Question

Class 9 Maths MCQ Questions of Areas of Parallelograms and Triangles with Answers?

Answer 1

We have given Class 9 Maths MCQ Questions of Areas of Parallelograms and Triangles with Answers to help students with understanding the concept well.  MCQ Questions for Class 9 Maths with Answers were arranged dependent on the most recent pattern of the exams.

Students can solve the chapter-wise Multiple Choice Questions at Sarthaks. Check the underneath Class 9 Maths MCQ Questions of Areas of Parallelograms and Triangles with Answers available here and pick the right answer from the given four choices.

Practice MCQ Questions for Class 9 Maths

1. What is the area of a parallelogram?

(a) 1/2 × Base × Altitude
(b) Base × Altitude
(c) 1/4 × Base × Median
(d) Base × Base

2. Two parallelograms are on the same base and between the same parallels. What is the ratio of their areas?

(a) 2:1                             
(b) 1:2                 
(c) 1 :1                
(d) 3:1                 

3. If ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 10 cm, AE = 6 cm and CF = 5 cm, then AD is equal to:

(a) 10cm
(b) 6cm
(c) 12cm
(d) 15cm

4. If P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD, then:

(a) ar(APB) > ar(BQC)
(b) ar(APB) < ar(BQC)
(c) ar(APB) = ar(BQC)
(d) None of the above

5. If ABCD and EFGH are two parallelograms between same parallel lines and on the same base, then:

(a) ar (ABCD) > ar (EFGH)
(b) ar (ABCD) < ar (EFGH)
(c) ar (ABCD) = ar (EFGH)
(d) None of the above

6.  A median of a triangle divides it into two

(a) Congruent triangles
(b) Isosceles triangles
(c) Right triangles
(d) Equal area triangles

7. If D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC). Then:

(a) DE is equal to BC
(b) DE is parallel to BC
(c) DE is not equal to BC
(d) DE is perpendicular to BC

8.  If Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Then,

(a) ar (AOD) = ar (BOC)
(b) ar (AOD) > ar (BOC)
(c) ar (AOD) < ar (BOC)
(d) None of the above

9. If Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(△AOD) = ar(△BOC). Then ABCD is a:

(a) Parallelogram
(b) Rectangle
(c) Square
(d) Trapezium

10. For two figures to be on the same base and between the same parallels, one of the lines must be.

(a) Making an acute angle to the common base
(b) The line containing the common base
(c) Perpendicular to the common base
(d) Making an obtuse angle to the common base

11. A triangle and a rhombus are on the same base and between the same parallels. Then the ratio of area of triangle to that rhombus is:

(a) 1 : 3
(b) 1 : 2
(c) 1 : 1
(d) 1 : 4

12. The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to

(a) ar (ABC)
(b) (1/2) ar(ABC)
(c) (1/3) ar(ABC)
(d) (1/4) ar(ABC)

13. The area of parallelogram ABCD for the following figure is

(a) BC × BN 
(b) AB × BM
(c) DC × DL
(d) AD × DL

14. The median of a triangle divides it into two

(a) congruent triangles.
(b) isosceles triangles.
(c) right angles.
(d) triangles of equal areas

15. The area of a right triangle is 30 sq cm. If the base is 5 cm, then the hypotenuse must be

(a) 12 cm
(b) 18 cm
(c) 13 cm
(d) 20 cm

16. If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of the parallelogram will be:

(a) 1:2
(b) 3:2
(c) 1:4
(d) 1:3

17. Parallelogram is a 

(a) Pentagon
(b) Quadrilateral
(c) Heptagon
(d) Octagon

18. Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is

(a) 1: 1
(b) 1: 2
(c) 2: 1
(d) 3: 1

19. The area of a parallelogram with base “b” and height “h” is

(a) b×h square units
(b) b² square units
(c) h² square units
(d) b+h square units

20. Triangles on the same base and between the same parallels are 

(a) Greater in area
(b) Equal in area 
(c) smaller in area
(d) None of the above

Answer 2

1. Answer: (b) Base × Altitude

2.  Answer: (c) 1 :1 

3. Answer: (c) 12cm

Explanation: Given,

AB = CD = 10 cm (Opposite sides of a parallelogram)

CF = 5 cm and AE = 6 cm

Now,

Area of parallelogram = Base × Altitude

CD × AE = AD × CF

10 × 6 = AD × 5

AD = 60/5

AD = 12 cm

4. Answer: (c) ar(APB) = ar(BQC)

Explanation: ΔAPB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.

ar(ΔAPB) = 1/2 ar(parallelogram ABCD) — 1

ar(ΔBQC) = 1/2 ar(parallelogram ABCD) — 2

From eq. 1 and 2:

ar(ΔAPB) = ar(ΔBQC)

5. Answer: (c) ar (ABCD) = ar (EFGH)

6. Answer: (d) Equal area triangles

Explanation: Suppose, ABC is a triangle and AD is the median.

AD is the median of ΔABC.

∴ It will divide ΔABC into two triangles of equal area.

∴ ar(ABD) = ar(ACD) — (i)

also,

ED is the median of ΔABC.

∴ ar(EBD) = ar(ECD) — (ii)

Subtracting (ii) from (i),

ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)

⇒ ar(ABE) = ar(ACE)

7. Answer: (b) DE is parallel to BC

Explanation: ΔDBC and ΔEBC are on the same base BC and also having equal areas.

∴ They will lie between the same parallel lines.

∴ DE || BC.

8. Answer: (a) ar (AOD) = ar (BOC)

Explanation: △DAC and △DBC lie on the same base DC and between the same parallels AB and CD.

ar(△DAC) = ar(△DBC)

⇒ ar(△DAC) − ar(△DOC) = ar(△DBC) − ar(△DOC)

⇒ ar(△AOD) = ar(△BOC)

9. Answer: (d) Trapezium

Explanation: ar(△AOD) = ar(△BOC)

ar(△AOD) = ar(△BOC)

⇒ ar(△AOD) + ar(△AOB) = ar(△BOC) + ar(△AOB)

⇒ ar(△ADB) = ar(△ACB)

Areas of △ADB and △ACB are equal.

Therefore, they must lie between the same parallel lines.

Therefore, AB ∥ CD

Hence, ABCD is a trapezium.

10. Answer: (b) The line containing the common base

11. Answer: (b) 1 : 2

Explanation: The ratio of area of a triangle and rhombus is 1:2. Since rhombus is a parallelogram and the triangle and rhombus lie on same base and between same parallel.
∴ar(triangle)= 1/2 ar(rhombus)
∴ The ratio =1:2

12. Answer: (b) (1/2) ar(ABC)

Explanation: Now, consider a triangle ABC as shown in the figure.

Let the area of the triangle be “A”.

By using Heron’s formula

Therefore, the area of triangle AEF = A/4 

Area of triangle CFG = A/4

Therefore, the area of parallelogram BEFG = Area of triangle ABC – Area of AEF – Area of CFG

= A – (A/4) – (A/4)

= (4A – A – A)/4

= 2A/4

=(1/2) A.

13. Answer: (c) DC × DL

Explanation: The area of a parallelogram for the given figure is DC × DL, as the formula for the area of the parallelogram is base×height.

14. Answer: (d) triangles of equal areas

15. Answer: (c) 13 cm

Explanation: We know that, area of triangle = ½ × base (small side) × height

30 = 1/2 × 5 cm × height

Height = (30 × 2)/5

Height = 60/5

Height = 12 cm

By the rule of Pythagoras theorem,

Hypotenuse² = height² + base²

Hypotenuse² = 12² + 5²

Hypotenuse² = 144 + 25

Hypotenuse² = 169

Hypotenuse = √169

Hypotenuse = 13 cm

16. Answer: (a) 1:2

17. Answer: (b) Quadrilateral

Explanation: Parallelogram is a quadrilateral, as it has four sides.

18. Answer: (a) 1: 1

Explanation: We know that the parallelogram on the equal bases and between the same parallels are equal in area. Hence, the ratio of their areas is 1 :1.

19. Answer: (a) b×h square units

Explanation: The area of a parallelogram with base “b” and height “h” is b×h.(i.e) A = base × height square units.

20.  Answer: (b) Equal in area 

Explanation: Triangles on the same base and between the same parallels are equal in area.

Click here to practice Areas of parallelograms MCQ Questions for Class 9 Maths