Gujarat Board Textbook Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles
GSEB Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles
Ex 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:
(i) APDC and quadrilateral ABCD lie on the same base DC and between the same parallels DC and AB.
(iii) TRQ and parallelogram PQRS lie on the same base QR and between the same parallels QR and PS.
(iv) Parallelograms ABCD and APQD lie on the same base AD and between the same parallels AD and BQ.
Ex 9.2
Question 1.
In figure, ABCD is a parallelogram, AE ⊥ DC andCF AD ⊥ AB = 16cm,AE = 8 cm and CF = 10 cm, find AD.
Solution:
ar( || gm ABCD) = AB x AE = 16 x 8 cm2
= 128 cm2 ………..(1)
ar( || gm ABCD) = AD x CF
= AD x 10 cm2 ……….(2)
From (1) and (2), we get
AD x 10 = 128
AD = 128/10
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) = 1/2 ar (ABCD).
OR
Prove that the area of the quadrilateral formed by joining the mid-points of the sides of a parallelogram is half the area of the parallelogram.
Solution:
Given: A parallelogram ABCD, in which E, F, G, and H are the mid-points of sides AB, BC, CD, and DA respectively.
To prove: ar (EFGH) = ar(ABCD)
Construction: Join HF.
Question 3.
P and Q are any two points lying on the sic DC and AD respectively of a parallelogram ABCD. Show that ar(ΔAPB) = ar(ΔBQC).
Solution:
APB and || gm ABCD are on the same AB and between the same parallels AB and DC.
∴ ar(ΔAPB) = 1/2 ar( || gm ABCD) ………..(1)
Again, ABQC and 11gm ABCD are on the same base BC and between the same parallels BC and AD.
∴ ar(BQC) = 1/2 ar( || gm ABCD) ………(2)
From (1) and (2),
ar[ΔAPB] = ar[ΔBQC]
Question 4.
In figure, P is a point in the interior of a parallelogram ABCD. Show that:
1. ar(ΔAPB) + ar(ΔPCD) = ar( || gm ABCD)
2. ar(ΔAPD) + ar(ΔPBC) = ar(ΔAPB) + ar(ΔPCD)
[Hint: Through P, draw a line parallel to AB]
Solution:
Construction: Through P, draw a line EF parallel to AB.
1. EF fi AB …….(1) (by construction)
∴ AD || BC
∴ Opposite sides of a parallelogram are parallel
∴ AE || BF ………..(2)
In view of (1) and (2),
Quadrilateral ABEF is a parallelogram.
A quadrilateral is a parallelogram if its opposite sides are parallel
Similarly, quadrilateral CDEF is a parallelogram.
∴ ΔAPB and fi gm ABFE are on the same base AB and between the same parallels AB and EF.
∴ ar(ΔAPB) = 1/2 ar(|| gm ABFE) …….(3)
Again, LSPCÐ and fi gin CDEF are on the same base DC and between the same parallels DC and EF.
∴ ar(ΔPCD) = 1/2 ar(|| gm CDEF) ………..(4)
Adding (3) and (4), we get
1/2 ar(ΔAPB) + 1/2 ar(ΔPCD)
= 1/2 ar(|| gm ABFE) + 1/2 ar(|| gm in CDEF)
= 1/2 [ar(|| gm ABFE) + ar(|| gm CDEF)
= 1/2 ar(|| gm ABCD) ……….(5)
2. ar(ΔAPD) + ar(ΔPBC)
= ar( || gm ABCD) – [arΔAPB) + ar(ΔPCD)]
= 2[ar(ΔAPB) + ar(ΔPCB)] – [ar(ΔAPB) + ar(ΔPCD)]
= ar(ΔPB) + ar(ΔPCD).
Alternative: Like in part (i), draw a line ∠M, through P, parallel to AD. Hence we can show
ar(ΔAPD) + ar(ΔPBC)
= ar(|| gm ABCD)
From eqn. (5) and (6),
ar(ΔAPD) + ar(ΔPBC)
= 1/2 ar(ΔAPB) + ar(ΔPCD)
Question 5.
In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
1. ar( || gm PQRS) = ar( || gm ABRS)
2. ar(ΔAXS) = 1/2 ar(|| gm PQRS)
Solution:
1. In ΔPSA and ΔQRB,
∠SPA = ∠RQB
Corresponding angles as PS || QR and transversal PB
∠PAS = ∠QBR
Corresponding angles from AS || BR and transversal PB
Also, PS = QR
[Opposite sides of || gm PQRS
ΔPSA = ΔQRB [By AAS rule]
∴ ar(ΔPSA) = ar(ΔQRB) ………..(1)
[∴ congruent figures have equal areas]
Now, ar(|| gm PQRS)
= ar(ΔPSA) + 1/2 ar(||gm AQRS)
= ar(ΔQRB) + 1/2 ar(||gm AQRS) [ using (1) ]
= ar(|| gm ABRS) …….(2)
2. ΔAXS and || gm ABRS are on the same base AS and between the same parallels AS and BR.
∴ ar(ΔAXS) = 1/2 ar( || gm ABRS)
= 1/2 ar( || gm PQRS) [(Using(2))]
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:
∴ ΔAPQ, ΔAPS, and ΔAQR lie between the same parallels.
∴ Their attitudes are the same. Let it be x. Then,
∴ SR = PQ (opposite sides of the parallelogram are equal)
From eqn. (1) and (2),
ar(ΔAPQ) = ar(ΔAPS) + ar(ΔAQR)
Therefore, either the former should sow wheat in ΔAPQ and pulses in the other two triangles APS and AQR or pulses in ΔAPQ and wheat in the other two triangles APS and AQR.
Ex 9.3
Question 1.
In the figure, E is any point on the median AD of a ΔABC. Show that ar(ΔABE) = ar(ΔACE).
Solution:
Given: E is any point on median AD of a ΔABC. In ΔABC,
∴ AD is a median.
∴ ar(ΔABD) = ar(ΔACD) ………(1)
A median of a triangle divides it into two triangles of equal areas.
In ΔEBC,
∴ ED is a median.
∴ ar(ΔEBD) = ar(ΔECD) ………(2)
A median of a triangle divides it into two triangles of equal areas.
Subtracting (2) from (1), we get
ar(ΔABD) ≅ ar(ΔEBD)
= ar(ΔACD) ≅ ar(ΔECD)
⇒ ar(ΔABE) = ar(ΔACE)
Question 2.
In a triangle ABC, E is the mid-point of median AD. Show that
ar(ΔBED) = 1/4 ar(ΔABC)
Solution:
In ΔABC,
Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:
Given: ABCD is a parallelogram whose diagonals AC and BD intersecting at O divide it into four triangles ΔOAB, ΔOBC, ΔOCD and ΔODA.
To Prove: ar(ΔOAB) = ar(ΔOBC)
= ar(ΔOCD) = ar(ΔODA)
Construction: Draw BE ⊥ AC
Proof: Now,
But OA = OC
∴ ar(ΔOAB) = ar(ΔOBC) …….(1)
Similarly,
ar(ΔOBC) = ar(ΔOCD) …….(2)
and, ar(ΔOCD) = ar(ΔODA) …….(3)
From (1), (2) and (3), we get ar(ΔOAB) = ar(ΔOBC)
= ar(ΔOCD) = ar(ΔODA)
Question 4.
In the 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(ΔABC) = ar(ΔABD).
Solution:
Line segment CD is bisected by AB at O.
∴ OC = OD
∴ BO is a median of ABCD, and
AO is a median of AACD.
∴ BO is a median of ABCD.
∴ ar(ΔOBC) = ar(ΔOBD) ………..(1)
∴ A median of a triangle divides it into two triangles of equal areas
Similarly, AO is a median of ΔACD.
∴ ar(ΔOAC) = ar(ΔOAD) ……….(2)
Adding (1) and (2), we get
ar(ΔOBC) + ar(ΔOAC)
= ar(ΔOBD) + ar(ΔOAD)
⇒ ar(ΔABC) = ar(ΔABD)
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) = 1/4 ar(ΔABC)
(iii) ar(□BDEF) = 1/2 ar(ΔABC)
Solution:
(i) In AABC, F is the mid-point of side AB and E is the mid-point of side AC.
∴ EF || BC
∴ In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side.
⇒ EF || BD ………(1)
Similarly, ED || BF ………(2)
In view of (1) and (2),
□BDEF is a parallelogram.
∴ A quadrilateral is a parallelogram if its opposite sides are parallel.
(ii) As in (i), we can prove that
□AFDE and □FDCE are parallelograms,
∴ FD is a diagonal of || gm BDEF.
∴ ar(ΔFBD) = ar(ΔDEF) ……(3)
Similarly, ar(ΔDEF) = ar(ΔFAE) ……(4)
and, ar(ΔDEF) = ar(ΔDCE) ……(5)
From (3), (4) and (5), we have
ar(ΔFBD) = ar(ΔDEF) = ar(ΔFAE) = ar(ΔDCE) ……..(6)
Now, ΔABC is divided into four non-overlapping triangles ΔFBD, ΔDEF, ΔFAE and ΔDCE.
∴ ar(ΔABC) = ar(ΔFBD) + ar(ΔDEF) + ar(ΔFAE) + ar(ΔDCE) = 4 ar(ΔDEF) [From (6)]
⇒ ar(ΔDEF) = 1/4 ar(ΔABC) ……..(7)
(iii) ar(□BDEF)
= ar(ΔFBD) + ar(ΔDEF)
= ar(ΔDEF) + ar(ΔDEF) [From (3)] = 2ar(ΔDEF)
= 2. 1/4 ar(ΔABC) [From (7)]
= 1/2 ar(ΔABC).
Question 6.
In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar(ΔDOC) = ar(ΔAOB)
(ii) ar(ΔDCB) = ar(ΔACB)
(iii) DA || CB or ABCD is a parallelogram
[Hint: From D and B, draw perpendiculars to AC.]
Solution:
Construction: Draw DE ⊥ AC and BF ⊥ AC.
(i) In ΔODE and DOBF,
∠OED = ∠OFB (each 90°)
∠DOE = ∠BOF (Vertically opposite angles)
OB = OB (given)
∴ ΔDOE = ΔBOF (By AAS congruency)
⇒ DE = BF
and OE = OF ……..(1) (CPCT)
Also, ar(ΔDOE) = ar(ΔBOF) ………(2)
Again, in right ΔDEC and ΔBFA,
Hyp. DC = Hyp. BA [Given]
DE = BF (Proved above)
∠DEC = ∠BFA (each 90°)
∴ ΔDEC = ΔBFA | RHS rule
∴ CE = AF ………(3)
Adding eqn. (2) and (4),
ar(ΔDOE) + ar(ΔCDE) = ar(ΔBOF) + ar(ΔABF)
⇒ ar(ΔDOC) – ar(ΔAOB)
(ii) From (i)
ar(ΔDOC) = ar(ΔAOB)
⇒ ar(ΔDOC) + ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB)
Adding equal areas on both sides
⇒ ar(ΔDCB) = ar(ΔACB)
(ii) In point (iii), ADCB and ΔACB having same area lie on the same base AD, therefore, DA || CB
Adding eqn. (1) and (3), we get
OE = CE = OF + AF
⇒ OC = OA
Also, OB = OD (given)
⇒ ABCD is a parallelogram
(∴ If diagonals of quadrilateral bisect each other, ten it is a parallelogram)
Question 7.
D and E are points on sides AB and AC respectively of ΔABC such that ar(ΔDBC) = ar(ΔEBC). Prove that DE || BC.
Solution:
Given: D and E are points on side AB and AC respectively of ΔABC, such that ar(ΔDBC) = ar(ΔEBC)
To prove: DE || BC
Proof: ΔDBC and AEBC are on the same base BC and have equal areas.
∴ Their altitudes must be the same.
∴ DE || BC
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(ΔABE) = ar(ΔACF).
Solution:
XF || BC [Given XY || BC]
and CF || BX [∴ CF || AB (Given)]
∴ □BCFX is a ||gm.
A quadrilateral is a parallelogram I if its opposite sides are parallel
∴ BC = XF
Opposite sides of a parallelogram are equal.
⇒ BC = XY + YF ………(1)
Again,
∴ EY || BC [Given]
and BE || CY [∴ BE || AC (Given)]
∴ ΔBCYE is a parallelogram.
A quadrilateral is a parallelogram I if its opposite sides are parallel.
∴ BC = YE
Opposite sides of a parallelogram are equal.
⇒ BC = XY + XE ………(2)
From (1) and (2),
XY + YF = XY + XE
⇒ YF = XE
⇒ XE = YF ………(3)
∴ ΔAEX and AAYF have equal bases
(∴ XE = YF) on the same line EF and have a common vertex A.
∴ Their altitudes are also the same.
∴ ar(ΔAEX) = ar(ΔAFY) ……..(4)
∴ ΔABEX and ACFY have equal bases (∴ XE = YF) on the same line EF and are between the same parallels EF and BC (∴ XY || BC).
∴ ar(ΔBEX) = ar(ΔCFY) ………(5)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Adding the corresponding sides of (4) and (5), we get
ar(ΔAEX) + ar(ΔBEX) = ar(ΔAFY) + ar(ΔCFY)
⇒ ar(ΔABE) = ar(ΔACF)
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. Show that ar(|| gm ABCD) = ar(|| gm PBQR)
[Hint: Join AC and PQ. Now compare ar(ACQ) and ar(APQ)]
Solution:
To Prove: ar(|| gm ABCD) = ar(|| gm PBQR)
Construction: Join AC and PQ
Proof: AC is a diagonal of || gm ABCD
∴ ar(ΔABC) = 1/2 ar(||gm ABCD) ……..(1)
PCQ is a diagonal of || gm BQRP
ar(ΔBPQ) = 1/2 ar(||gm BQRP) ……..(2)
Also, ar(ΔACQ) = ar(ΔAPQ)
ΔACQ and ΔAPQ are on the same base AQ and between the same. parallels (AQ and CP are equal in area
⇒ ar(ΔACQ) – ar(ΔABQ) = ar(ΔAPQ) – ar(ΔABQ)
Subtracting the same areas from both sides
⇒ ar(ΔABC) = ar(ΔBPQ)
⇒ 1/2 ar(|| gm ABCD) = 1/2 ar(|| gm PBQR) [From (1) and (2)]
⇒ ar(|| gm ABCD) = ar(|| gm PBQR)
Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(ΔAOD) = ar(ΔBOC)
Solution:
ΔABD and ΔABC are on the same base AB and between the same parallels AB and DC.
∴ ar(ΔABD) = ar(ΔABC)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
ar(ΔABD) – ar(ΔAOB) = ar(ΔABC) – ar(ΔAOB)
Subtracting the same areas from both sides.
⇒ ar(ΔAOD) = ar(ΔBOC)
Question 11.
In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that:
(i) ar(ΔACB) = ar(ΔACF)
(ii) ar(□AEDF) = ar(ΔBCDE)
Solution:
(i) ∴ ΔACB and ΔACF are on the same base AC and between the same parallels AC and BF.
[∴ AC || BF (Given)]
∴ ar(ΔACB) = ar(ΔACF)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
(ii) From (i),
ar(ΔACB) = ar(ΔACF)
⇒ ar(ΔACB) + ar(□AEDC) = ar(ΔACF) + ar(□AEDC)
Adding the same areas on both sides.
⇒ ar(ABCDE) = ar(□AEDF)
∴ ar(□AEDF) = ar(ABCDE)
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 Centre. 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 of the shape of a quadrilateral. Let the portion ADE be taken over by the Gram Panchayat of the village from one comer D to construct a Health Centre. Join AC. Draw a line through D parallel to AC to meet BC produced in P. Then Itwaari must be given the land ECP adjoining his plot so as to form a triangular plot ABP as then
ar(ΔADE) = ar(ΔPEC)
Proof: ∴ ΔDAP and ΔDCP are on the same base DP and between the parallels DP and AC.
∴ ar(ΔDAP) = ar(ΔDCP) ………..(1)
∴ Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
⇒ ar(ΔDAP) – ar(ΔDEP) = (ΔDCP) – ar(ΔDEP)
Subtracting the same areas from both sides.
⇒ ar(ΔADE) = ar(ΔPCE)
⇒ ar(ΔADE) + ar(□ABCE) = ar(ΔPCE) + ar(□ABCE)
Adding the same areas to both sides.
⇒ ar(□ABCD) = ar(ΔABP)
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(ΔACY).
[Hint: Join CX]
Solution:
Given: ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.
∴ ΔADX and ΔACX are on the same base AX and between the same parallels AB and DC.
∴ ar(ΔADX) = ar(ΔACX) …….(1)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area,
∴ ΔACX and ΔACY are on the same base AC and between the same parallels AC and XY.
∴ ar(ΔACX) = ar(ΔACY) …….(2)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
From (1) and (2), we get
ar(ΔADX) = ar(ΔACY)
Question 14.
In figure, AP || BQ || CR. Prove that ar(ΔAQC) = ar(ΔPBR).
Solution:
∴ ABAQ and ABPQ are on the same base BQ and between the same parallels BQ and AP.
∴ ar(ΔBAQ) = ar(ΔBPQ) ……..(1)
∴ Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Similarly, ΔBCQ and ΔBQR are on the same base BQ and between the same parallels BQ and CR.
∴ ar(ΔBCQ) = ar(ΔBQR) ……..(2)
Adding the corresponding sides of (1) and (2), we get
ar(ΔBAQ) + ar(ΔBCQ) = ar(ΔBPQ) + ar(ΔBQR)
∴ ar(ΔAQC) = ar(ΔPBR).
Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(ΔAOD) = ar(ΔBOC). Prove that ABCD is a trapezium.
Solution:
Given: ar(ΔAOD) = ar(ΔBOC)
⇒ ar(ΔAOD) + ar(ΔAOB) = ar(ΔBOC) + ar(ΔAOB)
Adding the same areas on both sides.
⇒ ar(ΔABD) = ar(ΔABC)
But ΔABD and ΔABC are on the same base AB.
∴ ΔABD and ΔABC will have equal corresponding altitudes.
∴ ΔABD and ΔABC will lie between the same parallels.
∴ AB || DC
∴ □ABCD is a trapezium.
A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel.
Question 16.
In figure, ar(ΔDRC) = ar(ΔDPC) and ar(ΔBDP) = ar(ΔARC). Show that both the quadrilateral ABCD and DCPR are trapeziums.
Solution:
Given: ar(ΔDRC) = ar(ΔDPC) ……..(1)
But ΔDRC and ΔDPC are on the same base DC.
ΔDRC and ΔDPC will have equal corresponding altitudes.
∴ ΔDRC and ΔDPC will lie between the same parallels.
∴ DC || RP
∴ □DCPR is a trapezium.
A quadrilateral is a trapezium if exactly one pair of opposite sides are parallel.
Again, ar(ΔBDP) = ar(ΔARC)
⇒ ar(ΔBDC) + ar(ΔDPC) = ar(ΔADC) + ar(ΔDRC)
⇒ ar(ΔBDC) = ar(ΔADC) [using (1)]
But ΔBDC and ΔADC are on the same base DC.
∴ ΔBDC and ΔADC will have equal corresponding altitudes.
∴ ΔBDC and ΔADC will lie between the same parallels.
∴ AB || DC
∴ □ABCD is a trapezium
A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel.
Ex 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:
Given: Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas.
To prove: The perimeter of the parallelogram ABCD is greater than that of rectangle ABEF. Proof: Here the parallelogram ABCD and the rectangle ABEF are on the same base AB and between the same parallels AB and FC. Then, perimeter of the parallelogram ABCD = 2(AB + AD)
and, the perimeter of the rectangle ABEF = 2(AB + AF).
In ΔADF, ∠AFD = 90°
∴ ∠ADF is an acute angle. (< 90°) [Angle sum property of a triangle ] ⇒ ∠AFD > ∠ADF
⇒ AD > AF
The side opposite to the greater angle of a triangle is longer.
⇒ AB + AD > AB + AF
⇒ 2(AB + AD) > 2(AB + AF)
∴ Perimeter of the parallelogram ABCD > Perimeter of the rectangle ABEF.
Question 2.
In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ΔABD) = ar(ΔADE) = ar(ΔAEC). 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 ΔABC into n triangles of equal areas.]
Solution:
Let AM ⊥ BC. Then, the height of each of ΔABD, ΔADE and ΔAEC is the same.
∴ ar(ΔABD) = 1/2 x BD x AM ……(1)
ar(ΔADE) = 1/2 x DE x AM ……(2)
and, ar(ΔAEC) = 1/2 x EC x AM ……(3)
∴ BD = DE = EC
∴ From (1), (2) and (3),
ar(ΔABD) = ar(ΔADE) = ar(ΔAEC)
Yes, the field of Budhia has been actually divided into three parts of equal area.
Question 3.
In figure, ABCD, DCFE and ABFE are parallelograms. Show that
ar(ΔADE) = ar(ΔBCF)
Solution:
∴ ABCD is a || gm.
∴ AD || BC …….(1)
[Opposite sides of a || gm are equal Again, DCFE is a || gm.]
∴ DE || CF ……(2)
[Opposite sides of a || gm are equal Also, ABFE is a || gm]
∴ AE = BF ……(3)
[Opposite sides of a || gm are equal From (1), (2) and (3)]
ΔADE ≅ ΔBCF [By SSS congruency]
∴ ar(ΔADE) = ar(ΔBCF)
∴ Two congruent triangles are equal in areas.
Question 4.
In the 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(ΔBPC) = ar(ΔDPQ).
Solution:
Join AC.
∴ AQAC and AQDC are on the same base QC and between the same parallels AD and QC.
∴ ar(ΔQAC) = ar(ΔQDC) ……(1)
Two triangles on the same base (or equal bases) and between the same parallels are equal in areas.
⇒ ar(ΔQAC) ≅ ar(ΔQPC) = ar(ΔQDC) ≅ ar(ΔQPC)
Subtracting the same areas from both sides.
⇒ ar(ΔPAC) = ar(ΔQDP) ……(2)
∴ ΔPAC and ΔPBC are on the same base PC and between the same parallels AB and DC.
∴ ar(ΔPAC) = ar(ΔPBC) …….(3)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
From (2) and (3), we get
ar(ΔPBC) = ar(ΔQDP)
∴ ar(ΔBPC) = ar(ΔDPQ)
Question 5.
In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersect BC at F, show that:
(i) ar(ΔBDE) = 1/4 ar(ΔABC)
(ii) ar(ΔBDE) = 1/2 ar(ΔBAE)
(iii) ar(ΔABC) = 2ar(ΔBEC)
(iv) ar(ΔBFE) = ar(ΔAFD)
(v) ar(ΔBFE) = 2ar(ΔFED)
(vi) ar(ΔFED) = 1/8 ar(AAFC)
[Hint: Join EC and AD. Show that BE || AC and DE || AB, etc.]
Solution:
∴ ΔABC is an equilateral triangle.
∴ ∠ABC = ∠BCA = ∠CAB = 60° ……(1)
∴ ΔBDE is an equilateral triangle.
∴ ∠BDE = ∠DEB = ∠EBD = 60° ……(2)
⇒ ∠ABC = ∠BDE
AB || DE …….(3)
∴ Alternate interior angles. Similarly, ∠ACB = ∠EBC
∴ AC || BE …….(4) (each 60°)
Now, ΔCBA and ΔCEA are on the same base AC and between the same parallels.
∴ ar(ΔCBA) = ar(ΔCEA)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
⇒ ar(ΔABC) = ar(ΔCDA) + ar(ΔCED) + ar(ΔADE) …….(5)
In ΔABC, AD is a medan.
∴ ar(ΔABD) = ar(ΔACD) = 1/2 ar(ΔABC) …….(6)
A median of a triangle divides it into two triangles of equal area.
In ΔEBC, ED is a median.
∴ ar(ΔECD) = ar(ΔEBD) = 1/2 ar(ΔEBC) ……(7)
∴ A median of a triangle divides it into two triangles of equal area.
∴ ΔDEA and ADBE are on the same base DE and between the same parallels AB and DE.
∴ ar(ΔDEA) = ar(ΔDBE) ………(8)
∴ Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Using (6), (7) and (8), in (5) gives
ar(ΔABC) = 1/2 ar(ΔABC) + ar(ΔBDE) + ar(ΔBDE)
⇒ 1/2 ar(ΔABC) = 2ar(ΔBDE)
⇒ ar(ΔBDE) = 1/4 ar(ΔABC)
(ii) ∴ ΔBAE and ΔBCE are on the same base BE and between the same parallels BE and AC.
∴ ar(ΔBAE) = ar(ΔBCE)
[∴ Two triangles on the same base (or equal bases) and between the same parallels are equal in area]
⇒ 1/2 ar(ΔBAE) = 2ar(ΔBDE) | From (7)
⇒ ar(ΔBDE) = 1/4 ar(ΔBAE)
(iii) 2ar(ΔBEC) = 2ar(ΔBDE) [From (7)]
= 4ar(ΔBDE)
= ar(ΔABC) [From (i)]
(iv) ∴ ΔEBD and ΔEAD are on the same base ED and between the same parallels AB and DE.
∴ ar(ΔEBD) = ar(ΔEAD)
[∴ Two triangles on the same base (or equal bases) and between the same parallels are equal in area]
⇒ ar(ΔEBD) – ar(ΔEFD) = ar(ΔEAD) – ar(ΔEFD)
[Subtracting the same areas from both sides]
⇒ ar(ΔBFE) = ar(ΔAFD)
(v) ar(ABDE) = 1/4 ar(ΔABC) [From (i)]
= 1/4 .2 ar(ΔABD)
= 1/4 ar(ΔABD)
∴ Bases of ΔBDE and ΔABD are the same.
∴ Altitude of ABDE = 1/2 Altitude of ΔABD ……..(9)
ar(ΔBEF) = ar(ΔAFD) ……..(10) [From (iv)]
But Altitude of ΔBDE = Altitude of ΔBEF (Considering common vertex)
And, altitude of ΔABD = Altitude of ΔAFD (Considering common vertex A)
Altitude of ΔBEF = 1/2 Altitude of ΔAFD ………(11)
From (10) and (11),
BF = 2FD ……(12)
In ΔBFE and ΔFED,
∴ BF = 2FD
and alt. (ΔBFE) = alt. (ΔFED)
∴ ar(ΔBFE) = 2ar(ΔFED).
Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(ΔAPB) x ar(ΔCPD) = ar(ΔAPD) x ar(ΔBPC)
[Hint: From A and C, draw perpendiculars to BD]
Solution:
Construction: From A and C, draw perpendiculars AE and CF respectively to BD.
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:
1. ar(APRQ) = 1/2 ar(ΔARC)
2. ar(ARQC) = 3/8 ar(ΔABC)
3. ar(ΔPBQ) = ar(ΔARC)
Solution:
Construction: Join AQ and CP.
Question 8.
In figure, ABC is a right triangle right angled at A. BCDE, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:
(i) ΔMBC – ΔABD
(ii) ar(BYXD) = 2ar(ΔMBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ΔFCB = ΔACE
(v) ar(CYXE) = 2ar(ΔFCB)
(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.
Solution:
(i) In ΔMBC and ΔABD,
MB = AB …(1) [Sides of a square]
BC = BD …(2) [Sides of a square]
∠MBA = ∠CBD [Each = 90°]
⇒ ∠MBA + ∠ABC = ∠CBD + ∠ABC
[Adding ∠ABC to both sides]
⇒ ∠MBC = ∠ABD …….(3)
In view of (1), (2) and (3),
⇒ ΔMBC = ΔABD [By congruence SAS rule]
(ii) ar(BYXD) = 2 ar(ΔABD) (∴ BD || AX and common base is BD)
= ar(BYXD) = 2ar(ΔMBC)
[From part (i), ar(ΔMBC) = ar(ΔABD)]
(iii) ∴ ar(BYXD) = 2ar(ΔABD)
and ar(ΔBMN) = 2ar(ΔMBC)
(∴ MB || NC, and common base is MB)
= 2ar(ΔABD) | From (i)
∴ ar(BYXD) = ar(ABMN)
(iv) In ΔFCB and ΔACE,
FC = AC [Sides of a square]
CB = CE [Sides of a square]
∠FCA = ∠BCE [Each = 90°]
⇒ ∠FCA + ∠ACB = ∠BCE + ∠ACB
(Adding the same on both sides)
⇒ ∠FCB = ∠ACE
∴ ΔFCB = ΔACE [SAS congruence rule]
(v) ar(CYXE) = 2ar(ΔACE) = 2ar(ΔFCB) [From (iv)]
∴ AFCB = AACE
∴ ar(AFCB) = ar(AACE)
Congruent As have equal areas
(vi) ∴ ar(CYXE) = 2ar(AACE) = 2ar(AFCB) And, ar(ACFG) = 2ar(AFCB)
(∴ BG || CF and CF is the common base)
∴ ar(CYXE) = ar(ACFG)
(vii) ar(BCED = ar(CYXE) + ar(BYXD)
= ar(ACFG) + ar(ABMN) [From (iii) and (vi)]
= ar(ABMN) + ar(ACFG)