R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 11 – Areas of Parallelograms and Triangles
Chapter 11 – Areas of Parallelograms and Triangles Exercise Ex. 11
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.
Fig (i)
Fig (ii)
Fig (iii)
Fig (iv)
Fig (v)
Fig (vi)
Following
figures lie on the same base and between the same parallels:
Figure
(i): No
Figure
(ii): No
Figure
(iii): Yes, common base – AB, parallel lines – AB and DE
Figure
(iv): No
Figure
(v): Yes, common base – BC, parallel lines – BC and AD
Figure
(vi): Yes, common base – CD, parallel lines – CD and BP
If P and Q are any two points
lying respectively on the sides DC and AD of a parallelogram ABCD then show
that ar(ΔAPB) = ar(ΔBQC).
Since
ΔAPB and
parallelogram ABCD are on the same base AB and between the same parallels AB
and DC, we have
Similarly,
ΔBQC and
parallelogram ABCD are on the same base BC and between the same parallels BC
and AD, we have
From (i) and (ii),
A(ΔAPB) = A(ΔBQC)
In the adjoining figure, MNPQ and
ABPQ are parallelograms and T is any point on the side BP. Prove that
(i) Parallelograms MNPQ and ABPQ are
on the same base PQ and between the same parallels PQ and MB.
(ii) ΔATQ and parallelogram ABPQ are on
the same base AQ and between the same parallels AQ and BP.
In the adjoining 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(Δ ABC) = ar(Δ
ABD).
We
know that median of a triangle divides it into two triangles of equal area.
Now,
AO is the median of ΔACD.
⇒ A(ΔCOA) = A(ΔDOA) ….(i)
And,
BO is the median of ΔBCD.
⇒ A(ΔCOB) = A(ΔDOB) ….(ii)
Adding
(i) and (ii), we get
A(ΔCOA) + A(ΔCOB) = A(ΔDOA) + A(ΔDOB)
⇒ A(ΔABC) = A(ΔABD)
D and E are points on sides AB and
AC respectively of Δ ABC such that ar(ΔBCD) = ar(ΔBCE).
Prove that DE ∥ BC.
Since
ΔBCD and ΔBCE are equal
in area and have a same base BC.
Therefore,
Altitude
from D of ΔBCD = Altitude from E of ΔBCE
⇒ ΔBCD and ΔBCE are between
the same parallel lines.
⇒ DE ∥ BC
P is any point on the diagonal AC
of parallelogram ABCD. Prove that ar(ΔADP) = ar(ΔABP).
Construction:
Join BD.
Let
the diagonals AC and BD intersect at point O.
Diagonals
of a parallelogram bisect each other.
Hence,
O is the mid-point of both AC and BD.
We
know that the median of a triangle divides it into two triangles of equal
area.
In
ΔABD, OA is the
median.
⇒ A(ΔAOD) = A(ΔAOB) ….(i)
In
ΔBPD, OP is the
median.
⇒ A(ΔOPD) = A(ΔOPB) ….(ii)
Adding
(i) and (ii), we get
A(ΔAOD) + A(ΔOPD) = A(ΔAOB) + A(ΔOPB)
⇒ A(ΔADP) = A(ΔABP)
In a trapezium ABCD, AB ∥
DC and M is the midpoint of BC. Through M, a line PQ ∥
AD has been drawn which meets AB in P and DC produced in Q, as shown in the
adjoining figure. Prove that ar(ABCD) = ar(APQD).
In
ΔMCQ and ΔMPB,
∠QCM = ∠PBM (alternate angles)
CM
= BM (M is the mid-point of
BC)
∠CMQ = ∠PMB (vertically opposite angles)
∴ ΔMCQ ≅ ΔMPB
⇒ A(ΔMCQ) = A(ΔMPB)
Now,
A(ABCD)
= A(APQD) + A(DMPB) – A(ΔMCQ)
⇒ A(ABCD) =
A(APQD)
ABCD is a parallelogram in which
BC is produced to P such that CP = BC, as shown in the adjoining figure. AP
intersect CD at M. If ar(DMB) = 7 cm^{2}, find the area of parallelogram
ABCD.
In
ΔADM and ΔPCM,
∠ADM = ∠PCM (alternate angles)
AD
= CP (AD = BC = CP)
∠AMD = ∠PMC (vertically opposite angles)
∴ ΔADM ≅ ΔPCM
⇒ A(ΔADM) = A(ΔPCM)
And,
DM = CM (c.p.c.t.)
⇒ BM is the
median of ΔBDC.
⇒ A(ΔDMB) = A(ΔCMB)
⇒ A(ΔBDC) = 2 × A(ΔDMB) = 2 × 7 = 14 cm^{2}
Now,
A(parallelogram
ABCD) = 2 × A(ΔBDC) = 2 × 14 = 28 cm^{2}
In a parallelogram ABCD, any point
E is taken on the side BC. AE and DC when produced meet at a point M. Prove
that
ar(ΔADM) – ar(ABMC)
Construction:
Join AC and BM
Let
h be the distance between AB and CD.
In a triangle ABC, the medians BE
and CF intersect at G. Prove that ar(ΔBCG) = ar(AFGE).
Construction: Join EF
Since the line segment
joining the mid-points of two sides of a triangle is parallel to the third
side,
FE ∥ BC
Clearly, ΔBEF and ΔCEF are on the
same base EF and between the same parallel lines.
∴ A(ΔBEF) = A(ΔCEF)
⇒ A(ΔBEF) – A(ΔGEF) = A(ΔCEF) – A(ΔGEF)
⇒ A(ΔBFG) = A(ΔCEG) …(i)
We know that a median of
a triangle divides it into two triangles of equal area.
⇒ A(ΔBEC) = A(ΔABE)
⇒ A(ΔBGC) + A(ΔCEG) = A(quad.
AFGE) + A(ΔBFG)
⇒ A(ΔBGC) + A(ΔBFG) = A(quad.
AFGE) + A(ΔBFG) [Using (i)]
⇒ A(A(ΔBGC) = A(quad. AFGE)
ΔDBC and ΔEBC are on the
same base and between the same parallels.
⇒ A(ΔDBC) = A(ΔEBC) ….(i)
BE
is the median of ΔABC.
In the adjoining figure, CE ∥
AD and CF ∥ BA. Prove that ar(ΔCBG) = ar(ΔAFG).
ΔBCF and ΔACF are on the
same base CF and between the same parallel lines CF and BA.
∴ A(ΔBCF) = A(ΔACF)
⇒ A(ΔBCF) – A(ΔCGF) = A(ΔACF) – A(ΔCGF)
⇒ A(ΔCBG) = A(ΔAFG)
In a trapezium ABCD, AB ∥
DC, AB = a cm, and DC = b cm. If M and N are the midpoints of the nonparallel
sides, AD and BC respectively then find the ratio of ar(DCNM) and ar(MNBA).
Construction:
Join DB. Let DB cut MN at point Y.
M
and N are the mid-points of AD and BC respectively.
⇒ MN ∥ AB ∥ CD
In
ΔADB, M is the
mid-point of AD and MY ∥ AB.
∴ Y is the
mid-point of DB.
Similarly,
in ΔBDC,
Now,
MN = MY + YN
Construction:
Draw DQ ⊥ AB. Let DQ cut MN at point P.
Then,
P is the mid-point of DQ.
i.e.
DP = PQ = h (say)
ABCD is a trapezium in which AB ∥
DC, AB = 16 cm and DC = 24 cm. If E and F are respectively the midpoints of
AD and BC, prove that
Construction:
Join AC. Let AC cut EF at point Y.
E
and F are the mid-points of AD and BC respectively.
⇒ EF ∥ AB ∥ CD
In
ΔADC, E is the
mid-point of AD and EY ∥ CD.
∴ Y is the
mid-point of AC.
Similarly,
in ΔABC,
Now,
EF = EY + YF
Construction:
Draw AQ ⊥ DC. Let AQ cut EF at point P.
Then,
P is the mid-point of AQ.
i.e.
AP = PQ = h (say)
In the adjoining figure, D and E
are respectively the midpoints of sides AB and AC of ΔABC.
If PQ ∥
BC and CDP and BEQ are straight lines then prove that ar(ΔABQ)
= ar(ΔACP).
Since
D and E are the mid-points of AB and AC respectively,
DE
∥ BC ∥ PQ
In
ΔACP, AP ∥ DE and E is the mid-point of AC.
⇒ D is the
mid-point of PC (converse of
mid-point theorem)
In
ΔABQ, AQ ∥ DE and D is the mid-point of AB.
⇒ E is the
mid-point of BQ (converse of
mid-point theorem)
From
(i) and (ii),
AP
= AQ
Now,
ΔACP and ΔABQ are on the
equal bases AP and AQ and between the same parallels BC and PQ.
⇒ A(ΔACP) = A(ΔABQ)
In the adjoining figure, ABCD and
BQSC are two parallelograms. Prove that ar(ΔRSC) = ar(ΔPQB).
In
ΔRSC and ΔPQB,
∠CRS = ∠BPQ (RC ∥ PB,
corresponding angles)
∠RSC = ∠PQB (RC ∥ PB, corresponding angles)
SC
= QB (opposite sides of a
parallelogram BQSC)
∴ ΔRSC ≅ ΔPQB (by AAS congruence criterion)
⇒ A(ΔRSC) = A(ΔPQB)
Find the area of a figure formed
by joining the midpoints of the adjacent sides of a rhombus with diagonals 12
cm and 16 cm.
In the adjoining figure, ABCD is a
trapezium in which AB ∥ DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB
and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD.
M is the midpoint of the side AB
of a parallelogram ABCD. If ar(AMCD) = 24 cm^{2}, find ar(Δ
ABC).
Construction:
Join AC
Diagonal
AC divides the parallelogram ABCD into two triangles of equal area.
⇒ A(ΔADC) = A(ΔABC) ….(i)
ΔADC and
parallelogram ABCD are on the same base CD and between the same parallel
lines DC and AM.
Since
M is the mid-point of AB,
A(AMCD)
= A(ΔADC) + A(ΔAMC)
Chapter 11 – Areas of Parallelograms and Triangles Exercise Ex. 11A
In the adjoining figure, show that ABCD is a parallelogram. Calculate the area of ||gm ABCD.
_{}
_{}
In the adjoining figure , ABCD is a quadrilateral in which diag. BD =14cm . If such that AL=8 cm. and CM =6 cm, find the area of quadrilateral ABCD.
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In the adjoining figure , ABCD is a trapezium in which AB||DC and its diagonals AC and BD intersects at O. prove that
_{ }
_{In the adjoining figure , DE||BC. Prove that }
_{ }
_{ }
Prove that the median divides a triangle into two triangles of equal area.
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Show that the diagonal divides a parallelogram into two triangles of equal area.
_{ }
_{In adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O. If BO=OD, prove that }
_{}
_{}
The vertex A of _{ is joined to a point D on the side BC. The mid-point of AD is E. prove that}
_{ }
_{}
_{}
_{D is the mid-point of side BC of and E is the midpoint of BD. If O is midpoint of AE, prove that }
_{}
_{In the adjoining figure , ABCD is a quadrilateral. A line through D , parallel to AC , meets BC produced in P. prove that }
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_{}
In the adjoining figure , _{are on the same base BC with A and D on opposite sides of BC such that . Show that BC bisects AD.}
_{}
_{ }
_{P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of ||gm ABCD. Show that PQRS is a parallelogram and also that }
_{Ar(||gm PQRS)=x ar (||gm ABCD)}
_{}
In a parallelogram ABCD, it is
being given that AB= 10 cm and the altitudes corresponding to the sides AB and AD are DL
=6 cm and BM =8cm, respectively
Find AD.
_{}
_{}
The base BC of _{ is divided at D such that . }
_{Prove that .}
_{}
_{The given figure shows the pentagon ABCDE. EG, drawn parallel to DA, meetsBA producedat G ,andCF, drawnparallel to DB, meets AB produced at F. }
_{Showthat }
_{}
_{In the adjoining figure , the point D divides the side BC of in the ratio m:n. Prove that }
_{}
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_{}
Find
the area of the trapezium whose
parallel sides are 9 cm and 6 cm respectively and the distance between these sides
is 8 cm.
_{}
_{}
Calculate the area of quadrilateral ABCD givenin Fig (i)
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Calculatethe area of trapezium PQRS , givenin Fig.(ii)
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_{}
BD is one of the diagonals of a quad. ABCD . If , show that
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Chapter 11 – Areas of Parallelograms and Triangles Exercise MCQ
Out of the following
given figures which are on the same base but not between the same parables?
Two parallelograms are on equal bases and between the same
parallels. The ratio of their areas is
(a) 1
: 2
(b) 1
: 1
(c) 2
: 1
(d) 3 : 1
Correct
option: (b)
Parallelograms on equal bases and between the same
parallels are equal in area.
In the given figure, ABCD and ABPQ are two parallelograms
and M is a point on AQ and BMP is a triangle.
(a) true
(b) false
Correct
option: (a)
ΔBMP
and parallelogram ABPQ are on the same base BP and between the same parallels
AQ and BP.
Parallelograms ABPQ and ABCD are
on the same base AB and between the same parallels AB and PD.
The midpoints of the sides of a triangle along with any of
the vertices as the fourth point makes a parallelogram of area equal to
(a)
(b)
(c)
(d) ar(ΔABC)
Correct
option: (a)
ΔABC
is divided into four triangles of equal area.
A(parallelogram AFDE) = A(ΔAFE)
+ A(DFE)
The lengths of the
diagonals of a rhombus are 12 cm and 16 cm. The area of the rhombus is
- 192
cm^{2} - 96
cm^{2} - 64
cm^{2} - 80
cm^{2}
Two parallel sides of a
trapezium are 12 cm and 8 cm long and the distance between them is 6.5 cm.
The area of the trapezium is
- 74
cm^{2} - 32.5
cm^{2} - 65
cm^{2} - 130
cm^{2}
In the given figure ABCD
is a trapezium such that AL ⊥ DC and BM ⊥ DC. If AB = 7
cm, BC = AD = 5 cm and AL = BM = 4 cm, then ar(trap. ABCD)= ?
- 24
cm^{2} - 40
cm^{2} - 55
cm^{2} - 27.5
cm^{2}
In a quadrilateral ABCD,
it is given that BD = 16 cm. If AL ⊥ BD and CM ⊥ BD such that
AL = 9 cm and CM = 7 cm, then ar(quad. ABCD) = ?
- 256
cm^{2} - 128
cm^{2} - 64
cm^{2} - 96
cm^{2}
ABCD is a rhombus in
which ∠C = 60°.
Then, AC : BD = ?
In the given figure ABCD
and ABFE are parallelograms such that ar(quad. EABC) = 17cm^{2} and
ar(‖gm ABCD) = 25
cm^{2}. Then, ar(ΔBCF) = ?
- 4
cm^{2} - 4.8
cm^{2} - 6
cm^{2} - 8
cm^{2}
ΔABC and ΔBDE are two
equilateral triangles such that D is the midpoint of BC. Then, ar(ΔBDE) : ar(ΔABC) = ?
(a) 1 : 2
(b) 1 : 4
(c)
(d) 3 : 4
In which of the
following figures, you find polynomials on the same base and between the same
parallels?
In a ‖gm ABCD, if P
and Q are midpoints of AB and CD respectively and ar(‖gm ABCD) = 16
cm^{2}, then ar(‖gm APQD) = ?
- 8
cm^{2} - 12
cm^{2} - 6
cm^{2} - 9
cm^{2}
The figure formed by
joining the midpoints of the adjacent sides of a rectangle of sides 8 cm and
6 cm is a
- rectangle
of area 24 cm^{2} - square
of area 24 cm^{2} - trapezium
of area 24 cm^{2} - rhombus
of area 24 cm^{2}
In ΔABC, if D is
the midpoint of BC and E is the midpoint of AD, then ar(ΔBED) = ?
The vertex A of ΔABC is joined
to a point D on BC. If E is the midpoint of AD, then ar(ΔBEC) = ?
In ΔABC, it is given that D is the midpoint of BC; E is the midpoint of BD and O is the midpoint of AE. Then, ar(ΔBOE) = ?
If a triangle and a
parallelogram are on the same base and between the same parallels, then the
ratio of the area of the triangle to the area of the parallelogram is
- 1
: 2 - 1
: 3 - 1
: 4 - 3
: 4
In the given figure ABCD
is a trapezium in which AB ‖ DC such that AB = a cm and DC = b
cm, If E and F are the midpoints of AD and BC respectively. Then, ar(ABFE) :
ar(EFCD) = ?
- a
: b - (a
+ 3b) : (3a + b) - (3a
+ b) : (a + 3b) - (2a
+ b) : (3a + b)
ABCD is a quadrilateral
whose diagonal AC divides it into two parts, equal in area, then ABCD is
- a
rectangle - a
‖gm - a
rhombus - all
of these
In the given figure, a ‖gm ABCD and a
rectangle ABEF are of equal area, Then,
- perimeter
of ABCD = perimeter of ABEF - perimeter
of ABCD < perimeter of ABEF - perimeter
of ABCD > perimeter of ABEF
In the given figure,
ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If
AD = cm, then area of the rectangle is
- 32
cm^{2} - 40
cm^{2} - 44
cm^{2} - 48
cm^{2}
The median of a triangle
divides it into two
- triangles
of equal area - congruent
triangles - isosceles
triangle - right
triangles
Which of the following
is a false statement?
- A
median of a triangle divides it into two triangles of equal areas. - The
diagonals of a ∥gm divide
it into four triangles of equal areas. - In
a ΔABC, if E
is the midpoint of median AD, then ar(ΔBED) =
- In
a trap. ABCD, it is given that AB ‖ DC and
the diagonals AC and BD intersect at O. Then, ar(ΔAOB) = ar(ΔCOD).
Which of the following
is a false statement?
- If
the diagonals of a rhombus are 18 cm and 14 cm, then its area is 126 cm^{2}. - A
parallelogram and a rectangle on the same base and between the same
parallels are equal in area. - If
the area of a ‖gm with
one side 24 cm and corresponding height h cm is 192 cm^{2}, then
h = 8 cm.
Look at the statements given below:
- A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
- In a ‖gm ABCD, it is given that AB = 10 cm. The altitudes DE on AB and BF on AD being 6 cm and 8 cm respectively, then AD = 7.5 cm.
Which is true?
- I only
- II only
- I and II
- II and III
Each question consists
of two statements, namely, Assertion (A) and Reason (R). For selecting the
correct answer, use the following code:
- Both
Assertion (A) and Reason (R) are true and Reason (R) is a correct
explanation of Assertion (A). - Both
Assertion (A) and Reason (R) are true but Reason (R) is not a correct
explanation of Assertion (A). - Assertion
(A) is true and Reason (R) is false. - Assertion
(A) is false and Reason (R) is true.
Assertion (A) | Reason (R) |
In Then, | Triangles on the same base and |
The correct answer is:
(a) / (b) / (c) / (d).
Each question consists
of two statements, namely, Assertion (A) and Reason (R). For selecting the
correct answer, use the following code:
- Both
Assertion (A) and Reason (R) are true and Reason (R) is a correct
explanation of Assertion (A). - Both
Assertion (A) and Reason (R) are true but Reason (R) is not a correct
explanation of Assertion (A). - Assertion
(A) is true and Reason (R) is false. - Assertion
(A) is false and Reason (R) is true.
Assertion (A) | Reason (R) |
If | Median of a triangle divides it |
The correct answer is:
(a) / (b) / (c) / (d).
Each question consists
of two statements, namely, Assertion (A) and Reason (R). For selecting the
correct answer, use the following code:
- Both
Assertion (A) and Reason (R) are true and Reason (R) is a correct
explanation of Assertion (A). - Both
Assertion (A) and Reason (R) are true but Reason (R) is not a correct
explanation of Assertion (A). - Assertion
(A) is true and Reason (R) is false. - Assertion
(A) is false and Reason (R) is true.
Assertion (A) | Reason (R) |
The | A diagonal of a ‖gm divides it |
The correct answer is:
(a) / (b) / (c) / (d).
Each question consists
of two statements, namely, Assertion (A) and Reason (R). For selecting the
correct answer, use the following code:
- Both
Assertion (A) and Reason (R) are true and Reason (R) is a correct
explanation of Assertion (A). - Both
Assertion (A) and Reason (R) are true but Reason (R) is not a correct
explanation of Assertion (A). - Assertion
(A) is true and Reason (R) is false. - Assertion
(A) is false and Reason (R) is true.
Assertion (A) | Reason (R) |
The |
The correct answer is:
(a) / (b) / (c) / (d).
Each question consists
of two statements, namely, Assertion (A) and Reason (R). For selecting the
correct answer, use the following code:
- Both
Assertion (A) and Reason (R) are true and Reason (R) is a correct
explanation of Assertion (A). - Both
Assertion (A) and Reason (R) are true but Reason (R) is not a correct
explanation of Assertion (A). - Assertion
(A) is true and Reason (R) is false. - Assertion
(A) is false and Reason (R) is true.
Assertion (A) | Reason (R) |
In | Area of a ‖gm = base × height. |
The correct answer is:
(a) / (b) / (c) / (d).
The area of
quadrilateral ABCD in the given figure is
- 57
cm^{2} - 108
cm^{2} - 114
cm^{2} - 195
cm^{2}
The area of trapezium
ABCD in the given figure is
- 62
cm^{2} - 93
cm^{2} - 124
cm^{2} - 155
cm^{2}
In the given figure,
ABCD is a ∥gm in which AB = CD = 5 cm and BD ⊥ DC such that
BD = 6.8 cm, Then the area of ‖gm ABCD = ?
- 17
cm^{2} - 25
cm^{2} - 34
cm^{2} - 68
cm^{2}
In the given figure,
ABCD is a ∥gm in which diagonals Ac and BD
intersect at O. If ar(‖gm ABCD) is 52 cm^{2},
then the ar(ΔOAB)=?
- 26
cm^{2} - 18.5
cm^{2} - 39
cm^{2} - 13
cm^{2}
In the given figure,
ABCD is a ∥gm in which DL ⊥ AB, If AB = 10
cm and DL = 4 cm, then the ar(‖gm ABCD) = ?
- 40
cm^{2} - 80
cm^{2} - 20
cm^{2} - 196
cm^{2}
The area of ∥gm ABCD is
(a) AB
×
BM
(b) BC
×
BN
(c) DC
×
DL
(d) AD
×
DL
Correct
option: (c)
Area of parallelogram ABCD = Base × Height = DC × DL