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)
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.)
(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