Chapter 11 : Areas of Parallelograms and Triangles  R S Aggarwal And V Aggarwal Solutions for Class 9 Maths CBSE
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Chapter 11  Areas of Parallelograms and Triangles Excercise MCQ
Out of the following given figures which are on the same base but not between the same parables?
In which of the following figures, you find polynomials on the same base and between the same parallels?
The median of a triangle divides it into two
 triangles of equal area
 congruent triangles
 isosceles triangle
 right triangles
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 ∥gm ABCD = Base × Height = DC × DL
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 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}
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 a trapezium ABCD we have AB ‖ DC and the diagonals AC and BD intersect at O. Then, ar(ΔAOD) = ar(ΔBOC)

Triangles on the same base and between the same parallels are equal in areas. 
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 ABCD is a rhombus whose one angle is 60°, then the ratio of the lengths of its diagonals is . 
Median of a triangle divides it into two triangles of equal area. 
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 diagonals of a ‖gm divide it into four triangles of equal area. 
A diagonal of a ‖gm divides it into two triangles of equal area. 
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 area of a trapezium whose parallel sides measure 25 cm and 15 cm respectively and the distance between them is 6 cm, is 120 cm^{2}. 

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 the given figure, ABCD is a ‖gm in which DE ⊥ AB and BE ⊥ AD. If AB = 16 cm, DE = 8cm and BF = 10cm, then AD is 12 cm.

Area of a ‖gm = base × height. 
The correct answer is: (a) / (b) / (c) / (d).
Chapter 11  Areas of Parallelograms and Triangles Excercise Ex. 11A
In the adjoining figure, show that ABCD is a parallelogram. Calculate the area of gm ABCD.
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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.
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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.
_{}
_{}
Calculatethe area of trapezium PQRS , givenin Fig.(ii)
_{}
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Calculate the area of quadrilateral ABCD givenin Fig (i)
_{}
_{}
BD is one of the diagonals of a quad. ABCD . If , show that
_{}
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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 ABDC and its diagonals AC and BD intersects at O. prove that _{}
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_{In the adjoining figure , DEBC. Prove that }
_{ }
_{ }
Prove that the median divides a triangle into two triangles of equal area.
_{}
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 midpoint of AD is E. prove that}
_{ }
_{}
_{}
_{D is the midpoint 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 }
_{}
_{}
_{}
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)}
_{}
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 }
_{ }
_{}
_{}
_{}
Chapter 11  Areas of Parallelograms and Triangles Excercise 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
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 midpoint of AB,
A(AMCD) = A(ΔADC) + A(ΔAMC)
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 midpoint 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 midpoint 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 midpoints 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 midpoints of AD and BC respectively.
⇒ MN ∥ AB ∥ CD
In ΔADB, M is the midpoint of AD and MY ∥ AB.
∴ Y is the midpoint of DB.
Similarly, in ΔBDC,
Now, MN = MY + YN
Construction: Draw DQ ⊥ AB. Let DQ cut MN at point P.
Then, P is the midpoint 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 midpoints of AD and BC respectively.
⇒ EF ∥ AB ∥ CD
In ΔADC, E is the midpoint of AD and EY ∥ CD.
∴ Y is the midpoint of AC.
Similarly, in ΔABC,
Now, EF = EY + YF
Construction: Draw AQ ⊥ DC. Let AQ cut EF at point P.
Then, P is the midpoint 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 midpoints of AB and AC respectively,
DE ∥ BC ∥ PQ
In ΔACP, AP ∥ DE and E is the midpoint of AC.
⇒ D is the midpoint of PC (converse of midpoint theorem)
In ΔABQ, AQ ∥ DE and D is the midpoint of AB.
⇒ E is the midpoint of BQ (converse of midpoint 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)
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