NCERT Solution: Areas of Parallelograms and Triangles
In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal.
∴ AD = BC ... (1)
Similarly, for parallelograms DCEF and ABFE, it can be proved that
DE = CF ... (2)
And, EA = FB ... (3)
In ΔADE and ΔBCF,
AD = BC [Using equation (1)]
DE = CF [Using equation (2)]
EA = FB [Using equation (3)]
∴ ΔADE ≅ BCF (SSS congruence rule)
⇒ Area (ΔADE) = Area (ΔBCF)
In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that
ar (BPC) = ar (DPQ).
[Hint: Join AC.]
Answer :
It is given that ABCD is a parallelogram.
AD || BC and AB || DC(Opposite sides of a parallelogram are parallel to each other)
Join point A to point C.
Consider ΔAPC and ΔBPC
ΔAPC and ΔBPC are lying on the same base PC and between the same parallels PC and AB. Therefore,
Area (ΔAPC) = Area (ΔBPC) ... (1)
In quadrilateral ACDQ, it is given that
AD = CQ
Since ABCD is a parallelogram,
AD || BC (Opposite sides of a parallelogram are parallel)
CQ is a line segment which is obtained when line segment BC is produced.
∴ AD || CQ
We have,
AC = DQ and AC || DQ
Hence, ACQD is a parallelogram.
Consider ΔDCQ and ΔACQ
These are on the same base CQ and between the same parallels CQ and AD. Therefore,
Area (ΔDCQ) = Area (ΔACQ)
⇒ Area (ΔDCQ) - Area (ΔPQC) = Area (ΔACQ) - Area (ΔPQC)
⇒ Area (ΔDPQ) = Area (ΔAPC) ... (2)
From equations (1) and (2), we obtain
Area (ΔBPC) = Area (ΔDPQ)
In the following 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
(i) ar(BDE) = 1/4 ar(ABC)
(i) Let G and H be the mid-points of side AB and AC respectively.
Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and also its length will be half of the length of BC (mid-point theorem).
=> GH = 1/2 BC and GH || BD
=> GH = BD = DC and GH || BD (D is the mid-point of BC)
Consider quadrilateral GHDB.
GH ||BD and GH = BD
Two line segments joining two parallel line segments of equal length will also be equal and parallel to each other.
Therefore, BG = DH and BG || DH
Hence, quadrilateral GHDB is a parallelogram.
We know that in a parallelogram, the diagonal bisects it into two triangles of equal area.
Hence, Area (ΔBDG) = Area (ΔHGD)
Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and their respective diagonals are dividing them into two triangles of equal area.
ar (ΔGDH) = ar (ΔCHD) (For parallelogram DCHG)
ar (ΔGDH) = ar (ΔHAG) (For parallelogram GDHA)
ar (ΔBDE) = ar (ΔDBG) (For parallelogram BEDG)
ar (ΔABC) = ar(ΔBDG) + ar(ΔGDH) + ar(ΔDCH) + ar(ΔAGH)
ar (ΔABC) = 4 x ar(ΔBDE)
Hence, ar(BDE) = 1/4 ar(ABC)
In the following 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
(iii) ar(ABC) = 2 ar(BEC)
Area (ΔBDE) - Area (ΔFED) = Area (ΔAED) - Area (ΔFED)
Area (ΔBEF) = Area (ΔAFD) (1)
Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)
Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (1)]
Area (ΔABD) = Area (ΔABE) (2)
AD is the median in ΔABC.

From (2) and (3), we obtain
2 ar (ΔBDE) = ar (ΔABE)
or ar(BDE) = 1/2 ar(BAE)
In the following 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
(iii) ar(ABC) = 2 ar(BEC)
ANSWER
ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)
ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)
Using equation (1), we obtain
ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC)
ar (ΔABD) = ar (ΔBEC)
1/2 ar(ΔABC) = ar(ΔBEC)
ar (ΔABC) = 2 ar (ΔBEC)
In the following 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
(iv) ar(BFE) = ar(AFD)
ANSWER
(iv)It is seen that ΔBDE and ar ΔAED lie on the same base (DE) and between the parallels DE and AB.
∴ar (ΔBDE) = ar (ΔAED)
⇒ ar (ΔBDE) - ar (ΔFED) = ar (ΔAED) - ar (ΔFED)
∴ar (ΔBFE) = ar (ΔAFD)
In the following 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 (v) ar(BFE) = 2 ar(FED)
ANSWER:
Let h be the height of vertex E, corresponding to the side BD in ΔBDE.
Let H be the height of vertex A, corresponding to the side BC in ΔABC.
In (i), it was shown that ar(BDE) = 1/4 ar(ABC)
In (iv), it was shown that ar (ΔBFE) = ar (ΔAFD)
=> ar (ΔBFE) = ar (ΔAFD)
= 2 ar (ΔFED)
Hence, ar(BFE) = 2 ar(FED)
In the following 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 (vi) ar(FED) = 1/8 ar(AFC)
ANSWER:
(vi) Area (AFC) = area (AFD) + area (ADC)