Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

Exercise 9.1

Question 1.
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Solution.
In Fig. (i), APDC and trapezium ABCD lies on the same’ base DC and between the same parallel lines AB and DC.
In Fig. (iii), ATRQ and parallelogram PQRS lies on the same base RQ and between the same parallel lies RQ and SP.
In Fig. (v), quadrilateral APCD and quadrilateral ABQD lies on the same base AD and between the same parallel lines AD and BQ.

 

Exercise 9.2

Question 1.

In the given figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB =16 cm,
AE = 8 cm and CF = 10 cm, find AD.
Solution.

Given, AB = 16 cm, AE = 8 cm and CF = 10 cm, D E
We know that
Area of a parallelogram = Base x Height = DC x AE
[ ∵  opposite sides of a parallelogram are equal i.e., AB = DC = 16 cm]
= 16 x 8 = 128 cm2…………….(i)
Area of a parallelogram = AD x CF
= AD x 10         …(ii)
From Eqs. (i) and (ii), we have
AD x 10 = 128
⇒ AD == 12.8 cm.
Hence, the value of AD is 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 (||gm EFGH) =  ar (||gm ABCD).
Solution.

Given: E, F, G, and H are respectively mid-points of the sides AB, BC, CD, and AD.
To prove : ar (||gm EFGH) =  (||gm ABCD)
Construction : Join HF.
Proof : Since, H and F are mid-points of AD and BC, respectively.
∴ DH = AD and CF =  BC ……..(i)
Now, as ABCD is a parallelogram.
∴  AD = DC and AD || BC
 AD = BC and DH || CF
⇒ DH = CF and DH || CF
So, HDCF is a parallelogram. [∵ a pair of opposite sides are equal and parallel]
Now, as the parallelogram HDCF and Δ HGF stand on the same base HF and lie between the same parallel lines DC and HF.
Similarly, ar (Δ HGF) =  ar ( ||gm  HBFH)……….(ii)
similarly, ar (Δ HGF) =  ar ( ||gm  ABFH)……….(iii)
On adding eqs. (ii) and (iii), we get
ar ( Δ HGF) + ar (Δ HEF) = i [ar (||gm HDCF) + ar (||gm ABFH)]
⇒ ar (EFGH) =  ar( ||gm ABCD)
Hence Proved

Question 3.
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (Δ APB) ar (Δ BQC).
Solution.

Given: In parallelogram ABCD, P and Q are any two points lying on the sides DC, and AD, respectively.
To prove : ar (Δ APB) = ar (Δ BQC)
Proof : Here, parallelogram ABCD and ABQC stand on the same base BC and lie between the same parallel BC and AD.
ar(ΔBQC) =  ar (||gm ABCD)        … (i)
Similarly, Δ APB and parallelogram ABCD stand on the same base AB and lie between the same parallel AB and CD.
∴ ar(Δ APB) =  ar (||gm ABCD)   … (ii)
From eqs. (i) and (ii), we get
ar(Δ APB) = ar(Δ BQC)
Hence Proved.

Question 4.
In the figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (Δ APB) + ar(Δ PCD) =  ar(||gm ABCD)
(ii) ar (Δ APD) + ar (Δ APBC) = ar (Δ APB) + ar(Δ PCD).
[Hint: Through P, draw a line parallel to AB]

Solution.

Given : ABCD is a parallelogram.
i.e., AB || CD and AD || BC
To prove :
(i) ar (Δ APB) + ar (Δ PCD) =  ar(||gm ABCD)
(ii) ar(Δ APD) + ar (Δ PBC) = ar (Δ APB) + ar (Δ PCD)
Proof:
(i) Through the point P, draw MR parallel to AB.
∵ MR || AB and AM || BR [∵ AD || BC]
So, ABRM is a parallelogram.
Now, as Δ APB and parallelogram ABRM are on the same base AB and between the same parallels AB and MR.
∴ ar (Δ APB) =  ar (||gm ABRM)
ar(Δ PCD) = ar (||gm MRCD)
Now, ar (ΔAPB) + ar (ΔPCD) =  ar (IP ABRM) +  ar (||gm  MRCD)
=  ar (||gm  ABCD) …(i)

(ii) Clearly, ar (||gm  ABCD) = ar (ΔAPD) +ar (ΔPBC) + ar (Δ APB) + ar(Δ PCD)
= ar (ΔAPD) + ar (ΔPBC) + ar (||gm ABCD)     [from eq. (i)]
∴ ar(Δ APD) + ar (Δ PBC) = ar (||gm  ABCD) –  ar (||gm  ABCD)
ar (||gm  ABCD)()
⇒ ar(Δ APD) + ar(Δ PBC) = – ar (||gm ABCD)   … (ii)
From eqs. (i) and (ii), we get
ar (Δ APD) + ar (Δ PBC) = ar (Δ APB) + ar (Δ PCD) Hence proved.

Question 5.
In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (||9m PQRS) = ar (||gm ABRS)
(ii) ar (Δ AXS) =  ar (||gm PQRS)

Solution.
Given: PQRS and ABRS both are parallelograms and X is any point on BR.
To Prove : (i) ar (||gm PQRS = ar (||gm ABRS)
(ii) ar (Δ AXS) =  ar (||gm PQRS)
Proof:
(i) Here, parallelograms PQRS and ABRS lie on the same base SR and between the same parallel lines SR and
∴ ar (||gm PQRS) = ar (||gm ABRS)   … (i)

(ii) Again, Δ AXS and parallelogram ABRS lie on the same base AS and between the same parallel lines AS and
∴  ar (Δ AXS) =  ar (||gm ABRS)      … (ii)
From eqs. (i) and (ii), we get
ar (Δ AXS) =  ar (||gm PQRS)
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?
Solution.

Given, PQRS is a parallelogram and A is any point on RS. Now, join PA and  PQ. Thus, the field will fie divided into three parts and each part is in the shape of triangle.
Since, the Δ APQ and parallelogram PQRS lie on the same base PQ and between same parallel lines PQ and SR.
ar(Δ APQ) =  ar( ||gm PQRS)    …(i)
Then, remaining
ar(Δ ASP) + ar(Δ ARQ) = ar(||gm PQRS)
Now, from Eqs. (i) and (ii), we get
ar(Δ  APQ) = ar(Δ ASP) + ar(Δ ARQ)
So, farmer has two options.
Either the farmer should sow wheat and pulses in Δ APS and Δ AQR or in ar [Δ AQP and (Δ APS and Δ AQR)] separately.

 

Exercise 9.3

Question 1.
In the given figure, E is any point on median AD  of a Δ ABC. Show that
ar (Δ ABE) = ar(Δ ACE).
Solution.

Given: AD is a median of Δ ABC and E is any point on AD.
To prove: ar (Δ ABE) – ar (Δ ACE)
Proof: Since AD is the median of Δ ABC.
∴ ar(Δ ABD) = ar (Δ ACD)    …(i)
[since a median of a triangle divides it into two triangles of equal areas]
Also, ED is the median of AF.RC
∴ ar (ABED) = ar (ACED)      … (ii)
[since a median of a triangle divides it into two triangles of equal areas] On subtracting eq. (ii) from eq. (i), we get
ar(Δ ABD) – ar (Δ BED) = ar (Δ ACD) – ar (ΔCED)
⇒  ar (Δ ABE) = ar (Δ ACE)
Hence proved

Question 2.
In a AABC, E is the mid-point of median AD. Show that
ar(ΔBED) =  ar (Δ ABC).
Solution.

Given: ABC is a triangle and E is the mid-point of the median AD.
To prove :     ar (Δ BED) =  ar (Δ ABC)
Proof : We know that the median divides a triangle into two triangles of equal areas.
∴ ar (Δ ABD) = ar(Δ  ADC)
⇒  ar (Δ ABD) =  ar (Δ ABC)  …(i)
In Δ ABD, BE is the median.
ar (ΔBED) = ar (ΔBAE)
⇒  ar(ΔBED) = ar (ΔABD)
⇒ ar(ΔBED) = . ar (ΔABC)
[from eq. (i)]
⇒ ar(ΔBED) = ar (ΔABC)
Hence proved.

Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution.

Given: ABCD is a parallelogram and its diagonals AC and BD intersect each other at O.
To prove: Diagonals AC and BD divide parallelogram ABCD into four triangles of equal areas.

i.e., ar (Δ OAB) = ar (Δ OBC) = ar (Δ OCD) = ar (Δ OAD)
Proof: We know that the diagonals of a parallelogram bisect each other, so we have
OA = OC and OB = OD.
Also, we know that a median of a triangle divides it into two triangles of equal areas.
Now, as in Δ ABC, BO is the median.
∴ ar (Δ OAB) = ar (Δ OBC)  … (i)
In Δ ABD, AO is the median.
∴  ar (Δ OAB) = ar (Δ OAD) … (ii)
Similarly, in Δ ACD, DO is the median.
ar (Δ AOD) = ar (Δ OCD)     …(iii)
From eqs. (i), (ii) and (iii), we get
ar (Δ OAB) = ar (Δ OBC) = ar (ΔOCD) = ar (Δ OAD) lienee proved.

Question 4.
In the given figure, ABC and ABD are two triangles on the same base AB. If the line-segment CD is bisected by AB at O, show that ar(Δ ABC )= ar( Δ ABD)
Solution.

Given: ABC and ABD are two triangles on the same base AB.
To prove : ar (ΔABC) = ar (Δ ABD)
Proof: Since the line segment CD is bisected by AB at O.
∴ OC = OD
In Δ ACD, we have OC = OD
So, AO is the median of ΔACD.
Also, we know that the median divides a triangle into two triangles of equal areas.
∴ ar (Δ AOC) = ar (ΔAOD)  … (i)
Similarly, in ABCD,
ar (ABOC) = ar (ABOD)    … (ii)
[Since, BO is the median of ABCD]
On adding eqs. (i) and (ii), we get ar (Δ AOC) + ar (Δ BOC) = ar (Δ AOD) + ar (Δ BOD)
⇒  ar (Δ ABC) = ar (Δ ABD)
Hence proved.

Question 5.
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.Show that
(i) BDEF is a parallelogram.
(ii) ar (ΔDEF) =  ar (ΔABC).
(iii) ar (||gm BDEF) = ar (AABC).
Solution.

Given: ABC is a triangle in which the mid-points of sides BC, CA and AB
(i) are respectively D, E, and F.
To prove :
BDEF is a parallelogram.
(ii) ar (ΔDEF) =  ar (AABC)
(iii) ar (||gm BDEF) =  ar (AABC)
(i) Proof:
Since, E and F are the mid-points of AC and AB.
BC || FE and FE =  BC =BD [by mid-point theorem]
=> BD || FE and BD = FE
Similarly, BF || DE and BF = DE
Hence, BDEF is a parallelogram.
[∵ a pair of opposite sides are equal and parallel]

(ii) Similarly, we can prove that both FDCE and AFDE are also parallelograms.
Now, BDEF is a parallelogram, so its diagonal FD divides it’s into two triangles of equal areas.

Question 6.
In the adjoining figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that
(i) ar (Δ DOC) = ar (Δ AOB).
(ii) ar (Δ DCE) = ar (Δ ACB.)”
(iii) DA || CB or ABCD is a parallelogram.

Solution.
Given: ABCD is a quadrilateral in which AB = CD and its diagonals AC and BD intersect at O such that
OB = OD.

Question 7.
D and E are points on sides AB and AC respectively of Δ ABC such that ar (Δ DBO = ar (Δ EBC). Prove that DE || BC.
Solution.

Given: A Δ ABC and D and E are points on sides AB and AC respectively, such that
ar (Δ DBC) = ar (Δ EBC).
To prove:   DE || BC
Proof: Here, ADBC and AEBC are equal in area and have same base BC.
∴ Altitude from D of Δ DBC Altitude from E of Δ EBC
Hence, Δ DBC and Δ AEBC are between the same parallels.
i.e., DE || BC

Question 8.
XY is a line parallel to side BC of an AABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar (Δ ABE) = ar (Δ ACF).
Solution.

Given A Δ ABC in which XY || BC, BE || AC,
i.e. BE || CY and CF || AB, i.e., CF || XB.
To prove :  ar (Δ ABE) = ar (Δ ACF)
Proof: Since, XY || BC and CY || BE. So, EYCB is a parallelogram.
Now, as Δ ABF and parallelogram EYCB lie on the same base BE and CA.
ar (AABE) =  ar (||gm  BCYE)     …(i)
Again, CF || BX and ZF || BC So, BCFX is a parallelogram.
Now, as Δ ACF and parallelogram BCFX lie on the same base CF and between the same parallel lines AB and FC.
ar (Δ ACF) =  ar (||gm BCFX)    …(ii)
Also, parallelograms BCFX and BCYE lie. on the same base BC and between the same parallels BC and EF.
ar (||gm BCFX) = ar (||gm BCYE) …(iii)
From eqs. (i), (ii) and (iii), we get
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 the figure).
Show that ar (||gm ABCD) = ar (||gm PBQR)
[Hint: Join AC and PQ. Now compare ar (Δ ACQ) and ar (Δ APQ).

Solution.
Given: Two parallelograms ABCD and PBQR.
To prove: ar Q F ABCD) = ar 0 F PBQR)
Construction: Join AC and PQ.
Proof: Since, PQ and AC are diagonals of parallelograms PBQR and ABCD, respectively,

Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that
ar  (Δ AOD) = ar (Δ BOC).

Solution.
Given: Diagonals AC and BD of a trapezium ABCD  with AB || DC intersect each other at 0.

Question 11.
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (Δ ACB) = ar (Δ ACF).
(ii) ar (||gm  AEDF) = ar (pentagon ABCDE). a b
Solution.

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 comers to construct a Health Center. Itwaari agrees to the above proposal with the condition that he should be given an 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?
Solution.

Let ABCD be the plot of land in the shape of a quadrilateral. Let the portion ADE be taken over by the Gram Panchayat of the village from one corner D to construct a Health Center. 
Join AC and draw a line through D parallel to AC to meet BC produced at P. Then, Itwaari must be given the land ECP adjoining his plot so as to form a triangular plot ABP.
Justification

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 (ΔADX) = ar (Δ ACY). [Hint: Join CX.]
Solution.

Ex 9.3 Class 9 Maths Question 14.
In figure, AP || BQ || CR. Prove that ar(Δ AQC) = ar(Δ PHR)

Solution.

Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that
ar (Δ AOD) = ar (Δ BOO. Prove that ABCD is a trapezium.
Solution.


Question 16.
In the figure, ar (Δ DRC) = ar (Δ DPC) and ar (Δ BDP) = ai(Δ ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Solution.

 

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.
Solution.
We have a parallelogram ABCD and rectangle ABEF such that
ar (parallelogram ABCD) = ar (rectangle ABEF)

Question 2.
In figure, D and E are two points on BC such that BD = DE = EC. Show that
ar (Δ ABD) = ar (Δ ADE) = ar (Δ  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 n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide Δ ABC into n triangles of equal areas.]

Solution.
Given : ABC is a triangle, D and E are two points on BC, such that BD = DE – EC.
To prove :     ar (Δ  ABD) = ar (Δ ADE) = ar (Δ AEC)
Proof: Let AO be the perpendicular to BC.
We know that,

Question 3.
In the figure, ABCD, DCFE and ABFE are parallelograms.Show that ar (Δ ADE) = ar (Δ BCF).
Solution.

Given : ABCD, DCFE and ABFE are parallelograms.
To prove : ar (Δ ADE) = ar (Δ BCE)

Question 4.
In the figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, then show that ar (Δ BPO = ar (Δ DPQ.) [Hint: Join AC]
Solution.

Question 5.
In the figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, then show that
(i) ar (Δ BDE) = ar (Δ ABO)
(ii) ar (Δ BDE) = ar (Δ BAE)
(iii) ar (Δ ABC) = 2 ar (Δ BEO
(iv) ar (Δ BFE) = ar (Δ AFD)
(v) ar (Δ BFE) = 2 ar (Δ FED)
(vi) ar (Δ FED) =  ar (Δ AFC).
[Hint: Join EC and AD. Show that BE || AC and DE || AB, etc.]

Solution.

Join AD and EC
Let x be the side of equilateral Δ ABC.




Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
Show that  ar (ΔAPB) x ar (Δ CPD)
= ar (Δ APD) x ar (Δ BPC). [Hint: From A and C, draw perpendiculars to BD].
Solution.

Question 7.
P and Q are respectively the mid-points of sides AB and BC of a Δ ABC and R is the mid-point of AP, show that
(i) ar (Δ PRQ) =  ar (Δ ARC)
(ii) ar (Δ RQC = ar (Δ ABC)
(iii) ar (Δ PBQ) = ar (Δ ARC).
Solution.



Question 8.
In the figure, ABC is a right angled 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 (Δ MBC)
(iii) ar (BYXD) = ar (Δ BMN)
(iv) ΔFCB ≅  AACE
(v) ar (CYXE) = 2 ar (Δ FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG).
[Note: Result (vii) is the famous Theorem of Pythagoras. You shall term a similar proof by this Theuonern is class X.]

Solution.
Given: ABC is a right  angled triangle in which ∠A = 90°. BCED, ACFG and ABMN are squares. Line segment AX ⊥ DE meets BC at Y