Class 9 Maths


Parallelograms

Exercise 9.4

Part 2

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)

parallelogram

Answer: AD = CQ (given)

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.)

triangle

(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) = `1/4`ar(ABC) proved

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

Answer: ∠ACB = 60°

∠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 = `1/2`BC = `1/2`AC

So, PE = `1/2`AD

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)

Answer: In right angle ΔABD

AD2 = AB2 - BD2

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)/2`a ……..(1)

Similarly, ar(ΔEFD) `=1/2` × FD × EP

`=1/2` × FD × `(sqrt3)/4`a …………….(2)

From equations (1) and (2)

Ar(ΔAFD) = 2 ar(ΔEFD)

From answer to question (d)

Ar(ΔAFD) = ar(ΔBEF)

Or, ar(ΔBFE) = 2 ar(ΔEFD)

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

Answer: ar(ΔAFC) = ar(ΔAFD) + ar(ΔADC)

= 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