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.
ANSWER
(i) Trapezium ABCD and ΔPDC lie on the same DC and between the same parallel lines AB and DC.
(ii) Parallelogram PQRS and trapezium SMNR lie on the same base SR but not between the same parallel lines.
(iii) Parallelogram PQRS and ΔRTQ lie on the same base QR and between the same parallel lines QR and PS.
(iv) Parallelogram ABCD and ΔPQR do not lie on the same base but between the same parallel lines BC and AD.
(v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and between the same parallel lines AD and BQ.
(vi) Parallelogram PQRS and parallelogram ABCD do not lie on the same base SR but between the same parallel lines SR and PQ.
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
Given,
AB = CD = 16 cm (Opposite sides of a parallelogram)
CF = 10 cm and AE = 8 cm
Now,
Area of parallelogram = Base × Altitude
= CD × AE = AD × CF
⇒ 16 × 8 = AD × 10
⇒ AD = 128/10 cm
⇒ AD = 12.8 cm
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.
To Prove,
ar (EFGH) = 1/2 ar(ABCD)
Construction,
H and F are joined.
Proof,
AD || BC and AD = BC (Opposite sides of a parallelogram)
⇒ 1/2 AD = 1/2 BC
Also,
AH || BF and and DH || CF
⇒ AH = BF and DH = CF (H and F are mid points)
Thus, ABFH and HFCD are parallelograms.
Now,
ΔEFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.
∴ area of EFH = 1/2 area of ABFH --- (i)
also, area of GHF = 1/2 area of HFCD --- (ii)
Adding (i) and (ii),
area of ΔEFH + area of ΔGHF = 1/2 area of ABFH + 1/2 area of HFCD
⇒ area of EFGH = area of ABFH
⇒ ar (EFGH) = 1/2 ar(ABCD)
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).
AnswerΔAPB and ||gm ABCD are on the same base AB and between same parallel AB and DC.
Therefore,
ar(ΔAPB) = 1/2 ar(||gm ABCD) --- (i)
Similarly,
ar(ΔBQC) = 1/2 ar(||gm ABCD) --- (ii)
From (i) and (ii),
we have ar(ΔAPB) = ar(ΔBQC)
In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]
Answer
(i) A line GH is drawn parallel to AB passing through P.
In a parallelogram,
AB || GH (by construction) --- (i)
Thus,
AD || BC ⇒ AG || BH --- (ii)
From equations (i) and (ii),
ABHG is a parallelogram.
Now,
In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH.
∴ ar(ΔAPB) = 1/2 ar(ABHG) --- (iii)
also,
In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.
∴ ar(ΔPCD) = 1/2 ar(CDGH) --- (iv)
Adding equations (iii) and (iv),
ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)}
⇒ ar(APB) + ar(PCD) = 1/2 ar(ABCD)
(ii) A line EF is drawn parallel to AD passing through P.
In a parallelogram,
AD || EF (by construction) --- (i)
Thus,
AB || CD ⇒ AE || DF --- (ii)
From equations (i) and (ii),
AEDF is a parallelogram.
Now,
In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF.
∴ ar(ΔAPD) = 1/2 ar(AEFD) --- (iii)
also,
In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.
∴ ar(ΔPBC) = 1/2 ar(BCFE) --- (iv)
Adding equations (iii) and (iv),
ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}
⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
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 (AXS) = 1/2 ar (PQRS)
Answer
(i) Parallelogram PQRS and ABRS lie on the same base SR and between the same parallel lines SR and PB.
∴ ar(PQRS) = ar(ABRS) --- (i)
(ii) In ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel lines AS and BR.
∴ ar(ΔAXS) = 1/2 ar(ABRS) --- (ii)
From (i) and (ii),
ar(ΔAXS) = 1/2 ar(PQRS)
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 (AXS) = 1/2 ar (PQRS)
Answer
The field is divided into three parts. The three parts are in the shape of triangle. ΔPSA, ΔPAQ and ΔQAR.
Area of ΔPSA + ΔPAQ + ΔQAR = Area of PQRS --- (i)
Area of ΔPAQ = 1/2 area of PQRS --- (ii)
Triangle and parallelogram on the same base and between the same parallel lines.
From (i) and (ii),
Area of ΔPSA + Area of ΔQAR = 1/2 area of PQRS --- (iii)
Clearly from (ii) and (iii),
Farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.
In Fig.9.23, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar(ACE).
Answer
Given,
AD is median of ΔABC. Thus, it will divide ΔABC into two triangles of equal area.
∴ ar(ABD) = ar(ACD) --- (i)
also,
ED is the median of ΔABC.
∴ ar(EBD) = ar(ECD) --- (ii)
Subtracting (ii) from (i),
ar(ABD) - ar(EBD) = ar(ACD) - ar(ECD)
⇒ ar(ABE) = ar(ACE)
Answer
ar(BED) = (1/2) × BD × DE
As E is the mid-point of AD,
Thus, AE = DE
As AD is the median on side BC of triangle ABC,
Thus, BD = DC
Therefore,
DE = (1/2)AD --- (i)
BD = (1/2)BC --- (ii)
From (i) and (ii),
ar(BED) = (1/2) × (1/2) BC × (1/2)AD
⇒ ar(BED) = (1/2) × (1/2) ar(ABC)
⇒ ar(BED) = 1/4 ar(ABC)
Answer
O is the mid point of AC and BD. (diagonals of bisect each other)
In ΔABC, BO is the median.
∴ ar(AOB) = ar(BOC) --- (i)
also,
In ΔBCD, CO is the median.
∴ ar(BOC) = ar(COD) --- (ii)
In ΔACD, OD is the median.
∴ ar(AOD) = ar(COD) --- (iii)
In ΔABD, AO is the median.
∴ ar(AOD) = ar(AOB) --- (iv)
From equations (i), (ii), (iii) and (iv),
ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)
So, the diagonals of a parallelogram divide it into four triangles of equal area.
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
In ΔABC,
AO is the median. (CD is bisected by AB at O)
∴ ar(AOC) = ar(AOD) --- (i)
also,
In ΔBCD,
BO is the median. (CD is bisected by AB at O)
∴ ar(BOC) = ar(BOD) --- (ii)
Adding (i) and (ii) we get,
ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD)
⇒ ar(ABC) = ar(ABD)
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(DEF) = 1/4 ar(ABC)
(iii) ar (BDEF) = 1/2 ar(ABC)
Answer
Answer
In ΔABC,
AO is the median. (CD is bisected by AB at O)
∴ ar(AOC) = ar(AOD) --- (i)
also,
In ΔBCD,
BO is the median. (CD is bisected by AB at O)
∴ ar(BOC) = ar(BOD) --- (ii)
Adding (i) and (ii) we get,
ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD)
⇒ ar(ABC) = ar(ABD)
In the given 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
Let us draw DN ⊥ AC and BM ⊥ AC.
(i) In ΔDON and ΔBOM,
∠ DNO = ∠ BMO (By construction)
∠ DON = ∠ BOM (Vertically opposite angles)
OD = OB (Given)
By AAS congruence rule,
ΔDON ≅ ΔBOM
∴ DN = BM ... (1)
We know that congruent triangles have equal areas.
∴ Area (ΔDON) = Area (ΔBOM) ... (2)
In ΔDNC and ΔBMA,
∠ DNC = ∠ BMA (By construction)
CD = AB (Given)
DN = BM [Using equation (1)]
∴ ΔDNC ≅ ΔBMA (RHS congruence rule)
⇒ Area (ΔDNC) = Area (ΔBMA) ... (3)
On adding equations (2) and (3), we obtain
Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)
Therefore, Area (ΔDOC) = Area (ΔAOB)
(ii) We obtained,
Area (ΔDOC) = Area (ΔAOB)
⇒ Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)
(Adding Area (ΔOCB) to both sides)
⇒ Area (ΔDCB) = Area (ΔACB)
(iii) We obtained,
Area (ΔDCB) = Area (ΔACB)
If two triangles have the same base and equal areas, then these will lie between the same parallels.
∴ DA || CB ... (4)
In ΔDOA and ΔBOC,
∠ DOA = ∠ BOC (Vertically opposite angles)
OD = OB (Given)
∠ ODA = ∠ OBC (Alternate opposite angles)
By ASA congruence rule,
ΔDOA ≅ ΔBOC
∴ DA = BC ... (5)
In quadrilateral ABCD, one pair of opposite sides is equal and parallel (AD = BC)
Therefore, ABCD is a parallelogram.
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.
∴ DE || BC
Answer
Given,
XY || BC, BE || AC and CF || AB
To show,
ar(ΔABE) = ar(ΔAC)
Proof:
EY || BC (XY || BC) --- (i)
also,
BE∥ CY (BE || AC) --- (ii)
From (i) and (ii),
BEYC is a parallelogram. (Both the pairs of opposite sides are parallel.)
Similarly,
BXFC is a parallelogram.
Parallelograms on the same base BC and between the same parallels EF and BC.
⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) --- (iii)
Also,
△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.
⇒ ar(△AEB) = 1/2ar(BEYC) --- (iv)
Similarly,
△ACF and parallelogram BXFC on the same base CF and between the same parallels CF and AB.
⇒ ar(△ ACF) = 1/2ar(BXFC) --- (v)
From (iii), (iv) and (v),
ar(△AEB) = ar(△ACF)
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 following figure). Show that
ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]
Answer
AC and PQ are joined.
ar(△ACQ) = ar(△APQ) (On the same base AQ and between the same parallel lines AQ and CP)
⇒ ar(△ACQ) - ar(△ABQ) = ar(△APQ) - ar(△ABQ)
⇒ ar(△ABC) = ar(△QBP) --- (i)
AC and QP are diagonals ABCD and PBQR. Thus,
ar(ABC) = 1/2 ar(ABCD) --- (ii)
ar(QBP) = 1/2 ar(PBQR) --- (iii)
From (ii) and (ii),
1/2 ar(ABCD) = 1/2 ar(PBQR)
⇒ ar(ABCD) = ar(PBQR)
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 (ΔDAC) - Area (ΔDOC) = Area (ΔDBC) - Area (ΔDOC)
⇒ Area (ΔAOD) = Area (ΔBOC)
In the given 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) 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)
Answer
Let ABCD be the plot of the land of the shape of a quadrilateral.
Construction,
Diagonal BD is joined. AE is drawn parallel BD. BE is joined which intersected AD at O. △BCE is the shape of the original field and △AOB is the area for constructing health centre. Also, △DEO land joined to the plot.
To prove:
ar(△DEO) = ar(△AOB)
Proof:
△DEB and △DAB lie on the same base BD and between the same parallel lines BD and AE.
ar(△DEB) = ar(△DAB)
⇒ ar(△DEB) - ar△DOB) = ar(△DAB) - ar(△DOB)
⇒ ar(△DEO) = ar(△AOB)
Answer
Given,
ABCD is a trapezium with AB || DC.
XY || AC
Construction,
CX is joined.
To Prove,
ar(ADX) = ar(ACY)
Proof:
ar(△ADX) = ar(△AXC) --- (i) (On the same base AX and between the same parallels AB and CD)
also,
ar(△ AXC)=ar(△ ACY) --- (ii) (On the same base AC and between the same parallels XY and AC.)
From (i) and (ii),
ar(△ADX)=ar(△ACY)
In Fig.9.28, AP || BQ || CR. Prove that ar(AQC) = ar(PBR).
Answer
Given,
AP || BQ || CR
To Prove,
ar(AQC) = ar(PBR)
Proof:
ar(△AQB) = ar(△PBQ) --- (i) (On the same base BQ and between the same parallels AP and BQ.)
also,
ar(△BQC) = ar(△BQR) --- (ii) (On the same base BQ and between the same parallels BQ and CR.)
Adding (i) and (ii),
ar(△AQB) + ar(△BQC) = ar(△PBQ) + ar(△BQR)
⇒ ar(△ AQC) = ar(△ PBR)
Answer
Given,
ar(△AOD) = ar(△BOC)
To Prove,
ABCD is a trapezium.
Proof:
ar(△AOD) = ar(△BOC)
⇒ ar(△AOD) + ar(△AOB) = ar(△BOC) + ar(△AOB)
⇒ ar(△ADB) = ar(△ACB)
Areas of △ADB and △ACB are equal. Therefore, they must lying between the same parallel lines.
Thus, AB ∥ CD
Therefore, ABCD is a trapezium.
Answer
Given,
ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC)
To Prove,
ABCD and DCPR are trapeziums.
Proof:
ar(△BDP) = ar(△ARC)
⇒ ar(△BDP) - ar(△DPC) = ar(△DRC)
⇒ ar(△BDC) = ar(△ADC)
ar(△BDC) = ar(△ADC). Therefore, they must lying between the same parallel lines.
Thus, AB ∥ CD
Therefore, ABCD is a trapezium.
also,
ar(DRC) = ar(DPC). Therefore, they must lying between the same parallel lines.
Thus, DC ∥ PR
Therefore, DCPR is a trapezium.
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 ... (1)
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 ... (2)
From equations (1) and (2), we obtain
AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD
In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal.
∴ AD = BC ... (1)
Similarly, for parallelograms DCEF and ABFE, it can be proved that
DE = CF ... (2)
And, EA = FB ... (3)
In ΔADE and ΔBCF,
AD = BC [Using equation (1)]
DE = CF [Using equation (2)]
EA = FB [Using equation (3)]
∴ ΔADE ≅ BCF (SSS congruence rule)
⇒ Area (ΔADE) = Area (ΔBCF)
In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that
ar (BPC) = ar (DPQ).
[Hint: Join AC.]
Answer :
It is given that ABCD is a parallelogram.
AD || BC and AB || DC(Opposite sides of a parallelogram are parallel to each other)
Join point A to point C.
Consider ΔAPC and ΔBPC
ΔAPC and ΔBPC are lying on the same base PC and between the same parallels PC and AB. Therefore,
Area (ΔAPC) = Area (ΔBPC) ... (1)
In quadrilateral ACDQ, it is given that
AD = CQ
Since ABCD is a parallelogram,
AD || BC (Opposite sides of a parallelogram are parallel)
CQ is a line segment which is obtained when line segment BC is produced.
∴ AD || CQ
We have,
AC = DQ and AC || DQ
Hence, ACQD is a parallelogram.
Consider ΔDCQ and ΔACQ
These are on the same base CQ and between the same parallels CQ and AD. Therefore,
Area (ΔDCQ) = Area (ΔACQ)
⇒ Area (ΔDCQ) - Area (ΔPQC) = Area (ΔACQ) - Area (ΔPQC)
⇒ Area (ΔDPQ) = Area (ΔAPC) ... (2)
From equations (1) and (2), we obtain
Area (ΔBPC) = Area (ΔDPQ)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(i) ar(BDE) = 1/4 ar(ABC)
(i) Let G and H be the mid-points of side AB and AC respectively.
Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and also its length will be half of the length of BC (mid-point theorem).
=> GH = 1/2 BC and GH || BD
=> GH = BD = DC and GH || BD (D is the mid-point of BC)
Consider quadrilateral GHDB.
GH ||BD and GH = BD
Two line segments joining two parallel line segments of equal length will also be equal and parallel to each other.
Therefore, BG = DH and BG || DH
Hence, quadrilateral GHDB is a parallelogram.
We know that in a parallelogram, the diagonal bisects it into two triangles of equal area.
Hence, Area (ΔBDG) = Area (ΔHGD)
Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and their respective diagonals are dividing them into two triangles of equal area.
ar (ΔGDH) = ar (ΔCHD) (For parallelogram DCHG)
ar (ΔGDH) = ar (ΔHAG) (For parallelogram GDHA)
ar (ΔBDE) = ar (ΔDBG) (For parallelogram BEDG)
ar (ΔABC) = ar(ΔBDG) + ar(ΔGDH) + ar(ΔDCH) + ar(ΔAGH)
ar (ΔABC) = 4 x ar(ΔBDE)
Hence, ar(BDE) = 1/4 ar(ABC)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(iii) ar(ABC) = 2 ar(BEC)
From (2) and (3), we obtain
2 ar (ΔBDE) = ar (ΔABE)
or ar(BDE) = 1/2 ar(BAE)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(iii) ar(ABC) = 2 ar(BEC)
ANSWER
ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)
ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)
Using equation (1), we obtain
ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC)
ar (ΔABD) = ar (ΔBEC)
1/2 ar(ΔABC) = ar(ΔBEC)
ar (ΔABC) = 2 ar (ΔBEC)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(iv) ar(BFE) = ar(AFD)
ANSWER
(iv)It is seen that ΔBDE and ar ΔAED lie on the same base (DE) and between the parallels DE and AB.
∴ar (ΔBDE) = ar (ΔAED)
⇒ ar (ΔBDE) - ar (ΔFED) = ar (ΔAED) - ar (ΔFED)
∴ar (ΔBFE) = ar (ΔAFD)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that (v) ar(BFE) = 2 ar(FED)
ANSWER:
Let h be the height of vertex E, corresponding to the side BD in ΔBDE.
Let H be the height of vertex A, corresponding to the side BC in ΔABC.
In (i), it was shown that ar(BDE) = 1/4 ar(ABC)
In (iv), it was shown that ar (ΔBFE) = ar (ΔAFD)
=> ar (ΔBFE) = ar (ΔAFD)
= 2 ar (ΔFED)
Hence, ar(BFE) = 2 ar(FED)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that (vi) ar(FED) = 1/8 ar(AFC)
ANSWER:
(vi) Area (AFC) = area (AFD) + area (ADC)
