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Areas Of Parallelograms And Triangles

Class 9th Mathematics CBSE Solution
Exercise 9.1
  1. Which of the following figures lie on the same base and between the same parallel. In such…
Exercise 9.2
  1. In Fig. 9.15, ABCD is a parallelogram, AE DC and CF AD. If AB = 16 cm, AE = 8 cm…
  2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram…
  3. P and Q are any two points lying on the sides DC and AD respectively of a…
  4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that…
  5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR.…
  6. A farmer was having a field in the form of a parallelogram PQRS. She took any…
Exercise 9.3
  1. In Fig.9.23, E is any point on median AD of ar ABC. Show that ar (ABE) = ar…
  2. In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) =1/4…
  3. Show that the diagonals of a parallelogram divide it into four triangles of…
  4. In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line-…
  5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC.…
  6. In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that…
  7. D and E are points on sides AB and AC respectively of ABC such that ar (DBC) =…
  8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB…
  9. The side AB of a parallelogram ABCD is produced to any point P. A line through A…
  10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at…
  11. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC…
  12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram…
  13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and…
  14. In Fig.9.28, AP || BQ || CR. Prove that: ar (AQC) = ar (PBR)
  15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that…
  16. In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the…
Exercise 9.4
  1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal…
  2. In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar…
  3. In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar…
  4. In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that…
  5. In Fig.9.33, ABC and BDE are two equilateraltriangles such that D is the…
  6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that…
  7. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and…
  8. In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are…

Exercise 9.1
Question 1.

Which of the following figures lie on the same base and between the same parallel. In such a case, write the common base and the two parallels.



Answer:

(i) Yes, It can be observed that trapezium ABCD and triangle PCD have a common base CD and these are lying between the same parallel lines AB and CD

(ii) No, It can be observed that parallelogram PQRS and trapezium MNRS have a common base RS. However, their vertices, (i.e., opposite to the common base) P, Q of parallelogram and M, N of trapezium, are not lying on the same line


(iii) Yes, It can be observed that parallelogram PQRS and triangle TQR have a common base QR and they are lying between the same parallel lines PS and QR


(iv) No, It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC. However, these do not have any common base


(v) Yes, It can be observed that parallelogram ABCD and parallelogram APQD have a common base AD and these are lying between the same parallel lines AD and BQ


(vi) No, It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However, these do not lie between the same parallel lines.




Exercise 9.2
Question 1.

In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.


Answer:

We know area of parallelogram = Base × Height

For ABCD,

AE ⊥ DC

Hence AE is the height and DC is the base.

Area of ABCD = DC × AE

As we know sides of parallelogram are equal.

So, DC=AB=16 cm

Hence,

Area of ABCD = 16×8 …. (1)

Also, for ABCD,

CF ⊥ AD

Area of ABCD = CF × AD

= 10 × AD …. (2)

From 1 and 2,

16×8 = 10 × AD

=12.8 cm


Question 2.

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


Answer:

Let us join HF


In parallelogram ABCD,


AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)


AB = CD (Opposite sides of a parallelogram are equal)


AD = BC


And AH || BF


AH = BF and AH || BF (H and F are the mid-points of AD and BC)


Therefore, ABFH is a parallelogram


Since,


ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF


Therefore,


Area of triangle HEF = × Area (ABFH) (i)


Similarly,


It can also be proved that


Area of triangle HGF = × Area (HDCF) (ii)


On adding (i) and (ii), we get


Area of triangle HEF + Area of triangle HGF = Area (ABFH) + Area (HDCF)


Area (EFGH)= [Area (ABFH) + Area (HDCF)]


Area (EFGH) = Area (ABCD)


Question 3.

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 ΔBQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC

If a triangle and parallelogram are on the same base and have the same altitude,
the area of the triangle will be half that of the parallelogram.

Area (ΔBQC) = Area (ABCD) (i)


Similarly,

If a triangle and parallelogram are on the same base and have the same altitude,
the area of the triangle will be half that of the parallelogram.

ΔAPB and parallelogram ABCD lie on the same base AB
and between the same parallel lines AB and DC


Area (ΔAPB) = Area (ABCD) (ii)


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


Area (ΔBQC) = Area (ΔAPB)


Question 4.

In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (APB) + ar (PCD) = ar(ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

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




Answer:

(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.

In parallelogram ABCD,


AB || EF (By construction) ... (i)


AD || BC (Opposite sides of a parallelogram)


So, AE || BF ... (ii)


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


Quadrilateral ABFE is a parallelogram.


It can be observed that ΔAPB and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF


Therefore,


Area of triangle APB = × Area (ABFE) ... (iii)


Similarly, we show that EFCD is a parallelogram,


For triangle PCD and parallelogram EFCD, lying on same base CD and same parallel lines,


Area of triangle PCD = × Area (EFCD).....(iv)


Adding (iii) and (iv), we get


Area of triangle APB + Area of triangle PCD = [Area (ABFE) + Area (EFCD)]


Area of triangle APB + Area of triangle PCD = Area (ABCD).... (v)


(ii) Let us draw a line segment MN, passing through point P and parallel to line segment AD


In parallelogram ABCD,


MN || AD (By construction) ... (vi)


ABCD is a parallelogram


AB || DC (Opposite sides of a parallelogram)


So, AM || DN (vii)


From equations (vi) and (vii), we get


MN || AD and


AM || DN


Therefore,


Quadrilateral AMND is a parallelogram.


It can be observed that ΔAPD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN


Area (ΔAPD) = Area (AMND) (viii)


Similarly,


For ΔPCB and parallelogram MNCB,


Area (ΔPCB) = Area (MNCB) (ix)


Adding equations (viii) and (ix), we get


Area (ΔAPD) + Area (ΔPCB) = [Area (AMND) + Area (MNCB)]


Area (ΔAPD) + Area (ΔPCB) = Area (ABCD) (x)


On comparing equations (v) and (x), we get


Area (ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area (ΔPCD)


Question 5.

In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
(i) ar (PQRS) = ar (ABRS)

(ii) ar (AX S) =ar (PQRS)



Answer:

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

Area (PQRS) = Area (ABRS) (i)

(ii) Consider ΔAXS and parallelogram ABRS

As these lie on the same base AS and are between the same parallel lines AS and BR,
(ii)

From equations (i) and (ii), we obtain



Hence proved.


Question 6.

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, it can be observed that point A divides the field into three parts. These parts are triangular in shape − ΔPSA, ΔPAQ, and ΔQRA




Area of ΔPSA + Area of ΔPAQ + Area of ΔQRA = Area of Parallelogram PQRS ..............eq(i)

We know that if a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram

Therefore,

Area (ΔPAQ) = Area (PQRS) .............eq(ii)

From equations (i) and (ii), we obtain

Area (ΔPSA) + Area (ΔQRA) + 1/2 Area (PQRS) = Area (PQRS)

Area (ΔPSA) + Area (ΔQRA) = 1/2 Area (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.



Exercise 9.3
Question 1.

In Fig.9.23, E is any point on median AD of ar Δ ABC. Show that ar (ABE) = ar (ACE).



Answer:

AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas

Area (ΔABD) = Area (ΔACD) (i)


ED is the median of ΔEBC


Area (ΔEBD) = Area (ΔECD) (ii)


On subtracting equation (ii) from equation (i), we get,


Area (ΔABD) − Area (EBD) = Area (ΔACD) − Area (ΔECD)


Area (ΔABE) = Area (ΔACE)


Question 2.

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


Answer:

Given: Δ ABC, with AD as median i.e., BD = CD and E is the mid-point of AD, i.e., AE = DE

To prove: ar (BED) = 1/4 ar (ABC).

Proof: AD is a median of Δ ABC and median divides a triangle into two triangles of equal area

∴ ar(ABD) = ar(ACD)

⇒ ar (ABD) = 1/2 ar (ABC) …(1)

In Δ ABD,

BE is median (As E is mid-point of AD)

median divides a triangle into two triangles of equal area

∴ ar(BED) = ar(BEA)

⇒ ar (BED) = 1/2 ar (ABD)

⇒ ar (BED) = 1/2 × 1/2 ar (ABC) (from (1): ar (ABD) = 1/2 ar (ABC))

⇒ ar (BED) = 1/4 ar (ABC)

Hence proved.


Question 3.

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


Answer:

We know that diagonals of a parallelogram bisect each other. Therefore, O is the mid-point of AC and BD

BO is the median in ΔABC. Therefore, it will divide it into two triangles of equal areas

Area (ΔAOB) = Area (ΔBOC) (i) In ΔBCD,

CO is the median


Area (ΔBOC) = Area (ΔCOD) (ii)

Similarly, Area (ΔCOD) = Area (ΔAOD) (iii)

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

Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD)

Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area


Question 4.

In Fig. 9.24, 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:


Consider ΔACD.

Line-segment CD is bisected by AB at O.Therefore, AO is the median of ΔACD


Area (ΔACO) = Area (ΔADO) (i)


Considering ΔBCD, BO is the median


Area (ΔBCO) = Area (ΔBDO) (ii)


Adding equations (i) and (ii), we get


Area (ΔACO) + Area (ΔBCO) = Area (ΔADO) + Area (ΔBDO)


Area (ΔABC) = Area (ΔABD)


Question 5.

D, E and F are respectively the mid-points of the sides BC, CA and AB of a Δ ABC. Show that:

(i) BDEF is a parallelogram.

(ii) ar (DEF) =ar (ABC)

(iii) ar (BDEF) =ar (ABC)


Answer:

(i) In triangle ABC it is given that,


EF is parallel to BC

And,

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of
those sides is parallel to the third side and is half the length of the third side.

EF = BC ( By using Mid – point theorem)

Also,

BD = BC (As D is the mid-point)

So,

BD = EF

BF and DE also parallel to each other

Therefore, the pair of opposite sides are equal and parallel in length.
Similarly BF = DE

Therefore,

BDEF is a parallelogram

(ii) Diagonal of a parallelogram divides it into two equal area

Therefore,

In parallelogram BDEF
DF is the diagonal

⇒ Area of triangle BFD = Area of triangle DEF ......(1)
In parallelogram DCEF

ED is the diagonal

⇒ Area of triangle AFE = Area of triangle DEF .....(2)

In parallelogram AFDE
EF is the diagonal.

⇒ Area of triangle CDE = Area of triangle DEF ......(3)

So from (1), (2) and (3)

Area of triangle BFD = Area of triangle AFE = Area of triangle CDE = Area of triangle DEF

Area of triangle ABC = Area of triangle BFD + Area of triangle AFE + Area of triangle CDE + Area of triangle DEF

4 (Area of triangle DEF) = Area of triangle ABC

Area of triangle DEF = × Area of triangle

(iii) Area of parallelogram BDEF = Area of triangle DEF + Area of triangle BDE

From part (ii)

Area of parallelogram BDEF = Area of triangle DEF + Area of triangle DEF


= 2 × Area of triangle DEF


= 2 × × Area of triangle ABC


= × Area of triangle ABC


Question 6.

In Fig. 9.25, 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:

Let us draw DN perpendicular to AC and BM perpendicular to AC


(i) In ΔDON and ΔBOM,


∠DNO = ∠BMO ( 90°)


∠DON = ∠BOM (Vertically opposite angles)


OD = OB (Given)


By AAS congruence rule,

ΔDON ΔBOM

DN = BM (BY CPCT) ..... (1)

We know that congruent triangles have equal areas.

Area (ΔDON) = Area (ΔBOM) ..... (2)

In ΔDNC and ΔBMA,

∠DNC = ∠BMA (Angles made by perpendicular)

CD = AB (Given)

DN = BM [Using equation (i)]

ΔDNC ΔBMA (RHS congruence rule)

Area (ΔDNC) = Area (ΔBMA) (iii)

Let us draw DN AC and BM AC

On adding equations (ii) and (iii), we obtain

Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)

Therefore,

Area (ΔDOC) = Area (ΔAOB)


(ii) TO prove :Area (ΔDCB) = Area (ΔACB)
we have proved,
Area (ΔDOC) = Area (ΔAOB)

Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)

(Adding Area (ΔOCB) to both sides)

Area (ΔDCB) = Area (ΔACB)


(iii)
We have proved above,
Area (ΔDCB) = Area (ΔACB)

If two triangles have the same base and equal areas, then these will lie between the same parallels

So, DA || CB (iv)

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

Therefore, ABCD is a parallelogram


Question 7.

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


Answer:


Since ΔBCE and ΔBCD are lying on a common base BC and also have equal areas,

ΔBCE and ΔBCD will lie between the same parallel lines

Therefore,


DE || BC
Hence proved


Question 8.

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:

It is given that:

In Δ ABC, XY||BC & BE||AC & CF||AB

To prove: Ar(Δ ABE) =Ar(Δ ACF)

Proof:

Let XY intersect AB & BC at points M and N respectively.

The figure of the question is:

Now, XY || BC

And, we know that the parts of the parallel lines are parallel

⇒ EN || BC .... (1)

Also, it is given that BE || AC

⇒ BE || CN .... (2)
From 1 and 2

BCNE is a parallelogram (because both the pair of opposite sides are parallel)

Now, the parallelogram BCNE & ΔABE have a common base i.e. BE & BE is parallel to AC.

⇒ ar(ΔABE) = ½ area of parallelogram BCNE ...(i)

Similarly, we BCFM is also a parallelogram.

And, the ΔACF and the parallelogram BCFM lie on the same base i.e. CF

⇒ ar(ΔACF) = ½ area of parallelogram BCFM ...(ii)

Now,

Parallelograms BCNE and BCFM are on the same base BC and between the same parallels BC and EF
hence there area will be equal.

⇒ Area (BCNE) = Area (BCFM)...(iii)

From the equations (i), (ii) and (iii), we can write that,

Ar(Δ ABE) = Ar(Δ ACF)


Question 9.

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 Fig. 9.26). Show that ar (ABCD) = ar (PBQR).


[Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ)]


Answer:

Let us join AC and PQ


ΔACQ and ΔAQP are on the same base AQ and between the same parallels AQ and CP.


Area (ΔACQ) = Area (ΔAPQ)


Area (ΔACQ) − Area (ΔABQ) = Area (ΔAPQ) − Area (ΔABQ)


Area (ΔABC) = Area (ΔQBP) (i)


Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,


Area (ΔABC) = Area (ABCD) (ii)


Area (ΔQBP) = Area (PBQR) (iii)


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


Area (ABCD) = Area (PBQR)


Area (ABCD) = Area (PBQR)


Question 10.

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 ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD

Area (ΔDAC) = Area (ΔDBC) [Area of triangles with same base and between the same parallels are equal in area.]


Area (ΔDAC) − Area (ΔDOC) = Area (ΔDBC) − Area (ΔDOC)


Area (ΔAOD) = Area (ΔBOC)


Question 11.

In Fig. 9.27, 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) ΔACB and ΔACF lie on the same base AC and are between the same parallels AC and BF

Area (ΔACB) = Area (ΔACF)


(ii) It can be observed that:


Area (ΔACB) = Area (ΔACF)


Area (ΔACB) + Area (ACDE) = Area (ACF) + Area (ACDE)


Area (ABCDE) = Area (AEDF)



Question 12.

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 soas 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 ΔAOB can be cut from the original field so that the new shape of the field will be ΔBCE



We have to prove that the area of ΔAOB (portion that was cut so as to construct Health Centre)

is equal to the area of ΔDEO (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 ΔDEB and ΔDAB lie on the same base BD and are between the same parallels BD and AE

So,

Area (ΔDEB) = Area (ΔDAB)


Area (ΔDEB) − Area (ΔDOB) = Area (ΔDAB) − Area (ΔDOB)


Area (ΔDEO) = Area (ΔAOB)


Question 13.

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 ΔADX and ΔACX lie on the same base AX and are between the same parallels AB and DC.


Area (ΔADX) = Area (ΔACX) (i)


ΔACY and ΔACX lie on the same base AC and are between the same parallels AC and XY


Area (ΔACY) = Area (ACX) (ii)


From equations (i) and (ii), we obtain


Area (ΔADX) = Area (ΔACY)


Question 14.

In Fig.9.28, AP || BQ || CR. Prove that:

ar (AQC) = ar (PBR)



Answer:

Since ΔABQ and ΔPBQ lie on the same base BQ and are between the same parallels AP and BQ

Area (ΔABQ) = Area (ΔPBQ) (i)


Again, ΔBCQ and ΔBRQ lie on the same base BQ and are between the same parallels BQ and CR


Area (ΔBCQ) = Area (ΔBRQ) (ii)


On adding equations (i) and (ii), we obtain


Area (ΔABQ) + Area (ΔBCQ) = Area (ΔPBQ) + Area (ΔBRQ)


Area (ΔAQC) = Area (ΔPBR)


Question 15.

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:


It is given that:

Area (ΔAOD) = Area (ΔBOC)

add area(AOB) to both sides,

Area (ΔAOD) + Area (ΔAOB) = Area (ΔBOC) + Area (ΔAOB)
⇒Area (ΔADB) = Area (ΔACB)

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

Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels i.e.,


AB || CD


Therefore, ABCD is a trapezium.


Question 16.

In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.



Answer:

It is given that:

Area (ΔDRC) = Area (ΔDPC)


As ΔDRC and ΔDPC lie on the same base DC and have equal areas, therefore, they must lie between the same parallel lines


DC || RP


Therefore, DCPR is a trapezium. It is


Also given that:


Area (ΔBDP) = Area (ΔARC)


Area (BDP) − Area (ΔDPC) = Area (ΔARC) − Area (ΔDRC)


Area (ΔBDC) = Area (ΔADC)


Since ΔBDC and ΔADC are on the same base CD and have equal areas, they must lie between the same parallel lines


AB || CD


Therefore,


ABCD is a trapezium.



Exercise 9.4
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:

As the parallelogram and the rectangle have the same base and equal area, therefore, these will also lie between the same parallels.

Consider the parallelogram ABCD and rectangle ABEF as follows



Here, it can be observed that parallelogram ABCD and rectangle ABEF are between the same parallels AB and CF


We know that opposite sides of a parallelogram or a rectangle are of equal lengths


Therefore,


AB = EF (For rectangle)


AB = CD (For parallelogram)


CD = EF


AB + CD = AB + EF (i)


Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest


AF < AD


And similarly,


BE < BC


AF + BE < AD + BC (ii)


From equations (i) and (ii), we obtain


AB + EF + AF + BE < AD + BC + AB + CD


Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD



Question 2.

In Fig. 9.30, 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?



[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into unequal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas]


Answer:

Let us draw a line segment AM perpendicular BC


We know that,


Area of a triangle = * Base * Altitude


Area (ΔADE) = * DE * AM


Area (ΔABD) = * BD * AM


Area (ΔAEC) = * EC * AM


It is given that DE = BD = EC


* DE * AM = * BD * AM = * EC * AM


Area (ΔADE) = Area (ΔABD) = Area (ΔAEC)


It can be observed that Budhia has divided her field into 3 equal parts



Question 3.

In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).



Answer:

It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal

AD = BC (i)


Similarly, for parallelograms DCEF and ABFE


DE = CF (ii)


And,


EA = FB (iii)


In ΔADE and ΔBCF,


AD = BC [Using equation (i)]


DE = CF [Using equation (ii)]


EA = FB [Using equation (iii)]


ΔADE BCF (SSS congruence rule)


Area (ΔADE) = Area (ΔBCF)



Question 4.

In Fig. 9.32, 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 (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) (i)


In quadrilateral ACDQ


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) (ii)


From equations (i) and (ii), we obtain


Area (ΔBPC) = Area (ΔDPQ)



Question 5.

In Fig.9.33, ABC and BDE are two equilateraltriangles such that D is the mid-point of BC. If AEintersects BC at F, show that


(i) ar (BDE) =ar (ABC)

(ii) ar (BDE) =ar (BAE)

(iii) ar (BFE) = ar (AFD)
(iv) ar (ABC) = 2 ar (BEC)

(v) ar (BFE) = 2 ar (FED)

(vi) ar (FED) =ar (AFC)




[Hint: Join EC and AD.

Show that BE || AC and DE || AB, etc]


Answer:

(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 = 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 × ar(ΔBDE)


Hence,


ar (ΔBDE) = ar (ABC)


(ii)Area (ΔBDE) = Area (ΔAED) (Common base DE and DE||AB)


Area (ΔBDE) − Area (ΔFED) = Area (ΔAED) − Area (ΔFED)


Area (ΔBEF) = Area (ΔAFD) (i)


Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)


Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (i)]


Area (ΔABD) = Area (ΔABE) (ii)


AD is the median in ΔABC


Area (ΔABD) = Area (ΔABC)


= Area (ΔBDE)


Area (ΔABD) = 2 Area (ΔBDE) (iii)


From (ii) and (iii), we obtain


2 ar (ΔBDE) = ar (ΔABE)


ar (ΔBDE) = ar (ΔABE)


(iii) ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)


ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)


Using equation (i), we obtain


ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC) ar (ΔABD) = ar (ΔBEC)


ar (ΔABC) = ar (ΔBEC)


ar (ΔABC) = 2 ar (ΔBEC)


(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)


(v)


An in equilateral triangle, median drawn is also perpendicular to the side,

Now,

ar Δ AFD = 1/2 × FD × AD ..... (i)

Draw EG ⊥ BC.

Draw EGBC

ar Δ FED = 1/2 × FD × EG ..... (ii)

Dividing eq. (i) by (ii), we get




As Altitude of equilateral triangle =side



As D is the mid-point of BC.




ar (ΔAFD) = 2 ar (ΔFED) ……(iii)


From part (iii)

ar (BFE) = ar (AFD)

So ar (ΔBFE) = 2 ar (ΔFED)


vi)

ar (Δ AFC) = ar (Δ AFD) + ar (Δ ADC) = 2 ar (Δ FED) +ar (Δ ABC) [using (v)

= 2 ar (Δ FED) + 1/2[4 ×ar (Δ BDE)] [Using result of part (i)]

= 2 ar (Δ FED) + 2 ar (Δ BDE) = 2 ar (Δ FED) + 2 ar (Δ AED)

[Δ BDE and Δ AED are on the same base and between same parallels]

= 2 ar (Δ FED) + 2 [ar (Δ AFD) + ar (Δ FED)]

= 2 ar (Δ FED) + 2 ar (Δ AFD) + 2 ar (Δ FED) [Using (viii)]

= 4 ar (Δ FED) + 4 ar (Δ FED)

⇒ ar (Δ AFC) = 8 ar (Δ FED)

⇒ ar (Δ FED) = 1/8[ar (ΔAFC)]


Question 6.

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


Answer:



Let us construct AM perpendicular to BD

Now,


Area (APB) x Area (CPD)


=


=


Area (APD) x Area (BPC)


=


=


Hence,


Area (APB) x Area (CPD) = Area (APD) x Area (CPB)


Question 7.

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) =ar (ARC)

(ii) ar (RQC) =ar (ABC)

(iii) ar (PBQ) = ar (ARC)


Answer:

Let ABC is a triangle.



P and Q are the midpoints of AB and BC respectively and R is the mid-point of AP.

Join PQ, QR, AQ, PC, RC as shown in the figure.

Now we know that the median of a triangle divides it into two triangles of equal areas.

In triangle CAP, CR is the median.

So Area(ΔCRA) = (1/2)× Area(ΔCAP) .......(1)

Again in triangle CAB, CP is the median.

So Area(ΔCAP) = (1/2)× Area(ΔCPB) .......... (2)

from equation 1 and 2, we get

Area(ΔCAP) = (1/2)×Area(ΔCPB) .............(3)

Again in triangle PBC, PQ is the median.

So (1/2)×Area(ΔCPB) = Area(ΔPBQ) ............(4)

From equation 3 and 4, we get

Area(ΔARC) = Area(ΔPBQ) ...............(5)

Now QP, And QR are the medians of triangle QAB and QAP respectively

So Area(ΔQAP) = Area(ΔQBP) ............(6)

and Area(ΔQAP) = 2× Area(ΔQRP) ............(7)

from equation 6 and 7, we get

Area(ΔPRQ) = (1/2)× Area(ΔPBQ) ............(8)

from equation 5 and 8, we get

Area(ΔPRQ) = (1/2)× Area(ΔARC)

Now CR is the median of triangle CAP

So Area(ΔARC) = (1/2)× Area(ΔCAP)

= (1/2)× (1/2)× Area(ΔABC)

= (1/4)× Area(ΔABC)

Again RQ is the median of triangle RBC

So Area(ΔRQC) = (1/2)× Area(ΔRBC)

= (1/2)× Area(ΔABC) - (1/2)× Area(ΔARC)

=(1/2)× Area(ΔABC) - (1/2)× (1/2)× (1/2)× Area(ΔABC)

=(1/2)× Area(ΔABC) - (1/8)× Area(ΔABC)

= (3/8)× Area(ΔABC)

So Area(ΔRQC) = = (3/8)× Area(ΔABC)


Question 8.

In Fig. 9.34, 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 ⊥ DE meets BC at Y. Show that:



(i) Δ MBC ≅Δ ABD

(ii) ar (BYXD) = 2 ar (MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) Δ FCB ≅Δ ACE

(v) ar (CYXE) = 2 ar (FCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)

Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X


Answer:

(i) Since, each angle of a square is 900

Hence, angle ABM = angle DBC = 900


ABM + ABC = DBC + ABC


MBC = ABD (1)


In ΔMBC and ΔABD,


Angle MBC = Angle ABD (From 1)


MB = AB (Sides of square)


BC = BD (Sides of square)


Therefore,


By SAS congruence rule,



(ii) Now, we have:



And we know that adjacent sides of square are perpendicular to each other. Hence,


BD perpendicular DE


And,


AX perpendicular DE


Also,


We know that two lines perpendicular to the same line are parallel to each other.


Therefore,


BD ‖ AX


Now,


= ( and ‖gm BXYD are on same base BD and between the same parallels BD and AX)


Area (BXYD) = 2 area (ΔABD)


Area (BXYD) = 2 area (ΔMBC) (2)


(iii) ΔMBC and parallelogram ABMN are on same base MB and between the same parallels MB and NC.


Therefore,


Area (ΔMBC) = � ar (ABMN)


2 Area (ΔMBC) = Area (ABMN) = Area(BXYD) (From 2) (3)


(iv) Since, each angle of square is 900


Therefore,


Angle FCA = Angle BCE = 900


FCA + ACB = BCE + ACB


Angle FCB = Angle ACE


Now,


In ΔFCB and ΔACE,


Angle FCB = Angle ACE


FC = AC (Sides of square)


CB = CE (Sides of square)


Therefore,


By SAS congruence,



(v) It is given that AX is perpendicular to DE and CE is perpendicular to DE


and,


We know that two lines perpendicular to the same line are parallel to each other.


Hence,


CE ‖ AX


Now,


ΔACE and parallelogram CYXE are on same base CE and between the same parallels CE and AX.


Therefore,


Area (ΔACE) = 1/2 Area (CYXE)


Area (CYXE) = 2 Area (ΔACE) (4)


We had proved that


ΔFCB ΔACE


Hence,


Area (ΔACE) =Area (ΔFCB) (5)


Comparing 4 and 5, we get:


CYXE = 2 Area (ΔFCB) (6)


(vi) ΔFCB and parallelogram ACFG are on same base CF and between the same parallels CF and BG


Hence,


Area (ΔFCB) = � Area (ACFG)


Area (ACFG) = 2 Area (ΔFCB) = Area (CYXE) (Using 6) (7)


(vii) From figure:


Area (BCED) = Area (BXYD) + Area (CYXE)


Area (BCED) = Area (ABMN) + Area (ACFG) (using 3 and 7)