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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)

Exercise 9.2

A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

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.

Exercise 9.3

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)

Exercise 9.3

In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = 1/4 ar(ABC).

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)

Exercise 9.3

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.

Total Questions :36