1 Marks Quetions

1. Find the area of parallelogram in the adjoining figure.

(a) 1759 foot

(b) 48 square

(c) 84 square foot

(d) 60 square foot

Ans. (d) 60 square foot

2. Find the measure of angle a

(a)

(b)

(c)

d)

Ans. (b)

3. A triangle has an area of 45 square foot. Base of the triangle is 9 foot. What is corresponding height of triangle

(a) 90 foot

(b) 5 foot

(c) 10 foot

(d) 40 square foot

Ans. (c) 10 foot.

4. What is area of parallelogram whose base=8 and corresponding altitude is 5

(a) 40

(b) 45

(c) 13

(d) 3

Ans. (a) 40

5. Parallelograms on the same base and between the same parallels have equal

(i) corresponding angle

(ii) area

(iii) congruent area

(iv) same parallel

Ans. (ii) area

6. Any side of a parallelogram is called

(i) Altitude

(ii) base

(iii) corres. Altitude

(iv) area

Ans. (ii) base

7. A diagonal of a parallelogram divides into ______ triangles of equal area

(i) 1

(ii) 2

(iii) 3

(iv) none of these

Ans. (ii) 2

8. Find the area of parallelogram, if Base = 3 and altitude is 4

(i) 7

(ii) 1

(iii) 12

(iv) none of these

Ans. (iii) 12

9. Find the area of 11||gm, if base = 8 cm and altitude = 10 cm,

(a) 80 sq.cm

(b) 80 cm

(c) 30 sq.cm

(d) 50 sq.cm

Ans. (a) 80 sq.cm

10. If Base = 9 and corresponding altitude = 4. Find area of ||gram

(a) 4

(b) 40

(c) 36

(d) none of these

Ans. (c) 36

11. If a triangle and a Parallelogram are on the same base and between the same parallel, the area of the triangle is equal to ______ that of ||gram.

(a)

(b)

(c)

(d) none of these

Ans. (a)

12. A parallelogram has an area of 36 square an and base of the parallelogram is 9 cm. what is the corresponding altitude of parallelogram?

(a) 6 cm.

(b) 5 cm.

(c) 4 cm.

(d) 3 cm.

Ans. (c) 4 cm.

13. A median of a triangle divides if into _______triangles of equal areas.

(a) 1

(b) same triangle

(c) 2

(d) none

Ans. (c) 2

14. The area of a rhombus is equal to _______ of the product of its two diagonals.

(a)

(b)

(c)

(d) none

Ans. (a)

15. Area of a triangle is half the product of any of its sides and the

(a) Corresponding altitude

(b) altitude

(c) median

(d) base

Ans. (a) Corresponding altitude

16. Given below are the measurements of a parallelogram. Find the missing measurement. Area = 90 square cm, Base = 5 cm, Height =?

(a) 18

(b) 450

(c) 85

(d) 15 cm

Ans. (a) 18

17. How many square feet are in a square yard

(a) 6

(b) 9

(c) 12

(d) 10

Ans. (d) 10

18. The perimeter of an equilateral triangle is 21 yard. what is the length of its each sides

(a) 7 yard

(b) 14 yard

(c) 8 yard

(d) 12 yard

Ans. (a) 7 yard

19. What is the area of a triangle with base 12 m and a height of 18 m

(a)

(b)

(c)

(d)

Ans. (c)

20. Find the area of parallelogram if base = 8 and corresponding Altitude = 4

(a) 12

(b) 32

(c) 4

(d) 8

Ans. (b) 32

2 Marks Quetions

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.

Ans. In figure (i): DPC and trap. ABCD are on the same base DC and between same parallel DC and AB.

In figure (iii): RTQ and parallelogram PQRS are on the same base QR and between same parallel QR and PS.

In figure (v): Parallelogram ABCD and parallelogram APQD are on the same base AD and between the same parallels AD and BQ.

2. In figure, ABCD is a parallelogram. AE DC and CF AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Ans. ABCD is a parallelogram.

DC = ABDC = 16 cm

AE DC[Given]

Now Area of parallelogram ABCD = Base Corresponding height

Using base AD and height CF, we can find,

Area of parallelogram

AD = = 12.8 cm

3. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = ar (ABCD).

Ans. Given: A parallelogram ABCD. E, F, G and H are mid-points of AB,

BC, CD and DA respectively.

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

Construction: Join HF

Proof: ar (GHF) = ar (gm HFCD)……….(i)

And ar (HEF) = ar (gm HABF)……….(ii)

[If a triangle and a parallelogram are on the same base and between the same parallel then the area of triangle is half of area of parallelogram]

Adding eq. (i) and (ii),

ar (GHF) + ar (HEF) = ar (gm HFCD) + ar (gm HABF)

ar (gm HEFG) = ar (gm ABCD)

4. In figure, PQRS and ABRS are parallelograms and X is any point on the side BR. Show that:

(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = ar (PQRS)

Ans. (i) Parallelogram PQRS and ABRS are on the same base SR and between the same parallels SR and PB.

ar (gm PQRS) = ar (gm ABRS)……….(i)

[ parallelograms on the same base and between the same parallels are equal in area]

(ii) AXS and gm ABRS are on the same base AS and between the same parallels AS and BR.

ar (AXS) = ar (gm ABRS)……….(ii)

Using eq. (i) and (ii),

ar (AXS) = ar (gm PQRS)

5. In figure, E is any point on median AD of a ABC. Show that ar (ABE) = ar (ACE).

B                               D

Ans. In ABC, AD is a median.

ar (ABD) = ar (ACD) ……….(i)

[ Median divides a into two s of equal area]

Again in EBC, ED is a median

ar (EBD) = ar (ECD)……….(ii)

Subtracting eq. (ii) from (i),

ar (ABD) – ar (EBD) = ar (ACD) – ar (ECD)

ar (ABE) = ar (ACE)

6. Show that DE||BC if ar ar

Ans. Since BCE and BCD are equal in area and have a same base BC

BCE and BCD are between the same Parallel lines.

DE||BC

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

Ans. Given: A ABC, AD is the median and E is the mid-point of median AD.

To prove: ar (BED) = ar (ABC)

Proof: In ABC, AD is the median.

ar (ABD) = ar (ADC)

[ Median divides a into two s of equal area]

ar (ABD) = ar (ABC)……….(i)

In ABD, BE is the median.

ar (BED) = ar (BAE)

ar (BED) = ar (ABD)

ar (BED) = ar (ABC) = ar (ABC)

8. D and E are points on sides AB and AC respectively of ABC such that ar (DBC) = ar (EBC). Prove that DEBC.

Ans. Given: ar (DBC) = ar (EBC)

Since two triangles of equal area have common base BC.

Therefore DEBC[ Two triangles having same base (or equal bases) and equal areas lie between the same parallel]

9. Diagonals AC and BD of a trapezium ABCD with ABDC intersect each other at O. Prove that ar(AOD) = ar (BOC).

Ans. ABD and ABC lie on the same base AB and between the same parallels AB and DC.

ar (ABD) = ar (ABC)

Subtracting ar (AOB) from both sides,

ar (ABD) – ar (AOB)

= ar (ABC) – ar (AOB)

ar (AOD) = ar (BOC)

10. In 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)

Ans. (i) Given that BFAC

ACB and ACF lie on the same base AC and between the same parallels AC and BF.

ar (ACB) = ar (ACF)……….(i)

(ii) Nowar (ABCDE) = ar (trap. AEDC) + ar (ABC)……….(ii)

ar (ABCDE) = ar (trap. AEDC) + ar (ACF) = ar (quad. AEDF)[Using (i)]

ar (AEDF) = ar (ABCDE)

11. In figure, APBQCR. Prove that ar (AQC) = ar (PBR).

Ans. ABQ and BPQ lie on the same base BQ and between same parallels AP and BQ.

ar (ABQ) = ar (BPQ)……….(i)

BQC and BQR lie on the same base BQ and between same parallels BQ and CR.

ar (BQC) = ar (BQR)……….(ii)

Adding eq (i) and (ii),ar (ABQ) + ar (BQC) = ar (BPQ) + ar (BQR)

ar (AQC) = ar (PBR)

12. In figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you know 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?

Ans. In ABC, points D and E divides BC in three equal parts such that BD = DE = EC.

BD = DE = EC = BC

Draw AFBC

ar (ABC) = ……….(i)

and ar (ABD) = ……….(ii)

= =

= ar (ABC)……….(iii)

And ar (AEC) = ar (ABC)……….(iv)

From (ii), (iii) and (iv),

ar (ABD) = ar (ADE) = ar (AEC)

13. In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

Ans. As we know that opposite sides of a parallelogram are always equal.

In parallelogram ABFE,AE = BF and AB = EF

In parallelogram DCFE,DE = CF and DC = EF

In parallelogram ABCD,AD = BC and AB = DC

Now in ADE and BCF,

AE = BF[Opposite sides of parallelogram ABFE]

DE = CF[Opposite sides of parallelogram DCFE]

And AD = BC[Opposite sides of parallelogram ABCD]

ADE BCF[By SSS congruency]

ar (ADE) = ar (BCF)

[ Area of two congruent figures is always equal]

14. Prove that ABCD is a parallelogram. If ABCD is a quadrilateral and BD is one of its diagonal.

Ans. Given quadrilateral ABCD in which AB = DC = 3, BD = 4 and

BD intersects AB and DC such that.

Thus, ABCD is a parallelogram.

15. In a parallelogram ABCD, AB= 20. The altitude DM to sides AB is 10 cm. Find area of parallelogram.

Ans. Area of parallelogram ABCD

= ABDM

=2010

=200 square cm.

16. If L be any Point on AB and the area of rectangle ABCD is 100 square cm. find area of .

Ans. Area of rectangle ABCD = 100 square cm

Area area rectangle ABCD

=square cm

=50 square cm

17. Find the area of parallelogram ABCD, BD is perpendicular on AB. AB = 7 and BD is 5.

Ans. Area of parallelogram = Base Corresponding Altitude =75

= 35 square cm

Area of parallelogram = 35 square cm

18. Show that ar (ABC) = ar (ABD). ABC and ABD are two triangles on the same base AB if line segment CD is bisected by AO at O

Ans. AO is the median of

19. Show that BDEF is parallelogram. If D, E and F the mid- points of the side BC, CA and AB of triangle ABC

Ans. Join DE, EF and FD

E and F are the mid-points of AC and AB

EF|| BC

EF || BD

DE|| BF

BDEF is a || gram.

20. Prove that ar ()= ar () if MN||PO

Ans.

21. Justify the line corresponding to side EF if ar in and altitude AB is 5 cm and

Ans. Given that

22. In a parallelogram PQRS, PQ = 6 cm and the corresponding altitude ST is 5 cm. find area of parallelogram.

Ans. Area of ||gm PQRS

= Base Altitude

= 65 (Square cm)

= 30 square cm

23. Show that the median of a triangle divides it into two triangles of equal area.

Ans. Given: A triangle PQR and PS is the median

To prove: ar (PQS) = ar (PSR)

Construction: Draw the altitude PT from vertex P on the base QR

Proof: Area of
And Area of

area of of

Hence proved.

24. The area of rectangle PQRS is 500 sq cm. if T be any Point on PQ, find area of

Ans. As of as rectangle PQRS
Square cm
=250 square cm

25. Prove that as if PS||RQ

Ans.

26. Show that ar (quad. ABCD)=BD (AM+CN) BD is one of the diagonals of a quadrilateral ABCD, AM and CN are the from A and C

Ans.

27. D, E, F are respectively the mid-points of the sides BC, CA and AB of Prove that ar (DEF) = ar (ABC).

Ans.
Now,

28. In a parallelogram PQRS, PS = 12. The altitude to side PS is equal to 12cm. find area of parallelogram PQRS

Ans. Area of parallelogram PQRS

=Base Corresponding Altitude

=12 12

=144 square cm

29. A line through D, Parallel to AC meets BC produced in P. prove that area .

Ans.

30. In a parallelogram PQRS, PQ = 13. The altitude corresponding to sides PQ is equal to 5 cm. find the area of parallelogram.

Ans. Area of parallelogram = base Altitude

=135

=65 cm

31. Prove that ar (AOD) = ar (BOC). Diagonals AC and BD of a trapezium ABCD with AB||DC intersect each other at O.

Ans. Ar

32. Prove that ar (AQC) = ar (PBR) if AP||BQ||CR.

Ans.

3 Marks Quetions

1. 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).

Ans. Given: ABCD is a parallelogram. P is a point on DC and Q is a point on AD.

To prove: ar (APB) = ar (BQC)

Construction: Draw PM  BC and QN  DC.

Proof: Since QC is the diagonal of parallelogram QNCD.

 ar (QNC) =  ar (gm QNCD) ……….(i)

Again BQ is the diagonal of parallelogram ABNQ.

 ar (BQN) =  ar (gm ABNQ) ……….(ii)

Adding eq. (i) and (ii),

 ar (QNC) + ar (BQN) =  ar (gm QNCD) +  ar (gm ABNQ)

 ar (BQC) =  ar (gm ABCD)  ………..(iii)

Again AP is the diagonal of gm AMPD.

 ar (APM) =  ar (gm AMPD) ……….(iv)

And PB is the diagonal of gm PCBM.

 ar (PBM) =  ar (gm PCBM) ……….(v)

Adding eq. (iv) and (v),

 ar (APM) + ar (PBM) =  ar (gm AMPD) +  ar (gm PCBM)

 ar (APB) =  ar (gm ABCD) ……….(vi)

From eq. (iii) and (vi),

 ar (BQC) = ar (APB) or ar (APB) = ar (BQC)

2. 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 field 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?

Ans.  When A is joined with P and Q; the field is divided into three parts viz. PAS, APQ and AQR.

APQ and parallelogram PQRS are on the same base PQ and between same parallels PQ and SR.

 ar (APQ) =  ar (gm PQRS)

It implies that triangular region APQ covers half portion of parallelogram shaped field PQRS.

So if farmer sows wheat in triangular shaped field APQ then she will definitely sow pulses in other two triangular parts PAS and AQR.

Or

When she sows pulses in triangular shaped field APQ then she will sow wheat in other two triangular parts PAS and AQR.

3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Ans. Let parallelogram be ABCD and its diagonals AC and BD intersect each other at O.

In ABC and ADC,

AB = DC [Opposite sides of a parallelogram]

BC = AD [Opposite sides of a parallelogram]

And AC = AC [Common]

 ABC CDA [By SSS congruency]

Since, diagonals of a parallelogram bisect each other.

 O is the mid-point of bisection.

Now in ADC, DO is the median.

 ar (AOD) = ar (COD) ……….(i)

[Median divides a triangle into two equal areas]

Similarly, in ABC, OB is the median.

 ar (AOB) = ar (BOC) ……….(ii)

And in AOB and AOD, AO is the median.

 ar (AOB) = ar (AOD) ……….(iii)

From eq. (i), (ii) and (iii),

ar (AOB) = ar (AOD) = ar (BOC) = ar (COD)

Thus diagonals of parallelogram divide it into four triangles of equal area.

4 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).

Ans. Draw CMAB and DNAB.

In CMO and DNO,

  CMO = DNO =  [By construction]

  COM = DON  [Vertically opposite]

   OC = OD [Given]

 CMO DNO  [By ASA congruency]

  AM = DN [By CPCT] ……(i)

Now ar (ABC) =  ……….(ii)

ar (ADB) =  ……….(iii)

Using eq. (i) and (iii),

ar (ADB) =  ……….(iv)

From eq. (ii) and (iv),

ar (ABC) = ar (ADB)

5. XY is a line parallel to side BC of triangle ABC. If BEAC and CFAB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).

Ans. ABE and parallelogram BCYE lie on the same base BE and between the same parallels BE and AC.

 ar (ABE) =  ar (gm BCYE)  ……….(i)

Also ACF and gm BCFX lie on the same base CF and between same parallel BX and CF.

 ar (ACF) =  ar (gm BCFX) ……….(ii)

But gm BCYE and gm BCFX lie on the same base BC and between the same parallels BC and EF.

 ar (gm BCYE) = ar (gm BCFX)  ……….(iii)

From eq. (i), (ii) and (iii), we get,

  ar (ABE) = ar (ACF)

6. The side AB of 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 (ABCD) = ar (PBQR).

Ans. Given: ABCD is a parallelogram, CPAQ and PBQR is a parallelogram.

To prove: ar (ABCD) = ar (PBQR)

Construction: Join AC and QP.

Proof: Since AQCP

 ar (AQC) = ar (AQP)

[Triangles on the same base and between the same parallels are equal in area]

Subtracting ar (ABQ) from both sides, we get

ar (AQC – ar (ABQ) = ar (AQP) – ar (ABQ)

 ar (ABC) = ar (QBP)  ……….(i)

Now ar (ABC) =  ar (gm ABCD)

[Diagonal divides a parallelogram in two parts of equal area]

And ar (PQB) =  ar (gm PBQR)

From eq. (i), (ii) and (iii), we get

ar (gm ABCD) = ar (gm PBQR)

7. A villager Itwaari has a plot of land of the shape of quadrilateral. The Gram Panchyat of two villages decided to take over some portion of his plot from one of the corners to construct a health centre. Itwaari agrees to the above personal 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.

Ans. Let Itwaari has land in shape of quadrilateral PQRS.

Draw a line through 5 parallel to PR, which meets QR produced at M.

Let diagonals PM and RS of new formed quadrilateral intersect each other at point N.

We have PRSM [By construction]

 ar (PRS) = ar (PMR)

[Triangles on the same base and same parallel are equal in area]

Subtracting ar (PNR) from both sides,

ar (PRS) – ar (PNR) = ar (PMR) – ar (PNR)

 ar (PSN) = ar (MNR)

It implies that Itwari will give corner triangular shaped plot PSN to the Grampanchayat for health centre and will take equal amount of land (denoted by MNR) adjoining his plot so as to form a triangular plot PQM.

8. ABCD is a trapezium with ABDC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

Ans. Join CX, ADX and ACX lie on the same base  

XA and between the same parallels XA and DC.

 ar (ADX) = ar (ACX) ……….(i)

Also ACX and ACY lie on the same base

AC and between same parallels CY and XA.

 ar (ACX) = ar (ACY) ……….(ii)

From (i) and (ii),

ar (ADX) = ar (ACY)

9. 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.

Ans. Given that ar (AOD) = ar (BOC)

Adding AOB both sides,

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

 ar (ABD) = ar (ABC)

Since if two triangles equal in area, lie on the same base then, they lie between same parallels. We have ABD and ABC lie on common base AB and are equal in area.

 They lie in same parallels AB and DC.

 ABDC

Now in quadrilateral ABCD, we have ABDC

Therefore ABCD is trapezium.[ In trapezium one pair of opposite sides is parallel]

10. In figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Ans. Given that DRC and DPC lie on the same base DC and ar (DPC) = ar (DRC) …..(i)

 DCRP

[If two triangles equal in area, lie on the same base then, they lie between same parallels]

Therefore, DCPR is trapezium. [ In trapezium one pair of opposite sides is parallel]

Also ar (BDP) = ar (ARC) ……….(ii)

Subtracting eq. (i) from (ii),

  ar (BDP) – ar (DPC) = ar (ARC) – ar (DRC)

 ar (BDC) = ar (ADC)

Therefore, ABDC [If two triangles equal in area, lie on the same base then, they lie between same parallels]

Therefore, ABCD is trapezium.

11. 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.

Ans. Given: Parallelogram ABCD and rectangle ABEF are on same base AB and between the same parallels AB and CF.

 ar (gm ABCD) = ar (rect. ABEF)

To prove: AB + BC + CD + AD > AB + BE + EF + AF

Proof: AB = CD [ opposites sides of a

parallelogram are always equal]

AB = EF [ opposites sides of a

  rectangle are always equal]

 CD = EF

Adding AB both sides,

  AB + CD = AB + EF ……….(i)

 Off all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest.

 BE < BC and AF < AD

 BC > BE and AD > AF

 BC + AD > BE + AF ……….(ii)

From eq. (i) and (ii),

  AB + CD + BC + AD = AB + EF + BE + AF

12. In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).

Ans. Join A and C.

APC and BPC are on the same base PC and between the same parallels PC and AB.

 ar (APC) = ar (BPC) ……….(i)

Now ACBD is a parallelogram.

AD = BC [opposite sides of a parallelogram are always equal]

Also BC = CQ [given]

 AD = CQ

Now AD  CQ [Since CQ is the extension of BC]

And AD = CQ

 ADQC is a parallelogram.

[ If one pair of opposite sides of a quadrilateral is equal and parallel then it is a parallelogram]

Since diagonals of a parallelogram bisect each other.

 AP = PQ and CP = DP

Now in APC and DPQ,

AP = PQ [Proved above]

APC = DPQ  [Vertically opposite angles]

PC = PD [Prove above]

 APC DPQ  ……….(ii)

 ar (APC) = ar (DPQ) [area of congruent figures is always equal]

From eq. (i) and (ii),

ar (BPC) = ar (DPQ)

13. PQRS is a quadrilateral and SQ is one of its diagonals. Show that PQRS is a Parallelogram and find its area too.

Ans.  We know that, area of ||gram PQRS. In which

 PQ=SR=3, SQ=4

 And

 PQ||SR

 PQ=SR=3

 ABCD is a ||gram

 Area of parallelogram

 =Basecorresponding Altitude

 =34

 =12 square units

14. In a parallelogram PQRS. The Altitude corresponding to sides PQ and PS are respectively. 7 cm and 8 cm find PS, if PQ=10 cm.

 Ans.  Area of ||gram PQRS

 =PQSM

 =107

 =70 square cm…………..(i)

 Area of Parallelogram PQRS

 =PSQL

 =(AD8) square cm……………(ii)

 From (i) and (ii)

 PS8=70

 PS=

 =8.75 cm

15. Area, base and corresponding altitude are and respectively. Find the area of parallelogram.

Ans.  Area of parallelogram

= Base  Corresponding Altitude

x=12

= (12-3)(12-4)

= (9)(16)

= 144 square units.

16. Find the altitude corresponding to side EF if area of . If  AB = 8 cm and altitude corresponding to AB is 5 cm. In  

 Ans.  

 Altitude corresponding to side EF is 4 cm

17. Prove that the area of a trapezium is half of the product of its height and the sum of the parallel sides.

Ans.  Join B and D. Draw BLDC (Produced)

18. Show that the area of a rhombus is half the product of the length of its diagonals.

Ans.  

 Adding (i) and (ii)

 =

 Hence, area of rhombus ABCD =

19. In parallelogram P is any point inside it. Prove that

Ans.  

20. Show that

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

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

If is any point on side BR of PQRS and ABRS.

Ans.  (i) ||gram PQRS and ABRS are on the same base SR and Between the same parallel PB and SR,

 So,

 (ii)

21. Show that the diagonals of a parallelogram divide if into four triangles of equal area.

 Ans. Given: A parallelogram ABCD and AC and BC are diagonals

 To prove: ar (ABO) = ar (COD) = ar (BCO) = ar (AOD)

 Proof: ar (ADB) = ar (ACB)

 Ar (ADC) = ar (BCD)

 In triangle ABC, BO is median

 In triangle ADC, OD is median

 From (i), (ii), (iii) and (iv)

 Ar(ABO) = ar (CDO) = ar (BCO) = ar (ADO)

 Hence proved.

22. Show that PQ divides the ||gram in two Part of equal area if diagonal of ||gram ABCD intersect Point O. through Point O, a line is drawn to intersect AD at P and BC at Q.

Ans.  To prove: Are (quadrilateral APQB) = ar (quadrilateral PQCD) = (ar ||gram ABCD)

Proof: (Vertically opposite angles)

ar (quad. ABQO) +

ar (quad. APQB) = ar (quad. PQCD)

23. Show that = are of  if E is any Point on its median AD.

Ans.  Join BE and CE

 ………….(i)

 Subtracting (ii) from (i)

24. The triangle PQR and PSR are equal in area, if PR and QS bisect at O.

Ans.   

Adding (i) and (ii)

25. Show that ar, if median of intersect at G.

Ans.  AD is median

 GD is median

 Subtracting (ii) and (i)

 From (iii) and (iv)

26. Show that ar (ABCD) = ar (BQRP), AQ is drawn Parallel to CP to intersect CB produced to Q and parallelogram BQRP is completed if P is any Point on AB produced.

Ans.  AC is diagonal of ||gram ABCD

 From (i) ,(ii) and (iii)

27. Show that area of  area of  . D is mid-point of AB, P is any point on BC. PQ is joined and line CQ is drawn parallel to PD to intersect AB at Q.

Ans. CD is median

 From (i)

28. E is the mid-point of median AD, show that ar..

Ans.  

 Similarly, in  is the median

29. Show that the line segments joining the mid-points of parallel sides of a trapezium divides it into two parts of equal area

Ans.  

Draw and

 DM=CN=h

 Area of trapezium APQD

 Area of trapezium PBCQ

 From (i) and (2)

 ar (trap. APQD)= ar (trap. PBCQ)

30. Prove that arif AB||DC and line parallel to AC intersects AB at and BC at Y

Ans.  Join

31. Prove that area of = area of quadrilateral AFGE if BE and CF medians intersect at G.

Ans.  In , BE is the median

 Area =area

 Area =area = area +

 Now, CF median of  ar ()=area ()

32. Show that ararea area  area if diagonals of quadrilateral AC and BD intersect at a Point E.

Ans.  Draw AMBD and also CNBD

4 Marks Quetions

1. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:

Ans. Given: A quadrilateral ABCD, in which diagonals

AC and BD intersect each other at point E.
To Prove:

Construction: From A, draw AM BD and from C, draw CN BD.
Proof: ar (ABE) = ……….(i)
And ar (AED) = ……….(ii)
Dividing eq. (ii) by (i), we get,

……….(iii)
Similarly ……….(iv)

From eq. (iii) and (iv), we get
=

2. P and Q are respectively the mid-points of sides AB and BC or a triangle ABC and R is the mid-point of AP, show that:
(i) ar (PRQ) = ar (ARC)
(ii) ar (RQC) = ar (ABC)
(iii) ar (PBQ) = ar (ARC)

Ans. (i) PC is the median of ABC.
ar (BPC) = ar (APC) ……….(i)
RC is the median of APC.
ar (ARC) = ar (APC) ……….(ii)
[Median divides the triangle into two triangles of equal area]
PQ is the median of BPC.
ar (PQC) = ar (BPC) ……….(iii)
From eq. (i) and (iii), we get,
ar (PQC) = ar (APC) ……….(iv)
From eq. (ii) and (iv), we get,
ar (PQC) = ar (ARC) ……….(v)
We are given that P and Q are the mid-points of AB and BC respectively.
PQ AC and PA = AC
ar (APQ) = ar (PQC) ……….(vi) [triangles between same parallel are equal in area]
From eq. (v) and (vi), we get
ar (APQ) = ar (ARC) ……….(vii)
R is the mid-point of AP. Therefore RQ is the median of APQ.
ar (PRQ) = ar (APQ) ……….(viii)
From (vii) and (viii), we get,
ar (PRQ) = ar (ARC)
(ii) PQ is the median of BPC
ar (PQC) = ar (BPC) = x ar (ABC) = ar (ABC) ……….(ix)
Also ar (PRC) = ar (APC) [Using (iv)]
ar (PRC) = x ar (ABC) = ar (ABC) ……….(x)
Adding eq. (ix) and (x), we get,

ar (PQC) + ar (PRC) = ar (ABC)

ar (quad. PQCR) = ar (ABC) ……….(xi)

Subtracting ar (PRQ) from the both sides,

ar (quad. PQCR) – ar (PRQ) = ar (ABC) – ar (PRQ)

ar (RQC) = ar (ABC) – ar (ARC) [Using result (i)]

ar (ARC) = ar (ABC) – ar (APC)

ar (RQC) = ar (ABC) – ar (APC)

ar (RQC) = ar (ABC) – ar (ABC) [PC is median of ABC]

ar (RQC) = ar (ABC) – ar (ABC)

ar (RQC) = x ar (ABC)

ar (RQC) = ar (ABC)

(iii) ar (PRQ) = ar (ARC) [Using result (i)]

2 ar (PRQ) = ar (ARC) ..(xii)

ar (PRQ) = ar (APQ) [RQ is the median of APQ] ……….(xiii)

But ar (APQ) = ar (PQC) [Using reason of eq. (vi)] ……….(xiv)

From eq. (xiii) and (xiv), we get,

ar (PRQ) = ar (PQC) ……….(xv)

But ar (BPQ) = ar (PQC) [PQ is the median of BPC] ……….(xvi)

From eq. (xv) and (xvi), we get,

ar (PRQ) = ar (BPQ) ……….(xvii)

Now from (xii) and (xvii), we get,

2 = ar (ARC) ar (BPQ) = ar (ARC)

3. In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AXDE meets BC at Y. Show that:
(i) MBC ABD
(ii) ar (BYXD) = 2 ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) FCB ACE
(v) ar (CYXE) = 2 ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)

Ans. (i) ABM = CBD =
Adding ABC both sides, we get,
ABM + ABC = CBD + ABC
MBC = ABD ……….(i)
Now in MBC and ABD,
MB = AB [equal sides of square ABMN]
BC = BD [sides of square BCED]
MBC = ABD [proved above]
MBC ABD [By SAS congruency]

(ii) From above, MBC ABD
ar (MBC) = ar (ABD) ar (MBC) = ar (trap. ABDX) – ar (ADX)
ar (MBC) = (BD + AX) BY – DX.AX
ar (MBC) = BD.BY + AX.BY – DX.AX
ar (MBC) = BD.BY + AX (BY – DX)
ar (MBC) = BD.BY + AX. 0 [BY = DX]
ar (MBC) = BD.BY
2 ar (MBC) = BD.BY 2 ar (MBC) = ar (rect. BYXD)
Hence ar (BYXD) = 2 ar (MBC)

(iii) Join AM. ABMN is a square.
Therefore, NA MB AC MB
Now AMB and MBC are on the same base and between the same parallels MB and AC.
ar (AMB) = ar (MBC) ……….(ii)
From result (ii), we have ar (BYXD) = 2 ar (MBC) ……….(iii)
Using eq. (ii) and (iii), we get, ar (BYXD) = 2 ar (AMB)
ar (BYXD) = ar (square ABMN)
[Diagonal AM of square ABMN divides it in two triangles of equal area]

(iv) In FCB and ACE,
FC = AC [sides of square ACFG]
BC = CE [sides of square BCED]
BCF = ACE [ACF = BCE = ]
Adding ACB both sides,
BCF + ACB = ACE + ACB BCF = ACE
FCB ACE [By SAS congruency]

(v) From (iv), we have, FCB ACE
ar (FCB) = ar (ACE) ar (FCB) = ar (trap. ACEX) – ar (AEX)
ar (FCB) = (CE + AX) CY – XE.AX
ar (FCB) = CE.CY + AX.CY – XE.AX
ar (FCB) = CE.CY + AX (CY – XE)
ar (FCB) = CE.CY + AX. 0 [CY = XE]
ar (FCB) = CE.CY
2 ar (FCB) = CE.CY 2 ar (FCB) = ar (rect. CYXE)
Hence ar (BYXD) = 2 ar (FCB)

(vi) Join AF. ACFG is a square.
FC AG FC AB
Now ACF and FCB are on the same base FC and between the same parallels FC and AB.
ar (ACF) = ar (FCB) ……….(v)
From result (v), we get, ar (CYXE) = 2 ar (FCB) ……….(vi)
Using eq. (v) in (vi), we get, ar (CYXE) = 2 ar (ACF)
Diagonal AF of square ACFG divides it in two triangles of equal area.
ar (CYXE) = ar (sq. ACFG) ……….(vii)

(vii) Adding eq. (iv) and (vii), we get,
ar (BYXD) + ar (CYXE) = ar (ABMN) + ar (ACFG)
ar (BCED) = ar (ABMN) + ar (ACFG)

4. Prove that the parallelogram which is a rectangle has the greatest area.

Ans. Let PQRS be a parallelogram in which PQ = a and PS = b and h be the altitude corresponding to base PQ

Area of parallelogram PQRS = Base corresponding Altitude = ah

is a right angled triangle being its hypotenuse.

But hypotenuse is the greatest side of

Area of (ah) of ||gram PQRS will be greatest when h is greatest

H = b, then PSPQ

The ||gram PQRS will be a rectangle.

Hence, the area of ||gram is greatest when it is a rectangle.

5. Prove that

(i) ar(BDE)=ar(ABC)

(ii) ar(BDE)=ar(BAE)

If ABC and DBE are two equilateral triangles such that D is the mid-point of BC and AE intersects BC at F.

Ans. Join EC

(i) let a be the side of equilateral

From (i) and (ii)

(ii)

6. Show that EFGH is a ||gram and its area is half of the area of ||gram ABCD. If E, F, G, H are respectively the mid points of the sides AB, BC, CD and DA.

Ans. Join AC and HF

E and F are the mid-points of AB and BC

EF=AC and EF||AC……….(i)

Similarly, GH = AC and GH||AC………(ii)

From (i) and (ii)

GH = EF and GH||EF

EFGH is a ||gram

ar ...(iii)

ar……(iv)

Adding (iii) and (iv),

ar

7. Show that ar (BPC) = ar (DPQ) if BC is produced to a point Q such that AD = CQ and AQ intersect DC at P

Ans. Join AC


AD=CQ


Hence, a pair of opposite side AD and CQ of the quadrilateral ADQC is equal and parallel.
In
AP=QP
CP=DP



From (i) and (ii)

8. If area of and two points A and B are positive real number K. find the lows of a point p

Ans. Let the perpendicular distance of P from AB be h



Since AB and K are given h is a fixed Positive real number. This means that P lies on a line Parallel to AB at a distance h from it.
Hence, the locus of P is a pair of lines at a distance parallel to AB.

9. In figure, P is a point in the interior of a parallelogram ABCD. Show that:

(i) ar (APB) + ar (PCD) = ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

Ans. (i) Draw a line passing through point P and parallel to AB which intersects AD at Q and BC at R respectively.

Now APB and parallelogram ABRQ are on the same base AB and between same parallels AB and QR.
ar (APB) = ar (gm ABRQ) ……….(i)
Also PCD and parallelogram DCRQ are on the same base AB and between same parallels AB and QR.
ar (PCD) = ar (gm DCRQ) ……….(ii)

Adding eq. (i) and (ii),

ar (APB) + ar (PCD) = ar (gm ABRQ) + ar (gm DCRQ)

ar (APB) = ar (gm ABCD) ……….(iii)

(ii) Draw a line through P and parallel to AD which intersects AB at M and DC at N.

Now APD and parallelogram AMND are on the same base AD and between same parallels AD and MN.

ar (APD) = ar (gm AMND) ……….(iv)

Also PBC and parallelogram MNCB are on the same base BC and between same parallels BC and MN.

ar (PBC) = ar (gm MNCB) ……….(v)

Adding eq. (i) and (ii),

ar (APD) + ar (PBC) = ar (gm AMND) + ar (gm MNCB)

ar (APD) = ar (gm ABCD) ……….(vi)

From eq. (iii) and (vi), we get,

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

or ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

Hence proved.

10. 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)

Ans. (i) F is the mid-point of AB and E is the mid-point of AC.

FEBC and FE = BD

[ Line joining the mid-points of two sides of a triangle is parallel to the third and half of it]

FEBD [BD is the part of BC]

And FE = BD

Also, D is the mid-point of BC.

BD = BC

And FEBC and FE = BD

Again E is the mid-point of AC and D is the mid-point of BC.

DEAB and DE = AB

DEAB [BF is the part of AB]

And DE = BF

Again F is the mid-point of AB.

BF = AB

But DE = AB

DE = BF

Now we have FEBD and DEBF

And FE = BD and DE = BF

Therefore, BDEF is a parallelogram.

(ii) BDEF is a parallelogram.

ar (BDF) = ar (DEF) ……….(i) [diagonals of parallelogram divides it in two triangles of equal area]

DCEF is also parallelogram.

ar (DEF) = ar (DEC) ……….(ii)

Also, AEDF is also parallelogram.

ar (AFE) = ar (DEF) ……….(iii)

From eq. (i), (ii) and (iii),

ar (DEF) = ar (BDF) = ar (DEC) = ar (AFE) ……….(iv)

Now, ar (ABC) = ar (DEF) + ar (BDF) + ar (DEC) + ar (AFE) ……….(v)

ar (ABC) = ar (DEF) + ar (DEF) + ar (DEF) + ar (DEF) [Using (iv) & (v)]

ar (ABC) = 4 x ar (DEF)

ar (DEF) = ar (ABC)

(iii) ar (gm BDEF) = ar (BDF) + ar (DEF) = ar (DEF) + ar (DEF) [Using (iv)]

ar (gm BDEF) = 2 ar (DEF)

ar (gm BDEF) = 2 x ar (ABC)

ar (gm BDEF) = ar (ABC)

11. In 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) DACB or ABCD is a parallelogram.

Ans. (i) Draw BM AC and DN AC.

In DON and BOM,

OD = OB [Given]

DNO = BMO = [By construction]

DON = BOM [Vertically opposite]

DON BOM [By RHS congruency]

DN = BM [By CPCT]

Also ar (DON) = ar (BOM) ……….(i)

Again, In DCN and ABM,

CD = AB [Given]

DNC = BMA = [By construction]

DN = BM [Prove above]

DCN BAM [By RHS congruency]

ar (DCN) = ar (BAM) ……….(ii)

Adding eq. (i) and (ii),

ar (DON) + ar (DCN) = ar (BOM) + ar (BAM)

ar (DOC) = ar (AOB)

(ii) Since ar (DOC) = ar (AOB)

Adding ar BOC both sides,

ar (DOC) + ar BOC = ar (AOB) + ar BOC

ar (DCB) = ar (ACB)

(iii) Since ar (DCB) = ar (ACB)

Therefore, these two triangles in addition to be on the same base CB lie between two same parallels CB and DA.

DACB

Now AB = CD and DACB

Therefore, ABCD is a parallelogram.

12. In figure, ABC and BDF are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:

(i) ar (BDE) = ar (ABC)

(ii) ar (BDE) = ar (BAE)

(iii) ar (ABC) = 2 ar (BEC)

(iv) ar (BFE) = ar (AFD)

(v) ar (BFE) = 2 ar (FED)

(vi) ar (FED) = ar (AFC)

Ans. Join EC and AD.

Since ABC is an equilateral triangle.

A = B = C =

Also BDE is an equilateral triangle.

B = D = E =

If we take two lines, AC and BE and BC as a transversal.

Then B = C = [Alternate angles]

BE AC

Similarly, for lines AB and DE and BF as transversal.

Then B = C = [Alternate angles]

BE AC

(i) Area of equilateral triangle BDE = (BD)² ……….(i)

Area of equilateral triangle ABC = (BC)² ……….(ii)

Dividing eq. (i) by (ii),

[ BD = DC]

ar (BDE) = ar (ABC)

(ii) In BEC, ED is the median.

ar (BEC) = ar (BAE) ……….(i)

[Median divides the triangle in two triangles having equal area]

Now BE AC

And BEC and BAE are on the same base BE and between the same parallels BE and AC.

ar (BEC) = ar (BAE) ……….(ii)

Using eq. (i) and (ii), we get

Ar (BDE) = ar (BAE)

(iii) We have ar (BDE) = ar (ABC) [Proved in part (i)] ……….(iii)

ar (BDE) = ar (BAE) [Proved in part (ii)]

ar (BDE) = ar (BEC) [Using eq. (iii)] ……….(iv)

From eq. (iii) and (iv), we het

ar (ABC) = ar (BEC)

ar (ABC) = 2 ar (BEC)

(iv) BDE and AED are on the same base DE and between same parallels AB and DE.

ar (BDE) = ar (AED)

Subtracting FED from both the sides,

ar (BDE) – ar (FED) = ar (AED) – ar (FED)

ar (BFE) = ar (AFD) ……….(v)

(v) An in equilateral triangle, median drawn is also perpendicular to the side,