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### Transcript of CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH ... CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA,...

• 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 (∆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

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

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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)

∴ 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

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

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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)

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

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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

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

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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),

| A quadrilateral is a parallelogram if its opposite sides are 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

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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 an