NCERT Solutions for Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

Q1 ) In Figure, ABCD is a parallelogram, \(AE\perp{DC}\) and \(CF\perp{AD}\).
If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Answer :

In parallelogram ABCD, CD = AB = 16 cm [Opposite sides of a parallelogram are equal]

We know that,
Area of a parallelogram
= Base x Corresponding altitude
\(\Rightarrow \)Area of parallelogram ABCD
= CD x AE ...(i)

Also, Area of parallelogram
= Base x Corresponding altitude
\(\Rightarrow \) Area of parallelogram ABCD
= AD x CF ...(ii)

Thus, from (i) and (ii), we get,
CD x AE = AD x CF
\(\Rightarrow \) 16cm x 8cm = AD x 10cm
\(\Rightarrow \) AD = \(\frac{16cm x 8cm}{10cm} \)
\(\Rightarrow \) AD = 12.8cm
\(\therefore\) the length of AD is 12.8 cm.

Q2 ) If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar (ABCD)

Answer :

Given: E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD


Construction: Let us join HF.

To prove: ar(EFGH) = \(\frac{1}{2} \) ar(ABCD)

Proof: In parallelogram ABCD,
AD = BC and AD || BC
( Opposite sides of a parallelogram are equal and parallel)
Multiplying both sides by \(\frac{1}{2} \) we get,
\(\Rightarrow \) \(\frac{1}{2} \) x AD = \(\frac{1}{2} \) x BC
\(\Rightarrow \) AH = BF and AH || BF
\(\therefore \) ABFH is a parallelogram.

Since, \(\triangle{HEF}\) and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
\(\therefore \) Area (\(\triangle{HEF}\)) = \(\frac{1}{2} \) Area (ABFH) ...(i)
Similarly, it can be proved that,

Area (\(\triangle{HGF}\)) = \(\Rightarrow \) Area (HDCF) ...(ii)
On adding equations (i) and (ii), we get,
\(\Rightarrow \) Area (\(\triangle{HEF}\)) + Area (\(\triangle{HGF}\)) = \(\frac{1}{2} \) Area (ABFH) + \(\frac{1}{2} \) Area (HDCF)
\(\Rightarrow \) Area [(\(\triangle{HEF}\)) + (\(\triangle{HGF}\))] = [Area (ABFH) + Area (HDCF)]
\(\therefore\) Area (EFGH) = \(\frac{1}{2} \) Area (ABCD)
Hence, proved.

Q3 ) P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
Show that ar(APB) = ar(BQC).

Answer :

It can be observed that \(\triangle{BQC}\) and paralleogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.

\(\triangle \)APB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.

ar(\(\triangle \)APB) = \(\frac{1}{2} \) ar(parallelogram ABCD) ...... (i)

Similarly,

ar(\(\triangle \)BQC) = \(\frac{1}{2} \) ar(parallelogram ABCD) ...... (ii)

From (i) and (ii), we have

ar(\(\triangle \) APB) = ar(\(\triangle \)BQC)
Hence, proved

Q4 ) In Figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (\(\triangle{APB}\) + \(\triangle{PCD}\)) = \(\frac{1}{2} \) ar (ABCD)
(ii) ar (\(\triangle{APD}\) + \(\triangle{PBC}\)) = ar(APB) + ar(PCD)

[Hint : Through P, draw a line parallel to AB.]

Answer :

Construction: Let us draw a line segment EF, passing through point P and parallel to line segment AB.

In parallelogram ABCD, AB || EF ...(i)(By construction)

Also, ABCD is a parallelogram
So, AD || BC
(Opposite sides of parallelogram)
Also, AE || BF ...(ii)
(Since, \(\frac{1}{2} \) AD = AE and \(\frac{1}{2} \) BC = BF)
From equations (i) and (ii), we get,
AB || EF and AE || BF

Therefore, quadrilateral ABFE is a parallelogram.

It can be observed that \(\triangle{APB}\) and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF.

Therefore, we can say that,
Area \(\triangle{APB}\) = \(\frac{1}{2} \) Area (ABFE) ...(iii)
Similarly,
for \(\triangle{PCD}\) and parallelogram EFCD,
\(\Rightarrow \) Area \(\triangle{PCD}\) = \(\frac{1}{2} \) Area (EFCD) ...(iv)
Now, adding eq. (iii) and (iv), we get,
\(\Rightarrow \) Area \(\triangle{APB}\) + Area \(\triangle{PCD}\) = \(\frac{1}{2} \) Area (ABFE) + \(\frac{1}{2} \) Area (EFCD)
\(\Rightarrow \) Area \(\triangle{APB}\) + Area \(\triangle{PCD}\) = \(\frac{1}{2} \)
[Area (ABFE) + Area (EFCD)]
\(\Rightarrow \) Area \(\triangle{APB}\) + Area \(\triangle{PCD}\) = \(\frac{1}{2} \) Area (ABCD) ...(v)

Hence, part i) is proved.



Construction: Let us draw a line segment MN, passing through point P and parallel to line segment AD.

In parallelogram ABCD, MN || AD ...(vi)(By construction)

Also, ABCD is a parallelogram
Thus, AB || DC
(Opposite sides of parallelogram)

Also, AM || DN ...(vii)
(Since, \(\frac{1}{2} \) AB = AM and \(\frac{1}{2} \) DC = DN)
From equations (vi) and (vii), we get,
MN || AD and AM || DN
\(\therefore \) quadrilateral AMND is a parallelogram.


It can be observed that \(\triangle{APD}\) and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.

Therefore, we can say that,
Area \(\triangle{APD}\) = \(\frac{1}{2} \) Area (AMND) ...(viii)

Similarly,
for \(\triangle{PCB}\) and parallelogram MNCB,
\(\Rightarrow \) Area \(\triangle{PCB}\) = \(\frac{1}{2} \) Area (MNCB) ...(ix)

Now, adding eq. (viii) and (ix), we get,

Area \(\triangle{APD}\) + Area \(\triangle{PCB}\) = \(\frac{1}{2} \) Area (AMND) + \(\frac{1}{2} \) Area (MNCB)
\(\Rightarrow \) Area \(\triangle{APD}\) + Area \(\triangle{PCB}\) = \(\frac{1}{2} \) [Area (AMND) + Area (MNCB)]
\(\Rightarrow \) Area \(\triangle{APD}\) + Area \(\triangle{PCB}\) = \(\frac{1}{2} \) Area (ABCD) ...(x)

Thus, comparing eq. (v) and (x), we get,

Area \(\triangle{APB}\) + Area \(\triangle{PCD}\) = Area \(\triangle{APD}\) + Area \(\triangle{PCB}\)

Hence, proved.

Q5 ) In Figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
i)ar(PQRS) = ar(ABRS)
ii) ar(AXS) = \(\frac{1}{2} \) ar(PQRS)

Answer :

i) It can be observed that paralleogram PQRS and ABRS lie on the same base SR and also, these lie in between the same parallel lines SR and PB.

\(\therefore \) Area (PQRS) = Area (ABRS) ...(i)

ii) Consider \(\triangle{AXS}\) and paralleogram ABRS.

As these lie on the same base and are between the same parallel lines AS and BR

\(\therefore \) Area (\(\triangle{AXS}\)) = \(\frac{1}{2} \) Area (ABRS) ...(ii)
From equations (i) and (ii), we get,
ar(AXS) = \(\frac{1}{2} \) ar(PQRS)
Hence, proved.

Q6 ) A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q.
In how many parts the fields is divided? What are the shapes of these parts?
The farmer wants to sow wheat and pulses in equal portions of the field separately.
How should she do it?

Answer :

From the figure, we can say that it is that point A which divides the field into three parts.

These parts are triangular in shape i.e., \(\triangle{PSA}\), \(\triangle{PAQ}\) and \(\triangle{QRA}\)

Thus, Area \(\triangle{PSA}\) + Area \(\triangle{PAQ}\) + Area \(\triangle{QRA}\) = Area of Parallelogram PQRS ...(i)

Also, Area \(\triangle{PAQ}\) = (\(\frac{1}{2} \)) Area of Parallelogram PQRS ...(ii)

From equation (i) and (ii), we get,

Area \(\triangle{PSA}\) + Area \(\triangle{QRA}\) = \(\frac{1}{2}\) Area of Parallelogram PQRS ...(iii)

Clearly, it can be observed that the farmer must sow wheat in triangular part PAQ and pulses in other two triangular parts PSA and QRA or wheat in triangular parts PSA and QRA and pulses in triangular parts PAQ.

Q1 ) In Figure, E is any point on median AD of a \(\triangle{ABC}\). Show that ar (ABE) = ar (ACE)

Answer :

Given,
AD is median of \(\triangle \) ABC.

\(\therefore \) it will divide \(\triangle \)ABC into two triangles of equal area.

\(\therefore \) ar(ABD) = ar(ACD) .....(i)

also,

ED is the median of \(\triangle \)ABC.

\(\therefore \) ar(EBD) = ar(ECD) — (ii)

Subtracting (ii) from (i),

ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)

\(\Rightarrow \) ar(ABE) = ar(ACE)

Q2 ) In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = \(\frac{1}{4} \) ar (ABC).

Answer :

ar(BED) = \(\frac{1}{2} \)×BD×DE
(Since, E is the mid-point of AD,)
AE = DE
(Since, AD is the median on side BC of triangle ABC,)
BD = DC

DE = \(\frac{1}{2} \) AD — (i)
BD = \(\frac{1}{2} \)BC — (ii)
From (i) and (ii), we get,
ar(BED) = \(\frac{1}{2} \)×\(\frac{1}{2} \)BC × \(\frac{1}{2} \)AD
\(\Rightarrow \) ar(BED) = \(\frac{1}{2} \)×\(\frac{1}{2} \)ar(ABC)
\(\Rightarrow \) ar(BED) = \(\frac{1}{4} \) ar(ABC)

Hence, proved.

Q3 ) Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer :

O is the mid point of AC and BD. (diagonals of bisect each other)

In \(\triangle \) ABC, BO is the median.
\(\therefore \) ar(AOB) = ar(BOC) .... (i)
also,
In \(\triangle \) BCD, CO is the median.
\(\therefore \) ar(BOC) = ar(COD) ...... (ii)
In \(\triangle \) ACD, OD is the median.
\(\therefore \) ar(AOD) = ar(COD) ....... (iii) In \(\triangle \) ABD, AO is the median.
\(\therefore \) ar(AOD) = ar(AOB) — (iv)

From equations (i), (ii), (iii) and (iv), we get,
ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)

Hence, we get, the diagonals of a parallelogram divide it into four triangles of equal area.

Q4 ) In Figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).

Answer :

In \(\triangle{ACD}\),
Line-segment CD is bisected by AB at O.

So, AO is the median of \(\triangle{ACD}\).

\(\therefore \) Area (\(\triangle{ACO}\)) = Area (\(\triangle{ADO}\)) ...(i)

Similarly,
In \(\triangle{BCD}\), BO is the median.
\(\therefore \) Area (\(\triangle{BCO}\)) = Area (\(\triangle{BDO}\)) ...(ii)

Now, adding equations (i) and (ii), we get,

Area (\(\triangle{ACO}\)) + Area (\(\triangle{BCO}\)) = Area (\(\triangle{ADO}\)) + Area (\(\triangle{BDO}\))
\(\therefore \) Area (\(\triangle{ABC}\)) = Area (\(\triangle{ABD}\))
Hence, proved.

Q5 ) D, E and F are respectively the mid-points of the sides BC, CA and AB of a \(\triangle{ABC}\). Show that
i) BDEF is a parallelogram.
ii) ar(DEF) = \(\frac{1}{4} \) ar(ABC)
iii) ar(BDEF) = \(\frac{1}{2} \) ar(ABC)

Answer :

(i) In \(\triangle \) ABC,
EF || BC and EF = \(\frac{1}{2} \) BC (by mid point theorem)
also,
BD = \(\frac{1}{2} \) BC (D is the mid point)
So, BD = EF
also,
BF and DE are parallel and equal to each other.
\(\therefore \) the pair opposite sides are equal in length and parallel to each other.
\(\therefore \) BDEF is a parallelogram.


(ii) Proceeding from the result of (i),

BDEF, DCEF, AFDE are parallelograms.
Diagonal of a parallelogram divides it into two triangles of equal area.
\(\therefore \) ar(?BFD) = ar(?DEF) (For parallelogram BDEF) — (i)
also,
ar(\(\triangle \) AFE) = ar(\(\triangle \) DEF) (For parallelogram DCEF) — (ii)
ar(\(\triangle \) CDE) = ar(\(\triangle \) DEF) (For parallelogram AFDE) — (iii)

From (i), (ii) and (iii) \( ar(\triangle BFD) = ar(\triangle AFE) = ar(\triangle CDE) = ar(\triangle DEF) \)
\(\Rightarrow ar(\triangle BFD) +ar(\triangle AFE) +ar(\triangle CDE) +ar(\triangle DEF) = ar(\triangle ABC) \)
\(\Rightarrow 4 ar(\triangle DEF) = ar(\triangle ABC ) \)
\(\Rightarrow \) ar(DEF) = \(\frac{1}{4} \) ar(ABC)


(iii) Area (parallelogram BDEF) = ar(\(\triangle \)DEF) +ar(\(\triangle \) BDE)
\(\Rightarrow \) ar(parallelogram BDEF) = ar(\(\triangle \) DEF) +ar(\(\triangle \) DEF)
\(\Rightarrow \) ar(parallelogram BDEF) = 2× ar(\(\triangle \) DEF)
\(\Rightarrow \) ar(parallelogram BDEF) = 2× \(\frac{1}{4} \) ar(\(\triangle \) ABC)
\(\Rightarrow \) ar(parallelogram BDEF) = \(\frac{1}{4} \) ar(\(\triangle \) ABC)

Hence, proved.

Q6 ) In Figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
i) ar (DOC) = ar (AOB)
ii) ar (DCB) = ar (ACB)
iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]

Answer :

Construction: Let us draw \(DN\perp{AC}\) and \(BN\perp{AC}\).



i) In (\(\triangle{DON}\)) and (\(\triangle{BOM}\)),

\(\angle{DNO}\) = \(\angle{BMO}\) ...(By construction)
\(\angle{DON}\) = \(\angle{BOM}\)
(Vertically opposite angles)
Also, OD = OB ...(Given)
\(\therefore \) \(\triangle{DON}\) \(\displaystyle \cong\) \(\triangle{BOM}\)
(By AAS congruency test)
Thus, DN = BM ...(i)(CPCT)
But, we know that, congruent triangles have equal areas.
\(\therefore \) Area (\(\triangle{DON}\)) = Area (\(\triangle{BOM}\)) ...(ii)

Now, in (\(\triangle{DNC}\)) and (\(\triangle{BMA}\)),
\(\angle{DNC}\) = \(\angle{BMA}\) ...(By construction)
DN = BM ...(Using Equation (i))
Also, CD = AB ...(Given)
\(\therefore \) \(\triangle{DNC}\) \(\displaystyle \cong\) \(\triangle{BMA}\)
(By RHS congruency test)
\(\therefore\) Area (\(\triangle{DNC}\)) = Area (\(\triangle{BMA}\)) ...(iii)
On adding eq. (ii)and (iii), we get,
Area \(\triangle{DON}\) + Area \(\triangle{DNC}\) = Area \(\triangle{BOM}\) + Area \(\triangle{BMA}\)
\(\therefore \) Area \(\triangle{DOC}\) = Area \(\triangle{AOB}\)


ii)We have, Area \(\triangle{DOC}\) = Area \(\triangle{AOB}\)

Adding Area \(\triangle{OCB}\) on both sides, we get,

\(\Rightarrow \) Area \(\triangle{DOC}\) + Area \(\triangle{OCB}\) = Area \(\triangle{AOB}\) + Area \(\triangle{OCB}\)
\(\Rightarrow \) Area \(\triangle{DCB}\) = Area \(\triangle{ACB}\)


iii)Now, we have, Area \(\triangle{DCB}\) = Area \(\triangle{ACB}\)

We also know that, if two triangles have the same base and equal areas, then these will lie between the same parallels.

\(\Rightarrow \) DA || CB
In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD ) and the other part of opposite sides is parallel (DA || CB)

\(\therefore \) we can say that, ABCD is a parallelogram.
Hence, proved.

Q7 ) D and E are points on sides AB and AC respectively of \(\triangle{ABC}\) such that ar (DBC) = ar (EBC). Prove that DE || BC.

Answer :

\(\because \) \(\triangle{BCE}\) and \(\triangle{BCD}\) are lying on a common base BC and also have equal areas.

\(\therefore\) \(\triangle{BCE}\) and \(\triangle{BCD}\) will lie between the same parallel lines.
\(\Rightarrow \) DE || BC

Hence, it is proved that DE || BC

Q8 ) XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively,
show that ar (ABE) = ar (ACF)

Answer :

XY || BC = EY || BC ...(Given)
Also, BE || AC = BE || CY
\(\therefore\) we can say that, EBYC is a parallelogram.
Similarly,
XY || BC = XF || BC
Also, FC || AB = FC || XB
\(\therefore \) BCFX is a parallelogram.


Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF.

\(\therefore \) Area (EBCY) = (\(\frac{1}{2} \) ) Area (BCFX) ...(i)

Now, considering parallelogram EBYC and \(\triangle{AEB}\),

We can say that, these lie on the same base BE and are between the same parallels BE and AC.

\(\therefore \) Area (\(\triangle{AEB}\)) = (\(\frac{1}{2} \) ) Area (EBYC) ...(ii)

Also, parallelogram BCFX and \(\triangle{ACF}\) are on the same base CF and between the same parallels CF and AB.

Thus, Area (\(\triangle{ACF}\)) = (\(\frac{1}{2} \) ) Area (BCFX) ...(iii)

So, from eq.(i), (ii) and (iii), we get,

Area (\(\triangle{AEB}\)) = (\(\triangle{ACF}\))
Hence, proved.

Q9 ) The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Figure).
Show that ar (ABCD) = ar (PBQR).
[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Answer :

Construction: Let us join AC and PQ.



\(\triangle{ACQ}\) and \(\triangle{AQP}\) are the same base AQ and between the same parallels AQ and CP.

\(\therefore \) Area \(\triangle{ACQ}\) = Area \(\triangle{AQP}\)
\(\Rightarrow Area (\triangle{ACQ} ) - Area (\triangle{ABQ} ) = Area (\triangle{APQ} ) - Area (\triangle{ABQ})\)
\(\Rightarrow \) Area \((\triangle{ABC}\)) = Area (\(\triangle{QBP}\)) ...(i)


\(\because \) AC and PQ are diagonals of parallelograms ABCD and PBQR, respectively
\(\therefore \) Area (\(\triangle{ABC}\)) = (\(\frac{1}{2} \) ) Area (ABCD) ...(ii)

Also, Area (\(\triangle{QBP}\)) = (\(\frac{1}{2} \) ) Area (PBQR) ...(iii)

From equations (i),(ii) and (iii), we get,

(\(\frac{1}{2} \) ) Area (ABCD) = (\(\frac{1}{2} \) ) Area(PBQR)
\(\therefore \) Area (ABCD) = Area(PBQR)
Hence, proved.

Q10 ) Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

Answer :

It can be observed that \(\triangle{DAC}\) and \(\triangle{DBC}\) lie on the same base DC and between the same parallels AB and CD.

\(\therefore \) Area \( (\triangle{DAC}\)) = Area (\(\triangle{DBC}\))
\(\Rightarrow Area (\triangle{DAC}) - Area (\triangle{DOC}) = Area (\triangle{DBC}) - Area (\triangle{DOC})\)
Area (\(\triangle{AOD}\)) = Area (\(\triangle{BOC}\))
Hence, proved.

Q11 ) In Figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
i) ar (ACB) = ar (ACF)
ii) ar (AEDF) = ar (ABCDE).

Answer :

i) \(\triangle{ACB}\) and \(\triangle{ACF}\) lie on the same base AC and are between The same parallels AC and BF.

\(\therefore \) Area (\(\triangle{ACB}\)) = Area (\(\triangle{ACF}\))


ii) \(\because \) it can be observed that
Area (\(\triangle{ACB}\)) = Area (\(\triangle{ACF}\))

Now adding Area (ACDE) on both side,

\(\Rightarrow \) Area (\(\triangle{ACB}\)) + Area (ACDE) = Area (\(\triangle{ACF}\)) + Area (ACDE)

\(\Rightarrow \) Area (ABCDE) = Area (AEDF)
Hence, proved.

Q12 ) A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre.
Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot.
Explain how this proposal will be implemented.

Answer :

Let quadrilateral ABCD be the original shape of the field.



The proposal may be implemented as follows:

Join diagonal BD and draw a line parallel to BD through point A.
Let it meet the extended side CD of ABCD at point E.

Join BE and AD. Let them intersect each other at O.

Then, portion \(\triangle{AOB}\) can be cut from the original field so that the new shape of the field will be \(\triangle{BCE}\) as shown below.



We have to prove that the area of \(\triangle{AOB}\) (portion that was cut so as to construct Health Centre) is equal to the area of \(\triangle{DEO}\)(portion added to the field so as to make the area of the new field so formed equal to the area of the original field).

It can be observed that \(\triangle{DEB}\) and \(\triangle{DAB}\) lie on the same base BD and are between the same parallels BD and AE.

Area (\(\triangle{DEB}\)) = Area (\(\triangle{DAB}\))

Now subtracting area (\(\triangle DOB \) ) On both side

Area (\(\triangle{DEB}\)) - Area (\(\triangle{DOB}\)) = (\(\triangle{DAB}\)) - Area (\(\triangle{DOB}\))

\(\therefore \) Area (\(\triangle{DEO}\)) = Area (\(\triangle{AOB}\))

Q13 ) ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.
Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]

Answer :

It can be observed that, \(\triangle{ADX}\) and \(\triangle{ACX}\) lie on the same base AX and are between the same parallels AB and DC.

\(\therefore \) Area (\(\triangle{ADX}\)) = Area (\(\triangle{ACX}\)) ...(i)

Also, \(\triangle{ADY}\) and \(\triangle{ACX}\) lie on the same base AC and are between the same parallels AC and XY.

\(\therefore \) Area (\(\triangle{ACY}\)) = Arca (\(\triangle{ACX}\)) ...(ii)

From Equations (i) and (ii), we get,

Area (\(\triangle{ADX}\)) = Area (\(\triangle{ACY}\))

Hence, proved.

Q14 ) In Figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

Answer :

It can be observed that, \(\triangle{ABQ}\) and \(\triangle{PBQ}\) lie on the same base BQ and are between the same parallels AP and BQ.

\(\therefore \) Area (\(\triangle{ABQ}\)) = Area (\(\triangle{PBQ}\)) ...(i)

Also, \(\triangle{BCQ}\) and \(\triangle{BRQ}\) lie on the same BQ and are between the same parallels BQ and CR.

\(\therefore \) , Area (\(\triangle{BCQ}\)) = Arca (\(\triangle{BRQ}\)) ...(ii)

From Equations (i) and (ii), we get,
\(\Rightarrow \) Area (\(\triangle{ABQ}\)) + Area(\(\triangle{BCQ}\)) = Area (\(\triangle{PBQ}\)) + Area (\(\triangle{BRQ}\))

\(\Rightarrow \) Area (\(\triangle{AQC}\)) = Area (\(\triangle{PBR}\))
Hence, proved.

Q15 ) Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC).
Prove that ABCD is a trapezium.

Answer :

Given:
Area (\(\triangle{AOD}\)) = Area (\(\triangle{BOC}\))

\(\Rightarrow \) Area (\(\triangle{AOD}\)) + Area (\(\triangle{AOB}\)) = (\(\triangle{BOC}\)) + Area (\(\triangle{AOB}\))

\(\therefore \) Area (\(\triangle{ADB}\)) = Area (\(\triangle{ACB}\))

We know that triangles on the same base having area equal to each other lie between the same parallels.

\(\therefore \) these triangles, \(\triangle{ADB}\) and \(\triangle{ACB}\), are lying between the same parallels.i.e., AB || CD

\(\therefore \) ABCD is a trapezium.
Hence, proved.

Q16 ) In Figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Answer :

Given:
Area (\(\triangle{DRC}\)) = Area (\(\triangle{DPC}\)).

As, \(\triangle{DRC}\) and \(\triangle{DPC}\) lie on the same base DC and have equal area, therefore, they must lie between the same parallel lines.

\(\because \) DC || RP
\(\therefore \) DCPR is a trapezium.

Also, Area (\(\triangle{BDP}\)) = Area (\(\triangle{ARC}\)).

\(\Rightarrow \) Area (\(\triangle{BDP}\)) - Area (\(\triangle{DPC}\)) = Area (\(\triangle{ARC}\)) - Area (\(\triangle{DRC}\))

\(\therefore \) Area (\(\triangle{BDC}\)) = Area(\(\triangle{ADC}\))

\(\because \) \(\triangle{BDC}\) and \(\triangle{ADC}\) lie on the same base CD and have equal areas, they must lie between the same parallel lines.

\(\therefore \) AB || CD
Hence, it is proved that ABCD is a trapezium.

Q1 ) 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 :

We have a parallelogram ABCD and rectangle ABEF such that

ar(||gm ABCD) = ar( rect. ABEF)

AB = CD [Opposite sides of parallelogram]
and AB = EF [Opposite sides of a rectangle]
\(\Rightarrow \) CD = EF
\(\Rightarrow \) AB + CD = AB + EF … (1)

BE < BC and AF < AD [In a right triangle, hypotenuse is the longest side]

\(\Rightarrow \) (BC + AD) > (BE + AF) …(2)

From (1) and (2), we have

(AB + CD) + (BC+AD) > (AB + EF) +( BE + AF)
\(\Rightarrow \) (AB + BC + CD + DA) > (AB + BE + EF + FA)
\(\Rightarrow \) Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.

Q2 ) In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).

Answer :

Let us draw AF, perpendicular to BC such that AF is the height of \(\triangle \)ABD, \(\triangle \)ADE and \(\triangle \)AEC



\(\therefore \) area of triangle =( \(\frac{1}{2} \)) × base ×Height
\(\therefore \) Area of \(\triangle \) ABD= (\(\frac{1}{2} \) ) × BD × AF
\(\therefore \) Area of \(\triangle \) ADE = (\(\frac{1}{2} \) ) × DE × AF
\(\therefore \) Area of \(\triangle \) AEC = (\(\frac{1}{2} \) ) × EC × AF
Since , BD= DE= EC (Given )
\(\therefore \) Area of \(\triangle \) ABD= Area of \(\triangle \) ADE= Area of \(\triangle \) AEC

Q3 ) In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF).

Answer :

Since, ABCD is a parallelogram [Given]
\(\therefore \)Its opposite sides are parallel and equal.
\(\Rightarrow \) AD = BC …(1)

Now, \(\triangle \)ADE and \(\triangle \)BCF are on equal bases AD = BC [from (1)] and between the same parallels AB and EF.

So, ar(\(\triangle \)ADE) = ar(\(\triangle \)BCF).

Q4 ) In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(\(\triangle \)BPC) = ar(\(\triangle \)DPQ).[Hint Join AC.]

Answer :

We have a parallelogram ABCD and AD = CQ. Let us join AC.

We know that triangles on the same base and between the same parallels are equal in area.

Since, \(\triangle \)QAC and \(\triangle \)QDC are on the same base QC and between the same parallels AD and BQ.

\(\therefore ar(\triangle QAC) = ar(\triangle QDC) \)
Subtracting ar(\(\triangle \) QPC) from both sides, we have

\(ar(\triangle QAQ – ar(\triangle QPC) = ar(\triangle QDC) – ar(\triangle QPC) \)

\(\Rightarrow \)ar(\(\triangle PAQ = ar(\triangle QDP) \) …(1)

Since, \(\triangle \)PAC and \(\triangle \)PBC are on the same base PC and between the same parallels AB and CD.

\(\therefore ar(\triangle PAC) = ar(\triangle PBC) \)…(2)

From (1) and (2), we get

ar(\(\triangle PBC) = ar(\triangle QDP) \)

Q5 ) In 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) = \(\frac{1}{4} \) ar (ABC)
(ii) ar (BDE) = \(\frac{1}{2} \) ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = \(\frac{1}{8} \) ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]

Answer :

Let us join EC and AD. Draw EP \(\perp \) BC.
Let AB = BC = CA = a, then
BD = \(\frac{a}{2} \) = DE = BE



(i) area of \(\triangle ABC = \frac{\sqrt{3}}{4} × a^2 \)
And area of \(\triangle BDE = \frac{\sqrt{3}}{4} × \frac{a}{2} ^2 \)
= \(\frac{\sqrt{3}}{4} × \frac{a^2}{4} \)
\(\therefore \) ar (BDE) = \(\frac{1}{4} \) ar (ABC)


(ii) Since, \(\triangle \) ABC and \(\triangle \) BED are equilateral triangles.

\(\Rightarrow \) \(\angle \)ACB = \(\angle \)DBE = 60°
\(\Rightarrow \) BE || AC

\(\triangle \)BAE and \(\triangle \)BEC are on the same base BE and between the same parallels BE and AC.

\(\therefore \) ar(\(\triangle \)BAE) = ar(\(\triangle \)BEC)
\(\Rightarrow \) ar(\(\triangle \)BAE) = 2 ar(\(\triangle \)BDE)
[ DE is median of \(\triangle \)EBC. ]
\(\Rightarrow \) ar(\(\triangle \) BDE) = \(\frac{1}{2} \) ar(\(\triangle \)BAE)


(iii) ar(\(\triangle \)ABC) = 4 ar(\(\triangle \)BDE)
[Proved in (i) part]
ar(\(\triangle \)BEC) = 2 ar(\(\triangle \)BDE)
[ \(\because \)DE is median of \(\triangle \) BEC]
\(\Rightarrow \) ar(\(\triangle \)ABC) = 2 ar(\(\triangle \)BEC)


(iv) Since, \(\triangle \)ABC and \(\triangle \)BDE are equilateral triangles.

\(\Rightarrow \) \(\angle \)ABC = \(\angle \)BDE = 60°
\(\Rightarrow \) AB || DE

\(\triangle \)BED and \(\triangle \)AED are on the same base ED and between the same parallels AB and DE.

\(\therefore \) ar(\(\triangle \)BED) = ar(\(\triangle \)AED) Subtracting ar(AEFD) from both sides, we get

\(\Rightarrow \) ar(\(\triangle \)BED) – ar(\(\triangle \)EFD) = ar(\(\triangle \)AED) – ar(\(\triangle \)EFD)
\(\Rightarrow \) ar(\(\triangle \)BEE) = ar(\(\triangle \)AFD)


(v) In right angled \(\triangle \) ABD, we get
\( AD^2 = AB^2 - BD^2 = a^2 - (\frac{a}{2} )^2 = \frac{3a^2}{4} \)
AD = \(\frac{\sqrt{3} a}{2} \)
In right triangle PED
Similarly,
EP = \(\frac{\sqrt{3} a}{4} \)
Area of triangle AFD = (\(\frac{1}{2} \) ) × FD × AD
= (\(\frac{1}{2} \) ) × FD × \(\frac{\sqrt{3} a}{2} \).....(i) And area of triangle EFD = (\(\frac{1}{2} \)) × FD × EP
= (\(\frac{1}{2} \) ) × FD × \(\frac{\sqrt{3} a}{4} \) .....(ii)

From (1) and (2), we get

ar(\(\triangle \) AFD) = 2 ar((\(\triangle \)EFD)
ar(\(\triangle \)AFD) = ar(\(\triangle \)BEF) [From (iv) part]
\(\Rightarrow \) ar(\(\triangle \)BFE) = 2 ar(\(\triangle \)EFD)


(vi) ar(\(\triangle \)AFC) = ar(\(\triangle \)AFD) + ar(\(\triangle \)ADC)
= ar(\(\triangle \)BFE) + \(\frac{1}{2} \) ar(\(\triangle \)ABC) [From (iv) part]
= ar(\(\triangle \)BFE) + (\(\frac{1}{2} \) ) x 4 x ar(\(\triangle \)BDE) [From (i) part]
= ar(\(\triangle \)BFE) + 2ar(\(\triangle \)BDE)
= 2ar(\(\triangle \)FED) + 2[ar(\(\triangle \)BFE) + ar(\(\triangle \)FED)]
= 2ar(\(\triangle \)FED) + 2[2ar(\(\triangle \)FED) + ar(\(\triangle \)FED)]
[From (v) part]
= 2ar(\(\triangle \)FED) + 2[3ar(\(\triangle \)FED)]
= 2ar(\(\triangle \)FED) + 6ar(\(\triangle \)FED)
= 8ar(\(\triangle \)FED)
\(\Rightarrow \) ar(\(\triangle \)FED) = \(\frac{1}{8} \) ar(\(\triangle \)AFC)

Q6 ) Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB) x ar(CPD) = ar(APD) x ar(BPC). [Hint From A and C, draw perpendiculars to BD.]

Answer :

We draw AM \(\perp \) BD and CN \(\perp \) BD

Area of triangle = \(\frac{1}{2} \) × b × h

ar(APB) × ar(CPD )
= [\(\frac{1}{2} \) × BP × AM ]×[ \(\frac{1}{2} \) × PD × CN ]
= \(\frac{1}{4} \) × BP × AM × PD × CN

And ar(APD) × ar(BPC )
=[ \(\frac{1}{2} \) × PD × AM] × [\(\frac{1}{2} \) × BP × CN ]
= \(\frac{1}{4} \) × BP × AM × PD × CN


\(\therefore \) ar(APB) x ar(CPD) = ar(APD) x ar(BPC).

Q7 ) P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = \(\frac{1}{2} \) ar(ARC)
(ii) ar(PBQ) = ar(ARC)
(iii) ar(RQC) = \(\frac{3}{8} \) ar(ABC)

Answer :

We have a \(\triangle \)ABC such that P is the mid-point of AB and Q is the mid-point of BC.
Also, R is the mid-point of AP. Let us join AQ, RQ, PC and PC.


(i) In \(\triangle \) APQ , R is the mid point of AP(Given )
\(\therefore \) RQ is the median of \(\triangle \) APQ
\(\Rightarrow \) ar(PRQ) = \(\frac{1}{2} \) ar (APQ) ..... (1 )

In triangle ABQ ,P is the mid point of AB
\(\therefore \) QP is the median of \(\triangle \) ABQ
\(\Rightarrow \) ar(APQ) = \(\frac{1}{2} \) ar (ABQ) ..... (2 )

from (1) and (2) we get
\(\Rightarrow \) ar(PRQ) = \(\frac{1}{2} × \frac{1}{2} \) ar (ABQ)
\(\Rightarrow \) ar(PRQ) = \(\frac{1}{4} \) ar (ABQ) = \(\frac{1}{4} × \frac{1}{2} \) ar (ABC)
[ \(\because \) AQ is the Median of \(\triangle \) ABC ]
\(\therefore \) ar(PRQ) = \(\frac{1}{8} \) × ar (ABC).....(3)

Now ar(ARC) = \(\frac{1}{2} \) ×ar(APC)
[ \(\because \) CR is a Median of \(\triangle APC \) ]

\(\Rightarrow \) ar(ARC) = \(\frac{1}{2} \frac{1}{2} \) × ar(ABC)
[ \(\because \) CP is a Median of \(\triangle ABC \) ]
\(\therefore \) ar(ARC) = \(\frac{1}{4} \) × ar(ABC) .....(4)

Now from (3) we get
ar(PRQ) = \(\frac{1}{8} \) (ABC)
ar(PRQ) = \(\frac{1}{2} \frac{1}{4} \) ar(ABC)
ar(PRQ) = \(\frac{1}{2} \) (ARC)


(iii) In triangle RBC , RQ is a Median
\(\therefore \) ar(RQC) = ar(RBQ)
\(\Rightarrow \) ar(RQC) = ar(PRQ) + ar(BPQ)
\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} \)× ar(ABC) + ar(BPQ)
\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} \) × ar(ABC) + ar(BPQ)
\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} × ar(ABC) + \frac{1}{2} \) ar(BPQ)
[\(\because \) PQ is the Median of BPC ]
\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} × ar(ABC) + \frac{1}{2} \frac{1}{2} \) ar(ABC)
[\(\because \) CP is the Median of ABC ]
\(\Rightarrow \) ar(RQC) = ( \(\frac{1}{4} + \frac{1}{8} \) ar(ABC)

Thus , ar(RQC) = ( \(\frac{3}{8} \) ar(ABC)


(ii) QP is the Median of \(\triangle \) ABQ
\(\therefore \) ar\(\triangle \) PBQ = \(\frac{1}{2} \) ar(ABQ)
= \(\frac{1}{2}× \frac{1}{2} \) × ar(ABC)
= \(\frac{1}{4} \) × ar(ABC) = ar(ARC )

From equation (4)
Thus ar(PBQ) = ar(ARC)

Q8 ) In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX \(\perp \) DE meets BC at Y. Show that

(i) \(\triangle \)MBC \(\cong \) \(\triangle \)ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ax(ABMN)
(iv) \(\triangle \)FCB \(\cong \) \(\triangle \)ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)

Answer :

We have a right \(\triangle \)ABC such that BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively. Line segment AX \(\perp \) DE is also drawn such that it meets BC at Y.


(i) \(\angle \)CBD = \(\angle \)MBA [Each90°]
\(\therefore \angle CBD + \angle ABC = \angle MBA + \angle ABC \)

(By adding \(\angle \)ABC on both sides)
\(\Rightarrow \) \(\angle \)ABD = \(\angle \)MBC

In \(\triangle \)ABD and \(\triangle \)MBC, we have
AB = MB [Sides of a square]
BD = BC
\(\angle \)ABD = \(\angle \)MBC [Proved above]
\(\therefore \) \(\triangle \)ABD \(\cong \) \(\triangle \)MBC [By SAS congruency]


(ii) Since parallelogram BYXD and \(\triangle \)ABD are on the same base BD and between the same parallels BD and AX.

\(\therefore \) ar(\(\triangle \) ABD) = \(\frac{1}{2} \) ar(||gm BYXD)

But \(\triangle \)ABD \(\cong \) \(\triangle \)MBC [From (i) part]
[Since, congruent triangles have equal areas.]
\(\therefore \) ar(\(\triangle \)MBC) = \(\frac{1}{2} \) ar(||gm BYXD)
\(\Rightarrow \) ar(||gm BYXD) = 2ar (\(\triangle \)MBC)


(iii) Since, ar(||gm BYXD) = 2ar(\(\triangle \)MBC) …(1) [From (ii) part]
and ar(\(\square \) ABMN) = 2ar(\(\triangle \)MBC) …(2)
[ABMN and AMBC are on the same base MB and between the same parallels MB and NC]

From (1) and (2), we have

ar(BYXD) = ar(ABMN) .



(iv) \(\angle \)FCA = \(\angle \)BCE (Each 90°)
\(\Rightarrow \) \(\angle \)FCA+ \(\angle \)ACB = \(\angle \)BCE+ \(\angle \)ACB
[By adding \(\triangle \)ACB on both sides]
\(\Rightarrow \) \(\angle \)FCB = \(\angle \)ACE
In \(\triangle \)FCB and \(\triangle \)ACE, we have
FC = AC [Sides of a square]
CB = CE [Sides of a square]
\(\angle \)FCB = \(\angle \)ACE [Proved above]
\(\Rightarrow \) \(\triangle \)FCB \(\cong \) \(\triangle \)ACE [By SAS congruency]


(v) Since, ||gm CYXE and \(\triangle \)ACE are on the same base CE and between the same parallels CE and AX.

\(\therefore \) ar(||gm CYXE) = 2ar((\(\triangle \)ACE)
But \(\triangle \)ACE \(\cong \) \(\triangle \)FCB [From (iv) part]
[Since, congruent triangles are equal in areas.]

\(\therefore \) ar (||gm CYXE) = 2ar(\(\triangle \)FCB)


(vi) Since, ar(||gm CYXE) = 2ar (\(\triangle \)FCB) …(3) [From (v) part]

Also (quad. ACFG) and \(\triangle \) FCB are on the same base FC and between the same parallels FC and BG.

\(\Rightarrow \) ar(quad. ACFG) = 2ar((\(\triangle \)FCB) …(4)

From (3) and (4), we get
ar( CYXE) = ar(ACFG) …(5)



(vii) We have ar(BCED) = ar(CYXE) + ar(BYXD)
= ar(CYXE) + ar(ABMN)
[From (iii) part]
Thus, ar (BCED) = ar( ABMN) + ar(ACFG)
[From (vi) part]