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KC Sinha Mathematics Solution Class 9 Chapter 11 Area of parallelograms and triangles exercise 11.1

Page 11.29

EXERCISE 11.1


Type 1

 


QUESTION 1

Fill in the blanks:

(i) Areas of congruent rectangles are __

(ii) The area of a rhombus is __the product of the lengths of its diagonals.

(iii) The area of a trapezium is __ the product of the distance between parallel sides and sum of the parallel sides.

(iv) The areas of congruent triangles are __

(v) The areas of the triangle on the same base (or equal bases) and between the same parallel lines are

Sol :

 


QUESTION 2

(i) In ΔABC and ΔDBC, formed on the same side of the base BC,  AD||BC. If perpendiculars be dropped from A and D on the base BC, then what will be the relation between their lengths ?

(ii) What is the relation between the areas of the two triangles formed on the same base and between the same parallel lines ?

(iii) In rectangle ABCD , AB=6cm and AD=4cm , then what is the area of ΔPAB

(iii) In rectangle ABCD , AB=6cm and AD=4cm , then what is the area of ΔPAB

 

(iv) Diagonals of parallelogram divide it in four triangles. What is the relation between their ages ?

(v) In ΔABC , line segment AD is a median , then what is the relation between ar(ΔABD) and ar(ΔADC) ?

 In ΔABC , line segment AD is a median , then what is the relation between ar(ΔABD) and ar(ΔADC) ?
Sol :

 

 


TYPE 2

 


QUESTION 3

The area of a parallelogram ABCD is 40 sq cm . What will be the area of a triangle PCD (in sq cm) formed by taking  a point P on AB ?

Sol :

 


QUESTION 4

The area of a rectangle ABCD is 50 sq cm . What will be area of a triangle PCD (in sq cm) formed by taking a point P on AB ?

Sol :

 


QUESTION 5

In ΔABC , AC=6cm and the attitude corresponding to side AC=4 cm . In ΔDEF , EF=8cm . If ar(ΔABC)=ar(ΔDEF) , then find the altitude corresponding to side EF

Sol :

 


QUESTION 6

ABCD is a trapezium in which AB||CD . If AB=10 cm , CD=7 cm and area of the trapezium is 102 sq cm , then find the height of the trapezium .

Sol :

 


QUESTION 7

From the adjoining figure , find 

(i) Area of ΔBCN

(ii) Area of parallelogram ABCD

(iii) Area of trapezium ANCD

(iv) Area of rectangle AMCN

<fig to  be added>

Sol :

 


TYPE 3


QUESTION 8

In the adjoining figure , ABCD is a parallelogram in which line segment EH⊥DC, CF⊥AB produced and BG⊥DA. If AB=16cm , EH=8cm and BG=10 cm , then find AD

<fig to  be added>

[Hint: DC=AB=16 cm

ar(||gm ABCD)=DC×HE=AD×BG

16×8=AD×10⇒AD=\dfrac{16\times 8}{10}=12.8cm]

Sol :

 


QUESTION 9

In parallelogram ABCD , AB=10 cm and altitude corresponding to the sides AB and AD are 7 cm and 8 cm respectively. Find AD. 

Sol :

 


QUESTION 10

Show that the diagonals of a rectangle divide it in four triangle of equal area .

Sol :

 


QUESTION 11

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

Show that the diagonals of a parallelogram divide it in four triangles of equal area.
[Hint : ∴OA=OC and OB=OD

∴ In ΔADB , ar(ΔAOD)=ar(ΔAOB)

Similarly , in ΔABC ,

ar(ΔAOB)=ar(ΔBOC)

In ΔBCD, ar(ΔOBC)=ar(ΔOCD)

In ΔADC, ar(ΔCOD)=ar(ΔAOD)

ar(ΔAOB)=ar(ΔBOC)=ar(ΔOCD)=ar(ΔDOA)]

]

Sol :

 


QUESTION 12

In the given f‌igure, there is a point E on the median of ΔABC , then show that

ar(ΔABE)=ar(ΔACE)

In the given f‌igure, there is a point E on the median of ΔABC , then show that ar(ΔABE)=ar(ΔACE)
[Hint: Construction: Draw AM⊥BC

Proof: ar(ΔABD)=1/2 BD×AM

ar(ΔACD)=1/2 DC×AM

=1/2 BD×AM [∵ BD=DC]

∴ ar(ΔABD)=ar(ΔACD)..(i)

Similarly, ar(ΔEBD)=ar(ΔEDC)..(ii)

Subtract (ii) from(i) , we get

ar(ΔABE)=ar(ΔAEC)]

Sol :

 


QUESTION 13

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

Sol :

Let ABC be a triangle with median AD

To prove :  ar(ABD)=ar(ΔACD)

Construction : Draw AM⊥BC

Proof : ar(ΔABD)=1/2 BD×AM

ar(ΔACD)=1/2 DC×AM

=1/2 BD×AM [∵ BD=CD]

ar(ABD)=ar(ΔADC)

 


QUESTION 14

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(ΔBPC)=ar(ΔDPQ)

[Hint: Join AC now , since ΔAPC and ΔPBC are on the same PC and between the same parallels PC and AB]

∴ ar(ΔAPC)=ar(ΔBPC)..(i)

Again , ABCD is a parallelogram,

∴ AD=BC=CQ [∵ AD=CQ]

Also AD||CQ and AD=CQ

∴ ADQC is a parallelogram as one pair of opposite sides are equal and parallel

∴ AP=PQ and CP=DP [∵ Diagonals of a ||gm bisect each other]

Now consider ΔAPC and ΔDPQ,

∴ AP=PQ ; ∠APC=∠DPQ [Vertically opposite angles]

∴ ΔAPC≅ΔQPD [By SAS congruence rule]

∴ ar(ΔBCP)=ar(ΔDPQ)
Sol :

 


QUESTION 15

ABCD is a quadrilateral and its diagonals AC and BD intersect each other at point O . If BO=OD , then prove that ΔABC and ΔADC are equal in area.
[Hint: BO=DO ,

∴AO is a median

∴ ar(ΔABO)=ar(ΔADO)..(i)

Similarly, ar(ΔBOC)=ar(ΔCOD)..(ii)

On adding (i) and (ii) , we get

ar(ΔABC)=ar(ΔACD)  ]

<fig to  be added>

Sol :

 


QUESTION 16

P and Q are two points on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(ΔAPB)=ar(ΔBQC)

[Hint: ΔAPB and ||gm ABCD stand on the same base AB and between the same parallel lines AB and CD

∴ ar(ΔAPB)=1/2 ar(||gm ABCD)..(i)

Similarly, ar(ΔBQC)=1/2 ar(||gm ABCD)..(ii)

∴ ar(ΔAPB)=ar(ΔBQC)]

Sol :

 


QUESTION 17

ΔABC and ΔABD are on the common base AB. Line segment CD is bisected by AB at O . Show that both the triangles are equal in area

<fig to be added>

[Hint : Draw BM⊥CD

Proof: ar(ΔBOD)=1/2 OD×BM

ar(ΔOBC)=1/2 OC×BM

=1/2 ×OD×BM [∵ OC=OD]

∵ ar(ΔBOD)=ar(ΔOBC) ..(i)

Similarly , ar(ΔAOD)=ar(ΔAOC)..(ii)

Adding (i) and (ii) , we get

ar(ΔABD)=ar(ΔABC)]

Sol :

 


QUESTION 18

Prove that the line segment joining the mid-points of the opposite sides of a parallelogram divide it into four parallelograms of equal area.

Sol :

 


QUESTION 19

Diagonals of a parallelogram ABCD intersect at the point O.  A line through O intersect AB at X and CD at Y . Show that

ar(AXYD)=1/2 ar(||gm ABCD)

[Hint: Prove that ΔAOX≅ΔCOY]

Sol :

 


QUESTION 20

E , F , G and H are the mid-points of the sides AB, BC, DC and DA respectively of a parallelogram ABCD. Show that quadrilateral EFGH is a parallelogram and its area is half the area of parallelogram ABCD

[Hint: Join HF

Proof: Since DH= and || CF

∴ HDFC is a ||gm , similarly HA= and parallel to FB

∴ HABF is a ||gm

Now ΔHFG and ||gm HDCF stand on the same base HF and between the same parallels HF and DC

∴ ar(ΔHGF)=1/2 ar(HDCF)

Similarly ΔHEF and ||gm HABF stand on the same base HF and between the same parallels HF and DC

∴ ar(ΔHEF)=1/2 ar(HABF) ..(ii)

Adding (i) and (ii) , we get

ar(ΔHGF)+ar(HEF)=1/2 [ar(HDCF)+ar(ΔHABF)]

⇒ ar(EFGH)=1/2 ar(ABCD)]

]

Sol :

 


QUESTION 21

Prove that line segment joining the mid-points of the parallel sides of a trapezium divides its in two equal parts

[Hint: Join mid-point of a parallel side to remaining two vertices]

Sol :

 


QUESTION 22

Medians BE and CF of the ΔABC intersect at G. Prove that

ar(ΔGBC)=ar(AFGE)

[Hint: Join EF and prove that

ar(ΔBFG)=ar(ΔCEG)

ar(ΔBEC)=ar(ΔABE) [∵ BE is median]

ar(ΔBGC)+ar(ΔCEG)=ar(ΔBFG)=ar(AFGE)

ar(ΔGBC)=ar(AFGE)

 

Sol :

 


QUESTION 23

Prove that of all parallelograms with given sides , rectangle has the greatest area.

Sol :

 


QUESTION 24

D and E are the mid-points of sides AB and AC respectively of ΔABC and 

ar(ΔBCE)=ar(ΔBCD). Prove that DE||BC

[Hint: Draw DM⊥BC and EN⊥BC

∵ ar(ΔBCE)=ar(ΔBCD) [Given]

⇒ 1/2 BC×DM=1/2 BC×EN

⇒ DM=EN

Now

∠1=∠2=90° and these are pair of corresponding angles

∴ DM||EN

In quadrilateral DMNE ,

∴ DMNE is a parallelogram

∴ DE||MN or DE||BC ]

<fig to be added>

Sol :

 


QUESTION 25

If each diagonal of a quadrilateral divides it in two triangles of equal area  , then show that this quadrilateral is a parallelogram

Sol :

 


QUESTION 26

Diagonals AC and BD of a quadrilateral ABCD intersect at point O such that they (diagonals) divide the quadrilateral in four triangles of equal area. Show that quadrilateral ABCD is a parallelogram

<fig to be added>

[Hint: Given

ar(ΔAOB)=ar(ΔBOC)=ar(ΔCOD)=ar(ΔDOA)

ar(ΔAOB)=ar(ΔCOD)

∴ ar(ΔAOB)+ar(ΔBOC)=ar(ΔCOD)+ar(ΔBOC)

or ar(ΔABC)=ar(ΔBCD)

Now ΔABC and ΔBCD stand on the same base BC

∴ Perpendicular from A in ΔABC= Perpendicular from D and ΔBCD

∴ AD||BC , similarly AB||BD ]

Sol :

 


QUESTION 27

Two triangles ABC and DBC are on the same base BC and their vertices A and D lie on the opposite sides of BC so that ar(ΔABC)=ar(ΔBCD) . Show that line segment BC bisects AD

<fig to be added>

[Hint: Draw AE⊥BC and DF⊥BC

ar(ΔABC)=ar(ΔBDC)

1/2 ×BC×AE = 1/2 ×BC×DF

AE=DF

Now in ΔAEO and ΔDEO

∠AEO=∠DFO=90°

∠AOE=∠DOF [Vertically Opposite Angles]

∠EAO=∠FDO

ΔAEO≅DFO

AO=OD

So , BC bisects AD]  [CPCT]

Sol :

 


QUESTION 28

In figure , medians of the ΔABC intersect at point G . Show that

ar(ΔAGB)=1/3 ar(ΔABC)

<fig to be added>

[Hint: Join AG

Since GF is the median of ΔAGB

ar(ΔGAF)=ar(ΔGFB)

ar(ΔAGB)=2ar(ΔGFB)..(i)

Similarly , ar(ΔACG)=2ar(GEC)..(ii)

Since ΔBFC and ΔBEC stand on the same side base BC and between the same parallel lines FE and BC

∴ ar(ΔBFC)=ar(ΔBEC)

∴ ar(ΔBGF)+ar(ΔBGC)=ar(ΔBGC)+ar(ΔCGE)

∴ ar(ΔBGF)=ar(ΔCGE)

∴ 2ar(ΔBGF)=2ar(ΔCGE)

∴ 2ar(ΔABG)=ar(ΔACG) [From (i) and (ii)] ..(iii)

Similarly we can show that

ar(ΔABG)=ar(ΔBGC)

Now , ar(ΔABC) ..(iv)

=ar(ΔABG)+ar(ΔBGC)+ar(ΔAGC)

=ar(ΔABG)+ar(ΔABG)+ar(ΔABG)

[From (iii) and (iv)]

=3ar(ΔABG)

ar(ΔABG)=1/3 ar(ΔABC)

]

Sol :

 


QUESTION 29

In ΔABC  , D is the mid-point of the side AB. P is any point on BC. CQ is || PD and intersects AB and Q as shown in the adjoining figure. Prove that 

ar(ΔBPQ)=1/2 ar(ΔABC)

<fig to be added>

[Hint: Join PQ and CD , then we have

ar(ΔBCD)=ar(ΔACD) [CD is median of ΔABC]

ar(ΔBCD)=1/2 ar(ΔABC) ..(i)

Now , ar(ΔDPQ)=ar(ΔDPC)

[These lie on the same base DP and between the same parallels DP and QC]

ar(ΔDPQ)+ar(ΔDBP)=ar(ΔDPC)+ar(ΔDBP)

ar(ΔBPQ)=ar(ΔBCD)=1/2 ar(ΔABC) [From (i)]

 

Sol :

 


QUESTION 30

In a figure  , a line XY is parallel to the side BC of ΔABC , BE||AC and CF||AB and BE and CF intersect line XY at E and F respectively . Prove that

ar(ΔABE)=ar(ΔACF)

<fig to be added>

[Hint: Construction: Through A , draw PQ||BC

ar(||gm PBCA)=ar(||gm ABCQ)

[These ||gm stand on the same base BC between  the parallel lines BC and PQ]

]

 

Sol :

 


QUESTION 31

D is any point on the base BC of ΔABC. AD is extended to E such that AD=DE . Show that

ar(ΔBCE)=ar(ΔABC)

<fig to be added>

[Hint: join B and C to the point E

Since AD=DE , therefore , D is the mid-point of AE

In ΔACE , ar(ΔACE)=ar(ΔCDE)  [∵ ‎ CD is median]..(i)

Similarly , in ΔABE , ar(ΔABD)=ar(ΔBDE) [∵ BD is median] ..(ii)

Adding (i) and (ii) , we get

ar(ΔACD)+ar(ΔABD)=ar(ΔCDE)+ar(ΔBDE)

∴ ar(ΔABC)=ar(ΔBCE)]

Sol :

 


QUESTION 32

Area of parallelogram PQRS and PABC are equal . Prove that SA||BR

<fig to be added>

[Hint: Produce QR and CB]

Sol :

 


QUESTION 33

In figure , diagonals of parallelogram ABCD , intersect at point O . Form O , a lines is drawn which meets AD at P and BC at Q . Prove that PQ divides parallelogram ABCD intwo parts of equal area

<fig to be added>

[Hint: ΔAOP≅ΔCOQ

∴ ar(ΔABC)=ar(ΔADC) [AC is diagonals of ||gm ABCD]

or ar(ABQO)+ar(ΔQOC)=ar(COPD)+ar(ΔAOP)

or ar(ABQO)+ar(ΔAOP)=ar(COPD)+ar(ΔCOQ)

or ar(ABQP)=ar(QCDP)]

Sol :

 


QUESTION 34

In figure , ABCD is a quadrilateral and E is the mid-point of AC . Prove that

ar(ABED)=ar(BCDE)

<fig to be added>

[Hint: Since E is the mid-point of AC , so DE and BE are the medians of ΔACD and ΔABC respectively

Clearly , ar(ΔADE)=ar(ΔDCE)..(i)

and ar(ΔABE)=ar(ΔBEC)..(ii)

On adding (i) and (ii) , we get

ar(ΔADE)+ar(ΔABE)=ar(ΔDCE)+ar(ΔBEC)

⇒ ar(ABED)=ar(BCDE)

]

Sol :

 


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