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• NCERT Solutions for Class 9 Maths

Class 9 NCERT Solutions – Chapter 9 Areas of Parallelograms And Triangles – Exercise 9.2

Question 1: In given Fig. 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:

As Given in the Question,

AB = CD = 16 cm (Opposite sides of a parallelogram)

CF = 10 cm and AE = 8 cm

Now, As we have studied in this Chapter we know,

Area of Parallelogram = Base x Altitude

â‡’ CD Ã— AE = AD Ã— CF

â‡’ 16 Ã— 8 = AD Ã— 10

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:

As Given in the Question ,

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

To Prove,

ar (EFGH) = Â½ ar(ABCD)

First, we have to do some construction

join H to E

Proof:

As we know,

AD || BC and AD = BC (Opposite sides of a parallelogram)

â‡’ Â½ AD = Â½ BC

As H and F are the mid points of AD and BC

AH || BF and and DH || CF

Therefore,

AH = BF and DH = CF (H and F are mid points)

âˆ´ ABFH and HFCD are parallelograms.

As we know that, Î”EFH and parallelogram ABFH, both lie on the same base FH and Î”EFH lie in-between the same parallel lines AB and HF.

Therefore,

Area of EFH = Â½ Area of ABFH â€” (i)

And, Area of GHF = Â½ Area of HFCD â€” (ii)

Now, Adding (i) and (ii) we get,

Area of Î”EFH + Area of Î”GHF = Â½ Area of ABFH + Â½ Area of HFCD

â‡’ Area of EFGH = Area of ABFH

âˆ´ ar (EFGH) = Â½ ar(ABCD)

Question 3: 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:

Î”APB and parallelogram ABCD lie on the same base AB and Î”APB lie in-between same parallel AB and DC.

Now, As we know that

ar(Î”APB) = Â½ ar(parallelogram ABCD) â€” (i)

Similarly,

ar(Î”BQC) = Â½ ar(parallelogram ABCD) â€” (ii)

From (i) and (ii), we have

ar(Î”APB) = ar(Î”BQC)

Hence proved,

Question 4: In the Figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar(APB) + ar(PCD) = Â½ ar(ABCD)

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

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

Solution:

Given: P is a point in the interior of the parallelogram ABCD

To prove: Area(APB) + Area(PCD) = Â½ Area(ABCD)

Construction:

Through P, draw a line EF parallel to AB

Proof:

(i) In a parallelogram,

AB || EF (by construction) â€” (i)

âˆ´AD || BC â‡’ AE || BF â€” (ii)

From equations (i) and (ii),

ABFE is a parallelogram.

Now,

Î”APB and parallelogram ABFE are lying on the same base AB and Î”APB lie in-between the same parallel lines AB and EF.

âˆ´ ar(Î”APB) = Â½ ar(ABFE) â€” (iii)

also,

Î”PCD and parallelogram CDEF are lying on the same base CD and Î”PCD lie in-between the same parallel lines CD and EF.

âˆ´ ar(Î”PCD) = Â½ ar(CDEF) â€” (iv)

Adding equations (iii) and (iv) we get,

ar(Î”APB) + ar(Î”PCD) = Â½ [ar(ABFE)+ar(CDEF)]

â‡’ ar(APB)+ ar(PCD) = Â½ ar(ABCD)

Hence Proved

(ii) Construction:

Through P, draw a line GH parallel to AB

In the parallelogram,

AD || GH (by construction) â€” (i)

âˆ´AB || CD â‡’ AG || DH â€” (ii)

From equations (i) and (ii) we get,

AGDH is a parallelogram.

Now,

Î”APD and parallelogram AGHD are lying on the same base AD and Î”APD lie in-between the same parallel lines AD and GH.

âˆ´ar(Î”APD) = Â½ ar(AGHD) â€” (iii)

also,

Î”PBC and parallelogram BCHG are lying on the same base BC and Î”PBC lie in-between the same parallel lines BC and GH.

âˆ´ar(Î”PBC) = Â½ ar(BCHG) â€” (iv)

Adding equations (iii) and (iv) we get,

ar(Î”APD) + ar(Î”PBC) = Â½ {ar(AGHD) + ar(BCHG)}

â‡’ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

Hence Proved

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) = Â½ ar (PQRS)

Solution:

Given:  PQRS and ABRS are parallelograms and X is any point on side BR

To Prove: (i) ar (PQRS) = ar (ABRS)

(ii) ar (AXS) = Â½ ar (PQRS)

Proof:

(i) In Î”PSA and Î”QRB,

âˆ  SPA = âˆ  RQB  â€” (i) (Corresponding Angles from PS || QR and traversal PB

âˆ  PAS = âˆ  QBR â€” (ii) (Corresponding Angles from AS || BR and traversal PB

âˆ  PSA = âˆ  QRB â€” (iii) (Angle Sum Property of triangle)

Also, PS = QR  â€” (iv) (Opposite sides of Parallelogram PQRS)

In view of (i), (iii) and (iv),

Î”PSA â‰…  Î”QRB â€” (v) (By ASA Rule)

âˆ´ Area(Î”PSA) = Area(Î”QRB)  â€” (vi)

âˆ´ Congruent figures have Equal Area

Now, ar(PQRS) = Area(Î”PSA) + Area(AQRS)

= Area(Î”QRB) + Area(AQRS)   —–|| Using (vi)

= ar(ABRS)

âˆ´ar (PQRS) = ar (ABRS)

Hence Proved

(ii) Î”AXS and Parallelogram ABRS are on the Same base As and between the Same parallels AS and BR

âˆ´ Area(Î”AXS) = Â½ Area(Parallelogram ABRS)

= Â½ {Area(AQRS) + Area(Î”QRB)}

=  Â½ Area(Parallelogram PQRS)

Hence Proved

Question 6: 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 are 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 into three parts each in triangular shape.

Let, Î”PSA, Î”PAQ and Î”QAR be the triangles.

Area of (Î”PSA + Î”PAQ + Î”QAR) = Area of PQRS â€” (i)

Area of Î”PAQ = Â½ area of PQRS â€” (ii)

Here, the triangle and parallelogram are on the same base and in-between the same parallel lines.

From (i) and (ii) we get,

Area of Î”PSA + Area of Î”QAR = Â½ area of PQRS â€” (iii)

From (ii) and (iii), we can conclude that,

The farmer must sow wheat or pulses in Î”PAQ or either in both Î”PSA and Î”QAR.

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