Class 9 Maths

Parallelograms

Exercise 9.4

Part 1

Question 1: Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Answer: In this figure, ABCD is a parallelogram and EFCD is a rectangle. They are on the same base DC.

parallelogram

Height of parallelogram = FC

AB = DC (Opposite sides of parallelogram)

EF = DC (Opposite sides of rectangle)

From above two equations, it is clear that

EF = DC

This means EA = FB

Perimeter of ABCD = AB + BC + CD + AD

Perimeter of EFCD = EF + FC + CD + ED

= AB + CD + FC + ED

In ΔEAD

AD > ED (hypotenuse is the longest side)

In ΔFBC

BC > FC (hypotenuse is the longest side)

So, AB + CD + BC + AD > AB + CD + FC + ED

So, it is proved that perimeter of parallelogram is greater than perimeter of rectangle made on the same base.

Question 2: In this figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).

triangle

Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

Answer: ΔABD, ΔADE and ΔAEC all have same height = h

BD = DE = EC (Given)

So, bases are equal

So, area of anyone of the three triangles = `1/2xx\h\xxb`

So, ar(ABD) = ar(ADE) = ar(AEC) PROVED

Question 3: In this figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF)

parallelogram

Answer: In ΔADE and Δ BCF

AE = BF (Opposite sides of parallelogram ABFE)

AD = BC (Opposite sides of parallelogram ABCD)

DE = CF (Opposite sides of parallelogram DCFE)

So, from SSS theorem

ΔADE ≅ Δ BCF

Or, ar(ΔADE) = ar(Δ BCF) PROVED



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