCD is a trapezium withAB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ADX) = ar (ACY).
[Hint:Join CX.]
Answer - 13 : - 
It can be observed thatΔADX and ΔACX lie on the same base AX and are between the same parallels AB andDC.
∴Area (ΔADX) = Area (ΔACX) … (1)
ΔACY and ΔACX lie on thesame base AC and are between the same parallels AC and XY.
∴Area (ΔACY) = Area (ACX) … (2)
From equations (1) and(2), we obtain
Area(ΔADX) = Area (ΔACY)Inthe given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
Answer - 14 : - 
Since ΔABQ and ΔPBQ lie onthe same base BQ and are between the same parallels AP and BQ,
∴Area (ΔABQ) = Area (ΔPBQ) … (1)
Again, ΔBCQ and ΔBRQ lieon the same base BQ and are between the same parallels BQ and CR.
∴Area (ΔBCQ) = Area (ΔBRQ) … (2)
On adding equations (1)and (2), we obtain
Area (ΔABQ) + Area (ΔBCQ)= Area (ΔPBQ) + Area (ΔBRQ)
⇒ Area (ΔAQC) = Area (ΔPBR)
DiagonalsAC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) =ar (BOC). Prove that ABCD is a trapezium.
Answer - 15 : - 
It is given that
Area (ΔAOD) = Area (ΔBOC)
Area (ΔAOD) + Area (ΔAOB)= Area (ΔBOC) + Area (ΔAOB)
Area (ΔADB) = Area (ΔACB)
We know that triangles onthe same base having areas equal to each other lie between the same parallels.
Therefore, thesetriangles, ΔADB and ΔACB, are lying between the same parallels.
i.e., AB || CD
Therefore,ABCD is a trapezium.
Inthe given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that boththe quadrilaterals ABCD and DCPR are trapeziums.
Answer - 16 : - 
It is given that
Area (ΔDRC) = Area (ΔDPC)
As ΔDRC and ΔDPC lie onthe same base DC and have equal areas, therefore, they must lie between thesame parallel lines.
∴DC || RP
Therefore, DCPR is atrapezium.
It is also given that
Area (ΔBDP) = Area (ΔARC)
⇒Area (BDP) − Area (ΔDPC) = Area (ΔARC) − Area (ΔDRC)
⇒Area (ΔBDC) = Area (ΔADC)
Since ΔBDC and ΔADC are onthe same base CD and have equal areas, they must lie between the same parallellines.
∴AB || CD
Therefore,ABCD is a trapezium.