In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2 ar (PQRS)
Answer
(i) Parallelogram PQRS and ABRS lie on the same base SR and between the same parallel lines SR and PB.
∴ ar(PQRS) = ar(ABRS) --- (i)
(ii) In ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel lines AS and BR.
∴ ar(ΔAXS) = 1/2 ar(ABRS) --- (ii)
From (i) and (ii),
ar(ΔAXS) = 1/2 ar(PQRS)
In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2 ar (PQRS)
Answer
The field is divided into three parts. The three parts are in the shape of triangle. ΔPSA, ΔPAQ and ΔQAR.
Area of ΔPSA + ΔPAQ + ΔQAR = Area of PQRS --- (i)
Area of ΔPAQ = 1/2 area of PQRS --- (ii)
Triangle and parallelogram on the same base and between the same parallel lines.
From (i) and (ii),
Area of ΔPSA + Area of ΔQAR = 1/2 area of PQRS --- (iii)
Clearly from (ii) and (iii),
Farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.
In Fig.9.23, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar(ACE).
Answer
Given,
AD is median of ΔABC. Thus, it will divide ΔABC into two triangles of equal area.
∴ ar(ABD) = ar(ACD) --- (i)
also,
ED is the median of ΔABC.
∴ ar(EBD) = ar(ECD) --- (ii)
Subtracting (ii) from (i),
ar(ABD) - ar(EBD) = ar(ACD) - ar(ECD)
⇒ ar(ABE) = ar(ACE)
Answer
ar(BED) = (1/2) × BD × DE
As E is the mid-point of AD,
Thus, AE = DE
As AD is the median on side BC of triangle ABC,
Thus, BD = DC
Therefore,
DE = (1/2)AD --- (i)
BD = (1/2)BC --- (ii)
From (i) and (ii),
ar(BED) = (1/2) × (1/2) BC × (1/2)AD
⇒ ar(BED) = (1/2) × (1/2) ar(ABC)
⇒ ar(BED) = 1/4 ar(ABC)
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Answer
O is the mid point of AC and BD. (diagonals of bisect each other)
In ΔABC, BO is the median.
∴ ar(AOB) = ar(BOC) --- (i)
also,
In ΔBCD, CO is the median.
∴ ar(BOC) = ar(COD) --- (ii)
In ΔACD, OD is the median.
∴ ar(AOD) = ar(COD) --- (iii)
In ΔABD, AO is the median.
∴ ar(AOD) = ar(AOB) --- (iv)
From equations (i), (ii), (iii) and (iv),
ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)
So, the diagonals of a parallelogram divide it into four triangles of equal area.