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.

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

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).

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?

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)

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

Or, ar(ΔADE) = ar(Δ BCF) PROVED

Question 4: In this figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar(BPC) = ar(DPQ). (Hint: Join AC)

AD = BC (Opposite sides of parallelogram ABCD)

So, AD = CQ = BC

ar(ΔQAC) = ar(ΔQDC)

(Triangles on same base QC and between same parallels DA and QC)

Subtracting ΔQPC from both sides:

ar(ΔQAC - ΔQPC) = ar(ΔQDC - ΔQPC)

= ar(ΔAPC) = ar(ΔDPQ) ………..(1)

Now, ar(ΔPAC) = ar(ΔPBC) ………..(2)

(Triangles on the same base PC and between same parallels AB and PC)

From equations (1) and (2)

ar(ΔBPC) = ar(ΔDPQ) PROVED

Question 5: In this figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(Hint: Join EC and AD. Show that BE||AC and DE||AB, etc.)

(a) ar(BDF) = 1/4 ar(ABC)

Answer: Side of ΔABC = s

Side of ΔBDE = s/2 (Because BD = DC)

ar(ABC) =(sqrt3)/4xxs^2

ar(BDE) =(sqrt2)/4xx(s/2)^2=(sqrt3)/4xx(s^2)/4

(BD\E)/(AB\C)=1/4

Or, ar(BDE) = 1/4ar(ABC) proved

(b) ar(BDE) = 1/2 ar(BAE)

∠DBE = 60°

(angles of equilateral triangles)

So, AC||BE

So, ar(ΔBAE) = ar(BEC)

(Triangles on same base BE and between same parallels AC and BE)

Since BD = 1/2 BC

So, ar(ΔBDE) = 1/2 ar(ΔBEC)

Or, ar(ΔBDE) = 1/2 ar(ΔBAE) PROVED

(c) ar(ABC) = 2 ar(BEC)

Answer: Since BD = 1/2BC = 1/2AC

So, PE = 1/2AD

So, ar(ABC) = 1/2 AD × BC

ar(BEC) = 1/2xx1/2 AD × BC

Or, (ar(BE\C))/(ar(AB\C))=1/2

Or, ar(ABC) = 2 ar(BEC) PROVED

(d) ar(BFE) = ar(AFD)

Answer: ΔABC and ΔBDE are equilateral triangles.

So,∠ABC = &BDE = 60°

So, AB||DE

ΔBED and ΔAED are on the same base ED and between same parallels AB and DE

So, ar(ΔBED) = ar(ΔAED)

Subtracting ar(ΔEFD) from both sides,

Ar(ΔBED) – ar(ΔEFD) = ar(ΔAED) – ar(ΔEFD)

= ar(ΔBFE) = ar(ΔAFD) PROVED

(e) ar(BFE) = 2 ar(FED)

Or, AD2 a^2-(a^2)/4

=(4a^2-a^2)/4=(3a^2)/4

Or, AD =(sqrt3a)/2

In right angle ΔPED

EP2 = DE2 - DP2

Or, EP2 =(a/2)^2-(a/4)^2

=(a^2)/4-(a^2)/(16)=(3a^2)/(16)

Or, EP =(sqrt3a)/4

Now, ar(ΔAFD) = 1/2 × FD × AD

=1/2× FD × (sqrt3)/2a ……..(1)

Similarly, ar(ΔEFD) =1/2 × FD × EP

=1/2 × FD × (sqrt3)/4a …………….(2)

From equations (1) and (2)

Ar(ΔAFD) = 2 ar(ΔEFD)

Ar(ΔAFD) = ar(ΔBEF)

Or, ar(ΔBFE) = 2 ar(ΔEFD)

(f) ar(FED) = 1/8 ar(AFC)

= ar(ΔBFE) + 1/2 ar(ΔABC) (From question (d) )

= ar(ΔBFE) + 1/2 × 4 × ar(ΔBDE) (From question (a) )

= ar(ΔBFE) + 2 ar(ΔBDE)

= 2 ar(ΔFED) + 2(ar(ΔBFE) + ar(ΔFED))

= 2 ar(ΔFED) + 2(2 ar(ΔFED) + ar(ΔFED)) (From question (e))

= 2ar(ΔFED) + 2(3 ar(ΔFED))

= 2 ar(ΔFED) + 6 ar(ΔFED)

= 8 ar(ΔFED)

Or, ar(ΔFED) = 1/8 ar(ΔAFC) PROVED