**In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.**

Solution:

In parallelogram ABCD, CD = AB = 16cm

We know that,

area of parallelogram = base x corresponding altitude

Area of parallelogram ABCD = CD x AE

= 16 x 8 cm^{2}

= 128 cm^{2}

Also, area of parallelogram ABCD = AD x FC = (AD x 10)cm^{2}

⸫ AD x 10 = 128

AD = ^{128}/_{10} = 12.8 cm

**If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) =**^{1}/_{2}ar (ABCD)

Solution:

Given, ,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD.

Now, we need to show ar (EFGH) = ^{1}/_{2}ar (ABCD)

Let us join HF,

In parallelogram ABCD,

AD = BC and AD||BC (opposite sides of a parallelogram are equal and parallel)

AB = CD (ooposite sides of a parallelogram are equal)

i.e., ^{1}/_{2} AD = ^{1}/_{2} BC

and AH||BF

That implies, AH = BF and AH||BF (Since Ha and f are the midpoints of AD and BC)

Thus, ABFH is a parallelogram

Since ∆HEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF

⸫Area of ∆HEF = ^{1}/_{2} area of ABFH ………..(1)

Similarly,

Area of ∆HGF = ^{1}/_{2} area ofHDCF ………..(2)

Add equation (1) and equation (2), we get,

area of ∆HEF + Area of ∆HGF = ^{1}/_{2} area of ABFH + ^{1}/_{2} area of HDCF

= ^{1}/_{2} (area of ABFH + area of HDCF)

Therefore, area of EFGH = ^{1}/_{2} area of ABCD

**P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).**

Solution:

It is given that, P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

We need to prove, ar (APB) = ar (BQC).

It can be observed that, ∆BQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC

⸫area of ∆BQC = ^{1}/_{2} area of ABCD ……………(1)

Similarly, ∆APB and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AB and DC

⸫area of ∆APB = ^{1}/_{2} area of ABCD ……………(1)

On adding equation (1) and equation (2), we obtain,

area of ∆BQC + area of ∆APB = ^{1}/_{2} area of ABCD + ^{1}/_{2} area of ABCD

area of ∆BQC = area of ∆APB

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**n Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that**

**(i) ar (APB) + ar (PCD) = ^{1}/_{2}ar (ABCD)**

**(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)**

**[Hint : Through P, draw a line parallel to AB.]**

Solution:

(i) let us draw a line segment EF , passing through point P and parallel to line segment AB.

In parallelogram ABCD,

AB||EF (construction)………………(1)

ABCD is a parallelogram.

⸫AD||BF (opposite sides of a parallelogram)

⇒AE||BF ……………..(2)

From equation (1) and equation (2), we get,

AB||EF and AE||BF

Therefore, quadrilateral ABEF is pallelogram

It can be observed that ∆APB and parallelogram ABEF are lying on the same base AB and between the same parallel lines AB and EF

area of ∆APB = ^{1}/_{2} area of ABEF ……………(3)

Similarly, for ∆PCD and parallelogram EFCD,

Area of ∆PCD = ^{1}/_{2} area of EFCD ………………(4)

Adding equations (3) and (4), we obtain,

area of ∆APB + Area of ∆PCD = ^{1}/_{2} area of ABEF + ^{1}/_{2} area of EFCD

area of ∆APB + Area of ∆PCD = ^{1}/_{2} [area of ABEF + area of EFCD]

area of ∆APB + Area of ∆PCD = ^{1}/_{2} area of ABCD ……………..(5)

(ii) let us draw a line segment MN , passing through point P and parallel to line segment AD.

In parallelogram ABCD,

MN||AD (construction)………………(6)

ABCD is a parallelogram.

⸫AB||DC (opposite sides of a parallelogram)

⇒AM||DN ……………..(7)

From equation (6) and equation (7), we get,

MN||AD and AM||DN

Therefore, quadrilateral AMND is parallelogram

It can be observed that ∆APD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN

area of ∆APD = ^{1}/_{2} area of AMND ……………(8)

Similarly, for ∆PCB and parallelogram MNCB,

Area of ∆PCB = ^{1}/_{2} area of MNCB ………………(9)

Adding equations (8) and (9), we obtain,

area of ∆APD + Area of ∆PCB = ^{1}/_{2} area of AMND + ^{1}/_{2} area of MNCB

area of ∆APD + Area of ∆PCB = ^{1}/_{2} [area of ABCD] ……………….(10)

On comparing equations (5) and (10), we obtain,

Area of ∆APD + area of ∆PBC = area of ∆APB + area of ∆PCD

**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 (AX S) = ^{1}/_{2 }ar (PQRS)**

Solution:

(i) It can be observed that parallelogram PQRS and ABRS lie on the same base SR and also these lie in between same parallel lines SR and PB

Therefore, area of PQRS = area of ABRS …………….(1)

(ii) Consider ∆AXS and parallelogram ABRS

As these lie on the same base and are between the same parallel lines AS and BR

Therefore, area of ∆AXS = ^{1}/_{2} area of ABRS ………..(2)

from equation (1) and equation (2), we obtain,

area of ∆AXS = ^{1}/_{2} area of PQRS

**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?**

Solution:

The field is divided in three triangles.

Since triangle APQ and parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS.

∴ ar (APQ) = ^{1}/_{2} ar (PQRS)

⇒ 2ar (APQ) = ar(PQRS)

But ar (PQRS) = ar(APQ) + ar (PSA) + ar (ARQ)

⇒ 2 ar (APQ) = ar(APQ) + ar(PSA) + ar (ARQ)

⇒ ar (APQ) = ar(PSA) + ar(ARQ)

Hence, area of ∆APQ = area of ∆PSA + area of ∆ARQ.

To sow wheat and pulses in equal portions of the field separately, farmer sow wheat in ∆APQ and pulses in other two triangles or pulses in ∆APQ and wheat in other two triangles.

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