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 the following figure). Show that
ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]
Answer
AC and PQ are joined.
ar(△ACQ) = ar(△APQ) (On the same base AQ and between the same parallel lines AQ and CP)
⇒ ar(△ACQ) - ar(△ABQ) = ar(△APQ) - ar(△ABQ)
⇒ ar(△ABC) = ar(△QBP) --- (i)
AC and QP are diagonals ABCD and PBQR. Thus,
ar(ABC) = 1/2 ar(ABCD) --- (ii)
ar(QBP) = 1/2 ar(PBQR) --- (iii)
From (ii) and (ii),
1/2 ar(ABCD) = 1/2 ar(PBQR)
⇒ ar(ABCD) = ar(PBQR)
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
(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)
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)
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)