Download - CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH …...CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315 CONTACT FOR MATHEMATICS GROUP/HOME

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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) The figures (∆PDC, quadrilateral ABCD), (quadrilaterals APCD and ABCD), and (quadrilaterals PBCD and ABCD), lie on the same base DC and between the same parallels DC and AB. (iii) The figures (∆TRQ and parallelogram SRQP), (quadrilaterals TPQR and parallelogram SRQP), (quadrilateral STQR and SRQP), lie on the same base RQ and between the same parallels RQ and SP. (v) Quadrilaterals APCD and ABQD lie on the same base AD and between the same parallels AD and BQ.

2. In the figure, ABCD is a parallelogram, AE ⊥⊥⊥⊥ DC And CF ⊥⊥⊥⊥ AD. If AB = 16cm, AE

= 8cm and CF = 10 cm, find AD.

• Solution: ar(parallelogram ABCD) = AB × AE = 16 × 8 cm2 = 128 cm2 ….(1)

Also ar(parallelogram ABCD) = AD × CF = AD × 10 cm2 ….(2)

From (1) and (2), we get

AD × 10=128

⇒ AD = ⇒ AD = 12.8 cm

3. E, F, G and H are respectively the mid-points of the sides of a parallelogram

ABCD. Show that ar (EFGH) = ar(ABCD).

• Solution:

BADHANEDUCATION D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI

CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT

CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315

CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,

DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES

EMAIL:1 [email protected] web site www.badhaneducation.in

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

To Prove: ar(EFGH) = ar(ABCD)

Construction: Join OF, OG, OH and OE. Also, join diagonals AC and BD to intersect at O.

Proof: In ∆BCD,

F and G are the mid-points of BC and DC respectively.

∴ FG || BD …..(1)

In ∆ BAD,

∴ E and H are the mid-points of AB and AD respectively.

∴ EH || BD ..…..(2)

From (1) and (2),

EH || BD || FG Hence EH || FG..........(3)

Similarly, we can prove that

EF || HG ...........(4)

From (3) and (4),

Quadrilateral EFGH is a parallelogram (since opposite sides are parallel) A quadrilateral is a parallelogram if its opposite sides are equal







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F is the mid-point of CB and O is the mid-point of CA

∴ FO || BA

⇒ FO || CG -(5)

∴ BA || CG

and FO= BA

= CD

= CG ...(6) | ∴ G is the mid-point of CD

In view of (5) and (6), Quadrilateral OFCG is a parallelogram

| A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length OP = PC | Diagonals of a parallelogram bisect each other

∴ ∆ OPF and ∆€CPF have equal bases (∴ OP = PC) and have a common vertex F

∴ Their altitudes are also the same ∴ ar(∆ OPF) = ar(∆ CPF) Similarly, ar(∆ OQF) = ar(∆ BQF) Adding, we get

ar(∆ OPF) + ar(∆ OQF) = ar(∆ CPF) + ar(∆ BQF)

⇒ ar(parallelogram OQFP) = ar(∆ CPF) + ar(∆ BQF) ...(7)

Similarly, ar(parallelogram OPGS) = ar(∆ GPC) + ar(∆ DSG) ...(8)

ar(parallelogram OSHR) = ar(∆ DSH) + ar(∆ HAR) ...(9)

ar(parallelogram OREQ) = ar(∆ ARE) + ar(∆ EQB) ...(10)

Adding the corresponding sides of (7), (8), (9) and (10), we get

ar(parallelogram EFGH) = {ar(∆ CPF) + ar(∆ GPC)} + {ar(∆ DSG) + ar∆ DSH)} + {ar(∆ HAR) + ar(∆ ARE)} + {ar(∆ BQF) + r(∆ EQB)} = ar(∆ FCG) + ar(∆ GDH) + ar(∆ HAE) + ar(∆ EBF)







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= ar(parallelogram ABCD) - ar(parallelogram EFGH)

= 2 × ar(parallelogram EFGH) = ar(parallelogram ABCD)

= ar(parallelogram EFGH) = × ar(parallelogram ABCD)

4. 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).

• Solution: Given: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

To Prove: ar(∆ APB) = ar(∆ BQC).

Proof: ∴ ∆ APB and || gm ABCD are on the same base AB and between the same parallels AB and DC.

∴ ar(∆ APB)= ar(ll gm ABCD) ...(1)

∴ ∆ BQC and || gm ABCD are on the same base BC and between the same parallels BC and AD.

∴ ar(∆ BQC) = ar(ll gm ABCD) ...(2)

From (1) and (2),

ar(∆ APB) = ar(∆ BQC).

5. In the figure P is a point in the interior of a parallelogram ABCD. Show that







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(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]

• Solution: Given: P is a point in the interior of a parallelogram ABCD.

To Prove: (i) ar(APB) + ar(PCD) = ar(ABCD)

(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD) Construction: Through P, draw a line EF parallel to AB.

Proof: (i) EF || AB ...(1) | by construction

AD || BC | ∴ Opposite sides of a parallelogram are parallel

∴ AE || BF ...(2)

In view of (1) and (2),

Quadrilateral ABFE is a parallelogram

| A quadrilateral is a parallelogram if its opposite sides are parallel







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Similarly, quadrilateral CDEF is a parallelogram ∴ ∆ APB and parallelogram ABFE are on the same base AB and between the same parallels AB and EF.

∴ ar(∆ APB)= ar(parallelogram ABFE) ...(3)

∴ ∆ PCD and parallelogram CDEF are on the same base DC and between the same parallels DC and EF.

∴ ar(∆ PCD) = ar(ll gm CDEF) ...(4)

Adding (3) and (4), we get

ar(∆ APB) + ar(∆ PCD) = ar(parallelogram ABFE) + ar (parallelogram CDEF)

= [ar(parallelogram ABFE) + ar(parallelogram CDEF)]

= ar(parallelogram ABCD) ...(5)

ar(∆ APB) + ar(∆ PCD) = ar(parallelogram ABCD)

(ii) ar(∆ APD) + ar(∆ PBC) = ar(parallelogram ABCD) - [ar(∆ APB) + ar(∆ PCD)]

= 2 [ar(∆ APB) + ar(∆ PCD)] - [ar(∆ APB) + ar(∆€PCD)]

= ar(∆ APB) + ar(∆ PCD)

ar(∆€APD) + ar(∆€PBC) = ar(∆€APB) + ar(∆€PCD)

6. In the figure, PQRS and ABRS are parallelograms and X is any point on side

BR. Show that

(i) ar (PQRS) = ar(ABRS)

(ii) ar(AXS) = ar(PQRS)







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• Solution: Given: PQRS and ABRS are parallelograms and X is any point on side BR. To Prove: (i) ar(PQRS) = ar(ABRS)

(ii) ar(AXS) = ar(PQRS) Proof: (i) In D PSA and D QRB,

∠ SPA = ∠ RQB ...(1) | Corresponding angles from PS || QR and transversal PB

∠ PAS = ∠ QBR ...(2) | Corresponding angles from AS || BR and transversal PB

∠ PSA = ∠ QRB ...(3) | Angle sum property of a triangle

Also, PS = QR ...(4) | Opposite sides of parallelogram PQRS In view of (1), (3) and (4),

∆ PSA ≅ ∆ QRB ...(5) | By ASA Rule

∴ ar(∆ PSA) = ar(∆ QRB) ...(6) | ∴ Congruent figures have equal areas

Now, ar(PQRS) = ar(∆ PSA) + ar(AQRS) = ar(∆ QRB) + ar(AQRS) | Using (6) = ar(ABRS) (ii) ∆ AXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR.

∴ ar(∆ AXS) = ar(parallelogram ABRS)

= {ar(AQRS) + ar(∆ QRB)}

= {ar(AQRS) + ar(∆ PSA)} | Using (6)







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= ar(|| gm PQRS).

7. A farmer has a field in the form of a parallelogram PQRS. He took any point A

on RS and joined it to points P and Q. In how many parts can the fields be 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 he do it?

• Solution: The field is divided into three triangles. The three triangles are:

(i) ∆ APS (ii) ∆ APQ

(iii) ∆ AQR

If the farmer wants to sow wheat and pulses in equal portions of the field separately, then A

must be taken such that SA = SR. Also, B must be taken such that

PB = . Let us assume that SA = SR and PB = . Now, we are going to prove ar(∆€ASP) = ar(parallelogram ARBP) = ar(∆€BQR), so that the farmer can sow wheat and pulses in equal portions of the field seperately.

SR || PQ | ∴ PQRS is a parallelogram

∴ AR || PB …..(1)

Again AR = | by assumption SA = SR







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PB = | by assumption PB =

SR = PQ | ∴ Opposite sides of a parallelogram are equal ∴ AR = PB …..(2)


ARBP is a parallelogram.

| A quadrilateral is a parallelogram if one of its pair of opposite sides is parallel and of equal length.

∴ ∆ ASP, parallelogram ARBP and ∆ BQR are in between the same parallels PQ and SR.

∴ Their altitudes are equal. Let it be x.

Now, Area of ∆ ASP = = | by assumption

= ….(3)

Area of parallelogram ARBP = (AR)(x) = ...(4) | by assumption

Area of ∆ BQR = = | by assumption

=

= .…(5)

In view of (3), (4) and (5)

ar(∆ ASP) = ar(parallelogram ARBP) = ar(∆ BQR)

8. In the given figure, E is any point on the median AD of a ∆∆∆∆ABC. Show that

ar(ABE) = ar(ACE).







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• Solution: Given: E is any point on median AD of a ∆ ABC.

To Prove: ar(∆ ABC) = ar(∆ ACE).

Proof: In ∆ ABC,

AD is a median.

∴ ar(∆ ABD) = ar(∆ ACD) ….(1)

In ∆ EBC,

ED is a median.

∴ ar(∆ EBD) = ar(∆ ECD) …..(2)

Subtracting (2) from (1), we get

ar(∆ ABD) - ar(∆ EBD) = ar(∆ ACD) - ar(∆ ECD)

⇒ ar(∆ ABE) = ar(∆ ACE).

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

• Solution: Given: In a triangle ABC, E is the mid-point of median AD.

To Prove: ar(∆ BED) = ar(∆ ABC).

Proof: In ∆ ABC,

∴ AD is a median.







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ar(∆ ABD) = ar(∆ ACD) = ar(∆ ABC) …..(1)

|∴ median of a triangle divides it into two triangles of equal areas.

In ∆ ABD,

∴ BE is a median.

∴ ar(∆ BED) = ar(∆ BEA) = ar(∆ ABD)

⇒ ar(∆ BED) = ar(∆ ABD) = . ar(∆ ABC) | From(1)

= ar (∆ ABC).

10. 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 intersect at O.

Divide it into four trianlges. They are ∆ OAB, ∆ OBC, ∆ OCA and ∆ OAD.

To Prove: ar(∆ OAB) = ar(∆ OBC) = ar(∆ OCD) = ar(∆ ODA).

Construction: Draw BE ⊥ AC.

Proof: ABCD is a parallelogram

∴ OA = 0C and OB = OD | Diagonals of a parallelogram bisect each other







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Now, ar(∆ OAB) = =

and ar(∆ OBC) = =

But OA = OC

∴ ar(∆ OAB) = ar(∆ OBC) .….(1)

Similarly,

ar(∆ OBC) = ar(∆ OCD) …..(2)

and ar(∆ OCD) = ar(∆ ODA) .….(3)

From (1), (2) and (3),

ar(∆ OAB) = ar(∆ OBC) = ar(∆ OCB) = ar(∆ ODA).

11. 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).

• Solution: Given: ABC and DBC are two triangles on the same base AB. Line segment CD is bisected by AB at O.







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To Prove: ar(∆ ABC) = ar(∆ ABD).

Proof: ∴ Line segment CD is bisected by AB at O.

∴ OC = OD

∴ BO is a median of ∆ ABCD ∴and AO is a median of ∆ ACD.

∴ BO is a median of D BCD. ∴ ar(∆ OBC) = ar(∆ OBD) …………(1)

∴ AO is a median of D ACD ∴ ar(∆ OAC) = ar(∆ OAD) ………….(2)


ar(∆ OBC) + ar(∆ OAC) = ar(∆ OBD) + ar(∆ OAD) ⇒ ar(∆ ABC) = ar(∆ ABD).

12. D, E and F are respectively the mid-points of the sides BC, CA and AB of

a ∆∆∆∆ABC. Show that

(i) BDEF is a parallelogram.

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

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

• Solution: Given: D, E and F are respectively the mid-points of the sides BC, CA and AB of a AABC.

To Prove: (i) BDEF is a parallelogram

(ii) ar(∆ DEF) = ar(∆€ABC) 4

(iii) ar(BDEF) = ar(∆ ABC).

Proof: (i) In ∆ ABC,

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.







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⇒ EF || BD .......... (1)

Similarly, ED || BF .......... (2)


BDEF is a parallelogram. | A quadrilateral is a parallelogram if opposite sides are parallel (ii) As in (i), we can prove that

AFDE and FDCE are parallelograms. FD is a diagonal of the parallelogram BDEF.

∴ ar(∆ FBD) = ar(∆ DEF) ...(3)

Similarly, ar(∆ DEF) = ar(∆ FAE) ...(4)

and, ar(∆ DEF) = ar(∆ DCE) ...(5)

From (3), (4) and (5), we have

ar(∆ FBD) = ar(∆ DEF) = ar(∆ FAE) = ar(∆ DCE) ...(6)

∴ ∆ ABC is divided into four non-overlapping triangles ∆ FBD, ∆ DEF, ∆ FAE and ∆ DCE

∴ ar(∆ ABC) = ar(∆ FBD) + ar(∆ DEF) + ar(∆ FAE) + ar(∆ DCE)

= 4ar(∆ DEF) | From (6)

⇒ ar(∆ DEF) = ar(∆ ABC) ...(7)

(iii) ar(BDEF)

= ar(∆ FBD) + ar(∆ DEF) | From (3)

= ar(∆ DEF) + ar(∆ DEF)

= 2 ar(∆ DEF)







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= 2. ar(∆ ABC) | From (7)

= ar(∆ ABC)

13. In the 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]

• Solution: Given: Diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB = OD. To Prove: If AB = CD, then

(i) ar(∆ DOC) = ar(∆ AOB) (ii) ar(∆ DCB) = ar(∆ ACB) (iii) DA || CB or ABCD is a parallelogram.

Construction: Draw DE ⊥ AC and BF ⊥ AC.

Proof: (iii) In ∆ ADB,







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AO is a median

∴ ar(∆ AOD) = ar(∆ AOB) ....(1) | A median of a triangle divides it into two triangles of equal areas

In ∆CBD,

CO is a median.

∴ ar(∆ COD) = ar(∆ COB) ....(2) | A median of a triangle divides it into two triangles of equal areas


ar(∆ AOD) + ar(∆ COD) = ar(∆ AOB) + ar(∆ COB)

⇒ ar(∆ ACD) = ar(∆ ACB)

⇒ =

⇒ DE = BF ...(3)

In right as DEC and BFA,

Hyp. DC = Hyp. BA | given

DE = BF | From (3)

∴ ∆ DEC ≅ ∆ BFA | R. H. S. Rule

∴ ∠ DCE = ∠ BAF | CPCT

But these angles form a pair of equal alternate interior angles.

∴ DC || AB ...(4)

DC = AB and DC || AB ∴ ABCD is a parallelogram. | A quadrilateral is a parallelogram if a pair of opposite sides is parallel and equal

∴ DA || CB | ∴ Opposite sides of a parallelogram are parallel

(i) ABCD is a parallelogram







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∴ OC = OA ...(5) | ∴ Diagonals of a parallelogram bisect each other

ar(∆DOC) =

ar(∆ AOB) =

∴ DE = BF | From (3)

and OC = OA | From (5)

∴ ar(∆ DOC) = ar(∆ AOB). (ii) From (i), ar(∆ DOC) = ar(∆ AOB)

⇒ ar(∆ DOC) + ar(∆ OCB) = ar(∆ AOB) + ar(∆ OCB) | Add same area ar(∆ OCB) on both sides

⇒ ar(∆ DCB) = ar(∆ ACB)

14. D and E are points on sides AB and AC respectively of ∆∆∆∆ABC such that ar(DBC)

= ar(EBC). Prove that DE || BC.

• Solution: Given: D and E are points on sides AB and AC respectively of ∆ABC such that ar(∆ DBC) = ar(∆EBC).

To Prove: DE || BC

Proof: ∆ DBC and ∆ EBC are on the same base BC and have equal areas.

∴Their altitudes must be the same.

∴DE || BC







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15. 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).

• Solution: Given: XY is a line parallel to side BC of a triangle ABC. BE || AC and CF || AB meet XY at E and F respectively.

To Prove: ar(∆ ABE) = ar(∆ ACF).

Proof: XY || BC | given

and CF || BX | ∴ CF || AB (given)

∴€BCFX is a || gm

BC = XF | Opposite sides of a parallelogram are equal

BC = XY + YF …..(1)

Again,

XY || BC | ∴ BE || AC (given)

and BE || CY

∴ BCYE is a parallelogram

∴ 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







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∴ ∆ AEX and ∆ AYF have equal bases (XE = YF) on the same line EF and have a common vertex A.

∴ Their altitudes are also the same.

∴ ar(∆ AEX) = ar(∆ AFY) ...(4)

∴ ∆ BEX and ∆ CFY have equal bases (XE = YF) on the same line EF and are between the same parallels EF and BC (XY || BC).

∴ ar(∆ BEX) = ar(∆ CFY) ...(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(∆ AEX) + ar(∆ BEX) = ar(∆ AFY) + ar(∆ CFY)

⇒ ar(∆ ABE) = ar(∆ ACF).

16. 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)].

• Solution: Given: 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.

To Prove: ar(parallelogram ABCD) = ar(parallelogram PBQR).

Construction: Join AC and PQ.

Proof: AC is a diagonal of parallelogram ABCD

∴ ar(∆ ABC) = ar(parallelogram ABCD) …..(1)







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PQ is a diagonal of parallelogram BQRP

∴ ar(∆ BPQ) = ar(parallelogram BQRP) …….(2)

∆ ACQ and ∆ APQ are on the same base AQ and between the same parallels AQ and CP.

∴ ar(∆ ACQ) = ar(∆ APQ)

⇒ ar(∆ ACQ) - ar(∆ ABQ) = ar(∆ APQ) - ar(∆ ABQ)

| Subtract the same area ar(∆ ABQ) from both sides

⇒ ar(∆ ABC) = ar(∆ BPQ)

⇒ ar(parallelogram ABCD) = ar(parallelogram PBQR)

⇒ ar(parallelogram ABCD) = ar(parallelogram PBQR) | From (1) and (2)

17. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other

at O. prove that ar(AOD) = ar(BOC).

• Solution: Given: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. To Prove: ar(∆ AOD) = ar(∆ BOC). Proof: ∆ ABD and ∆ ABC are on the same base AB and between the same parallels AB and DC.







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∴ ar(∆ ABD) = ar(∆ ABC)

⇒ ar(∆ ABD) - ar(∆ AOB) = ar(∆ ABC) - ar(∆ AOB)

| Subtract ar(∆ AOB) both sides

⇒ ar(∆ AOD) = ar(∆ BOC).

18. In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC

produced at F. Show that

a. ar(ACB) = ar(ACF) b. ar(AEDF) = ar(ABCDE)

• Solution: Given: ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.

To Prove: (i) ar(∆ ACB) = ar(∆ ACF) (ii) ar(AEDF) = ar(ABCDE)

Proof: (i) ∆ ACB and AACF are on the same base AC and between the same parallels AC and BF.

[AC || BF (given)]







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∴ ar(∆ ACB) = ar(∆ ACF)

| Two triangles on the same base (or equal bases) and between the same parallels are equal in area (ii) From (i),

ar(∆ ACB) = ar(∆ ACF)

⇒ ar(∆ ACB) + ar(AEDC) = ar(∆ ACF) + ar(AEDC)

| Add ar(AEDC) on both sides

⇒ ar(ABCDE) = ar(AEDF)

⇒ ar(AEDF) = ar(ABCDE).

19. A villager Itwaari has a plot of land in 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 center. 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.

• 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 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(∆ADE) = ar(∆PEC).

Proof: ∆DAP and ∆DCP are on the same base DP and between the same parallels DP and AC.

∴ ar(∆ DAP) = ar(∆ DCP)

| Two triangles on the same base (or equal bases) and between the same parallels are equal in area







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⇒ ar(∆ DAP) - ar(∆ DEP) - ar(∆ DCP) - ar(∆ DEP) | Subtract ar(∆ DEP) from both sides

ar(∆ ADE) = ar(∆ PCE)

⇒ ar(∆ DAE) + ar(ABCE) = ar(∆ PCE) + ar(ABCE) | Add ar(ABCE) both sides

⇒€ar(ABCD) = ar(∆ ABP)

20. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X And

BC at y. Prove that (∆∆∆∆ ADX) = ar(∆∆∆∆ ACY). [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. To Prove: ar(∆ ADX) = ar(∆ ACY) Construction: Join CX

Proof: ∆ ADX and ∆ ACX are on the same base AX and between the same parallels AB and DC.

∴ ar(∆ ADX) = ar(∆ ACX)

...(1)

∆ ACX and ∆ ACY are on the same base AC and between the same parallels AC and XY.

∴ ar(∆ ACX) = ar(∆ ACY)

...(2)

From (1) and (2), we get







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ar(∆ ADX) = ar(∆ ACY).

21. In figure AP BQ CR. Prove that ar(AQC) = ar(PBR)

• Solution: Given: AP || BQ || CR.

To Prove: ar(∆ AQC) = ar(∆ PBR).

Proof: ∴ ∆ BAQ and ∆ BPQ are on the same base BQ and between the same parallels BQ and AP.

∴ ar(∆ BAQ) = ar(∆ BPQ) …..(1)

∴ ∆ BCQ and ∆ BQR are on the same base BQ and between the same parallels BQ and CR.

ar(∆ BCQ) = ar(∆ BQR) .…(2)

Adding the corresponding sides of (1) and (2), we get

ar(∆ BAQ) + ar(∆ BCQ) = ar(∆ BPQ) + ar(∆ BQR) ⇒ ar(∆ AQC) = ar(∆ PBR).

22. 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.

• Solution: Given: Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(∆ AOD) = ar(∆ BOC).







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To Prove: ABCD is a trapezium.

Proof: ar(∆ AOD) = ar(∆ BOC)

⇒ ar(∆ AOD) + ar(∆ AOB) = ar(∆ BOC) + ar(∆ AOB)

| Adding the same areas on both sides

⇒ ar(∆ ABD) = ar(∆ ABC)

But ∆ ABD and ∆ ABC are on the same base AB.

∴ ∆ ABD and ∆ ABC will have equal corresponding altitudes ∴ ∆ ABD and ∆ ABC 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

23. In figure ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the

quadrilaterals ABCD and DCPR are trapeziums.

• Solution: Given: ar(∆ DRC) = ar(∆ DPC) and ar(∆ BDP) = ar(∆ ARC).







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To Prove: Both the quadrilaterals ABCD and DCPR are trapeziums.

Proof: ar(∆ DRC) = ar(∆ DPC) ...(1) | given

But ∆ DRC and ∆ DPC are on the same base DC.

∴ ∆ DRC and ∆ DPC will have equal corresponding altitudes.

∴ ∆ DRC and ∆ DPC will lie between the same parallels.

∴ DC || RP

∴ DCPR is a trapezium.

Again, ar(∆ BDP) = ar(∆ ARC)

⇒ ar(∆ BDC) + ar(∆ DPC) = ar(∆ ADC) + ar(∆ DRC)

⇒€ar(∆ BDC) = ar(∆ ADC) | Using (1)

But ∆ BDC and ∆ ADC are on the same base DC. ∴€∆ BDC and ∆ ADC will have equal corresponding altitudes. ∴ ∆ BDC and ∆ ADC 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.







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