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(||^gmABCD)=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.

[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)

[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.

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 ||^gmABCD 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 :