RS Agarwal Solution | Class 9th | Chapter-11 |  Areas of Parallelograms and Triangles | Edugrown

Exercise MCQ

Question 1

Out of the following given figures which are on the same base but not between the same parables?

Solution 1

Question 2

In which of the following figures, you find polynomials on the same base and between the same parallels?

Solution 2

Question 3

The median of a triangle divides it into two

1. triangles of equal area
2. congruent triangles
3. isosceles triangle
4. right triangles

Solution 3

Question 4

The area of quadrilateral ABCD in the given figure is

1. 57 cm²
2. 108 cm²
3. 114 cm²
4. 195 cm²

Solution 4

Question 5

The area of trapezium ABCD in the given figure is

1. 62 cm²
2. 93 cm²
3. 124 cm²
4. 155 cm²

Solution 5

Question 6

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 = ?

1. 17 cm²
2. 25 cm²
3. 34 cm²
4. 68 cm²

Solution 6

Question 7

In the given figure, ABCD is a ∥gm in which diagonals Ac and BD intersect at O. If ar(‖gm ABCD) is 52 cm², then the ar(ΔOAB)=?

1. 26 cm²
2. 18.5 cm²
3. 39 cm²
4. 13 cm²

Solution 7

Question 8

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) = ?

1. 40 cm²
2. 80 cm²
3. 20 cm²
4. 196 cm²

Solution 8

Question 9

The area of ∥gm ABCD is

(a) AB × BM

(b) BC × BN

(c) DC × DL

(d) AD × DL

Solution 9

Correct option: (c)

Area of ∥gm ABCD = Base × Height = DC × DL Question 10

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 : 1Solution 10

Correct option: (b)

Parallelograms on equal bases and between the same parallels are equal in area. Question 11

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

Solution 11

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.

Question 12

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)

Solution 12

Correct option: (a)

ΔABC is divided into four triangles of equal area.

A(parallelogram AFDE) = A(ΔAFE) + A(DFE)

Question 13

The lengths of the diagonals of a rhombus are 12 cm and 16 cm. The area of the rhombus is

1. 192 cm²
2. 96 cm²
3. 64 cm²
4. 80 cm²

Solution 13

Question 14

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

1. 74 cm²
2. 32.5 cm²
3. 65 cm²
4. 130 cm²

Solution 14

Question 15

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)= ?

1. 24 cm²
2. 40 cm²
3. 55 cm²
4. 27.5 cm²

Solution 15

Question 16

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) = ?

1. 256 cm²
2. 128 cm²
3. 64 cm²
4. 96 cm²

Solution 16

Question 17

ABCD is a rhombus in which ∠C = 60°.

Then, AC : BD = ?

Solution 17

Question 18

In the given figure ABCD and ABFE are parallelograms such that ar(quad. EABC) = 17cm² and ar(‖gm ABCD) = 25 cm². Then, ar(ΔBCF) = ?

1. 4 cm²
2. 4.8 cm²
3. 6 cm²
4. 8 cm²

Solution 18

Question 19

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 : 4Solution 19

Question 20

In a ‖gm ABCD, if P and Q are midpoints of AB and CD respectively and ar(‖gm ABCD) = 16 cm², then ar(‖gm APQD) = ?

1. 8 cm²
2. 12 cm²
3. 6 cm²
4. 9 cm²

Solution 20

Question 21

The figure formed by joining the midpoints of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a

1. rectangle of area 24 cm²
2. square of area 24 cm²
3. trapezium of area 24 cm²
4. rhombus of area 24 cm²

Solution 21

Question 22

In ΔABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(ΔBED) = ?

Solution 22

Question 23

The vertex A of ΔABC is joined to a point D on BC. If E is the midpoint of AD, then ar(ΔBEC) = ?

Solution 23

Question 24

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) = ?

Solution 24

Question 25

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. 1 : 2
2. 1 : 3
3. 1 : 4
4. 3 : 4

Solution 25

Question 26

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) = ?

1. a : b
2. (a + 3b) : (3a + b)
3. (3a + b) : (a + 3b)
4. (2a + b) : (3a + b)

Solution 26

Question 27

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD is

1. a rectangle
2. a ‖gm
3. a rhombus
4. all of these

Solution 27

Question 28

In the given figure, a ‖gm ABCD and a rectangle ABEF are of equal area, Then,

1. perimeter of ABCD = perimeter of ABEF
2. perimeter of ABCD < perimeter of ABEF
3. perimeter of ABCD > perimeter of ABEF
4. 

Solution 28

Question 29

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

1. 32 cm²
2. 40 cm²
3. 44 cm²
4. 48 cm²

Solution 29

Question 30

Which of the following is a false statement?

1. A median of a triangle divides it into two triangles of equal areas.
2. The diagonals of a ∥gm divide it into four triangles of equal areas.
3. In a ΔABC, if E is the midpoint of median AD, then ar(ΔBED) =

4. In a trap. ABCD, it is given that AB ‖ DC and the diagonals AC and BD intersect at O. Then, ar(ΔAOB) = ar(ΔCOD).

Solution 30

Question 31

Which of the following is a false statement?

1. If the diagonals of a rhombus are 18 cm and 14 cm, then its area is 126 cm².
2. 
3. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
4. If the area of a ‖gm with one side 24 cm and corresponding height h cm is 192 cm², then h = 8 cm.

Solution 31

Question 32

Look at the statements given below:

1. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
2. 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.

3. 

Which is true?

1. I only
2. II only
3. I and II
4. II and III

Solution 32

Question 33

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. 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).Solution 33

Question 34

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. 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).Solution 34

Question 35

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. 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).Solution 35

Question 36

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. 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 cm2.

The correct answer is: (a) / (b) / (c) / (d).Solution 36

Question 37

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. 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).Solution 37

Exercise Ex. 11A

Question 2

In the adjoining figure, show that ABCD is a parallelogram. Calculate the area of ||gm ABCD.

Solution 2

Question 3

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.

Solution 3

Question 5

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.Solution 5

Question 6(ii)

Calculatethe area of trapezium PQRS , givenin Fig.(ii)

Solution 6(ii)

Question 6(i)

Calculate the area of quadrilateral ABCD givenin Fig (i)

Solution 6(i)

Question 8

BD is one of the diagonals of a quad. ABCD . If , show that 

Solution 8

Question 10

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.

Solution 10

Question 13

In the adjoining figure , ABCD is a trapezium in which AB||DC and its diagonals AC and BD intersects at O. prove that 

Solution 13

Question 14

_{In the adjoining figure , DE||BC. Prove that} 

Solution 14

Question 15

Prove that the median divides a triangle into two triangles of equal area.Solution 15

Question 16

Show that the diagonal divides a parallelogram into two triangles of equal area.Solution 16

Question 20

_{In adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O. If BO=OD, prove that}

Solution 20

Question 21

The vertex A of  _{is joined to a point D on the side BC. The mid-point of AD is E. prove that}

Solution 21

Question 22

_{D is the mid-point of side BC of   and E is the midpoint of BD. If O is midpoint of AE, prove that} 

Solution 22

Question 24

_{In the adjoining figure , ABCD is a quadrilateral. A line through D , parallel to AC , meets BC produced in P. prove that} 

Solution 24

Question 25

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.}

Solution 25

Question 28

_{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)}

Solution 28

Question 30

The base BC of   _{is divided at D such that .}

_{Prove that .}Solution 30

Question 32

_{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 

Solution 32

Question 34

_{In the adjoining figure , the point D divides the side BC of   in the ratio m:n. Prove that} 

Solution 34

Exercise Ex. 11

Question 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.

Fig (i)

Fig (ii)

Fig (iii)

Fig (iv)

Fig (v)

Fig (vi)

Solution 1

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 BPQuestion 4

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.Solution 4

Question 7

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.

Solution 7

Question 9

M is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD) = 24 cm², find ar(Δ ABC).Solution 9

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)

Question 11

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).Solution 11

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) Question 12

In the adjoining figure, MNPQ and ABPQ are parallelograms and T is any point on the side BP. Prove that

Solution 12

(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. 

Question 17

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).

Solution 17

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)Question 18

D and E are points on sides AB and AC respectively of Δ ABC such that ar(ΔBCD) = ar(ΔBCE). Prove that DE ∥ BC.Solution 18

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 Question 19

P is any point on the diagonal AC of parallelogram ABCD. Prove that ar(ΔADP) = ar(ΔABP).Solution 19

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)Question 23

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).

Solution 23

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)Question 26

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 cm², find the area of parallelogram ABCD.

Solution 26

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 cm²

Now,

A(parallelogram ABCD) = 2 × A(ΔBDC) = 2 × 14 = 28 cm² Question 27

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)Solution 27

Construction: Join AC and BM

Let h be the distance between AB and CD.

Question 29

In a triangle ABC, the medians BE and CF intersect at G. Prove that ar(ΔBCG) = ar(AFGE).Solution 29

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)Question 31

Solution 31

ΔDBC and ΔEBC are on the same base and between the same parallels.

⇒ A(ΔDBC) = A(ΔEBC) ….(i)

BE is the median of ΔABC.

Question 33

In the adjoining figure, CE ∥ AD and CF ∥ BA. Prove that ar(ΔCBG) = ar(ΔAFG).

Solution 33

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)Question 35

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).

Solution 35

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)

Question 36

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 Solution 36

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)

Question 37

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).

Solution 37

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)Question 38

In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that ar(ΔRSC) = ar(ΔPQB).

Solution 38

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)