1 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 (ΔPDC, 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 (ΔTRQ 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 cm 2 = 128 cm 2 ….(1) Also ar(parallelogram ABCD) = AD × CF = AD × 10 cm 2 ….(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 www.badhaneducation.in 09810144315
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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
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
• 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)
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
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
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
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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)
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
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∆ 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.
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
| 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
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,
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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
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
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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.
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