The document Ex 9.2 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.

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**Question 1. In the figure, ABCD is a parallelogram, AE âŠ¥ DC and CF âŠ¥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. Solution: **We have AE âŠ¥ DC and AB = 16 cm

âˆµ AB = CD [opp. sides of parallelogram ABCD]

âˆ´ CD = 16 cm

Now, area of parallelogram ABCD= CD x AE = 16 x 8 cm^{2} [âˆ´ AE = 8 cm (Given)]

= 128 cm2

Since, CF âŠ¥ AD

âˆ´ Area of parallelogram ABCD = AD x CF

â‡’ AD x CF = 128

â‡’ AD x 10 = 128 [âˆµ CF = 10 cm]

â‡’ AD = (128/10)

= 12.8

Thus, the required length of AD is 12.8 cm.

**Question 2. 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: **Let us join E and G.**REMEMBER**

If a D and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

âˆµ E and G are the mid-points of AB and CD respectively.

âˆ´ EG is parallel to BC or AD.

Also, ar (parallelogram EBCG) = ar (parallelogram AEGD)

= (1/2) ar (parallelogram ABCD) â€¦(1)

Now, Î”EFG and parallelogram EBCG are on the same base EG, and between the same parallels EG and BC.

âˆ´ ar (Î”EFG) = (1/2) ar (parallelogram EBCG) â€¦(2)

Similarly, ar (Î”EHG) = (1/2)

ar (parallelogram AEGD) â€¦(3)

Adding (2) and (3), we get

ar (Î”EFG) + ar (Î”EHG)

= (1/2) [ar (parallelogram EBCG) + ar (parallelogram AEGD)]

â‡’ ar (EFGH)

=(1/2) [ (1/2)ar (parallelogram ABCD) + (1/2)ar (parallelogram ABCD)]

[from (1)]

â‡’ ar (EFGH) =(1/2)[ar (parallelogram ABCD)]

Thus, ar (EFGH) = (1/2)ar (ABCD).**Note**: By joining H and F, we can also have

ar (Î”HEF) + ar (Î”HGF)

= (1/2) [(1/2)ar (parallelogram ABCD) + (1/2)ar (parallelogram ABCD)]

â‡’ ar (EFGH) =(1/2) ar (ABCD)

**Question 3. â€˜Pâ€™ and â€˜Qâ€™ are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (**Î”**APB) = ar(**Î”**BQC). Solution**: âˆµ ABCD is a parallelogram.

âˆ´ AB || CD and BC || AD.

Now, Î”APB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.

âˆ´ ar (Î”APB) =(1/2) ar (parallelogram ABCD) â€¦(1)

Also,Î”BQC and parallelogram ABCD are on the same base BC and between the same parallels BC and AD.

âˆ´ ar (Î”BQC) = (1/2) ar (parallelogram ABCD) â€¦(2)

From (1) and (2), we have ar (Î”APB) = ar (Î”BQC).**Question 4. In the figure, P is a point in the interior of a parallelogram ABCD. Show that **

**(i) ar (**Î”**APB) + ar (**Î”**PCD) = **(1/2)** ar (ABCD) (ii) ar (**Î”

**Solution: **(i) We have a parallelogram ABCD, i.e. AB || CD and BC || AD.

Let us draw EF || AB, through P.

Since AB || EF,

âˆ´ Î”APB and parallelogram AEFB are on the same base AB and between the same parallels AB and EF.

âˆ´ ar (Î”APB) = (1/2)ar (AEFB) â€¦(1)

Also, Î”(PCD) and parallelogram CDEF are on the same base CD and between the same parallels CD || EF.

ar (Î”PCD) = (1/2)ar (parallelogram CDEF) â€¦(2)

Adding (1) and (2), we have ar (Î”APB) + ar (Î”PCD)

= (1/2)ar (AEFB) + (1/2)ar (CDEF)

â‡’ ar (Î”APB) + ar (Î”PCD) = (1/2)ar (ABCD)

(ii) Let us join GH || BC, through P.

âˆ´ Î”APD and parallelogram ADGH are on the same base AD and between the same parallels AD and GH.

âˆ´ ar (Î”APD) = (1/2) ar (ADGH) â€¦(3)

Similarly, ar (Î”PBC) =(1/2) ar (BCGH) â€¦(4)

Adding (3) and (4), we have ar (Î”APD) + ar (Î”PBC)

=(1/2) ar (ADGH) +(1/2) ar (BCGH)

ar (Î”APD) + ar (Î”PBC) =(1/2) ar (ABCD) â€¦(5)

But, ar (Î”APB) + ar (Î”PCD) =(1/2) ar (ABCD) â€¦(6)

From (5) and (6), we have: ar (Î”APD) + ar (Î”PBC) = ar (Î”APB) + ar (Î”PCD)

**Question 5. In the figure, 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)

Solution:

(i) To prove that ar (PQRS) = ar (ABRS)

âˆµ Parallelogram PQRS and parallelogram ABRS are on the same base RS and between the same parallels RS and PAQB.

âˆ´ ar (PQRS) = ar (ABRS)

(ii) To prove that ar (AXS) = (1/2) ar (PQRS)

âˆµ Î”AXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR.

âˆ´ ar (Î”AXS) = (1/2) ar (ABRS) â€¦(1)

But ar (parallelogram PQRS) = ar (parallelogram ABRS) (Proved) â€¦(2)

From (1) and (2), we have ar (Î”AXS)

= (1/2) ar (parallelogram PQRS)

**Question 6. A farmer was having a field in the form of a parallelogram PQRS. She took any point â€˜Aâ€™ on RS and joined it to â€˜Pâ€™ and â€˜Qâ€™. In how many parts the field 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 farmer is having the field in the form of parallelogram PQRS, and a point A is situated at RS.

Let us join AP and AQ.

Obviously the field is divided into three parts i.e. in Î”APS, Î”PAQ and Î”QAR, i.e these parts are triangular in shape.

Since Î”PAQ and parallelogram PQRS are on the same base and between the same parallels PQ and RS.

âˆ´ ar (Î”PAQ) =(1/2) ar (parallelogram PQRS) â€¦(1)

â‡’ ar (parallelogram PQRS) - ar (Î”PAQ) = ar (parallelogram PQRS) - (1/2) ar (parallelogram PQRS)

â‡’ [ar (Î”APS) + ar (Î”QAR)] = (1/2) ar (parallelogram PQRS) â€¦(2)

From (1) and (2), we have ar (Î”PAQ) = ar [(Î”APS) + (Î”QAR)]

Thus, the farmer can sow wheat in (Î”PAQ) and pulses in [(Î”APS) + (Î”QAR)] OR

wheat in [(Î”APS) + (Î”QAR)] and pulses in (Î”PAQ).