Class 9 Maths Chapter 9 (Areas of parallelogram and triangles) MCQs are available here online. Students can prepare for the final exams by solving these objective questions to score good marks. The questions here are provided with answers and detailed explanations, as per the CBSE syllabus and NCERT curriculum. Students can solve the chapter-wise MCQs at BYJUâ€™S and also check Important Questions for Class 9 Maths.

## MCQs on Class 9 Areas of Parallelogram and Triangles

Solve the multiple-choice questions here and choose the correct answer from the given four options.

**1) If ABCD is a parallelogram, AE âŠ¥ DC and CF âŠ¥ 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

Answer:** c**

Explanation: Given,

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

CF = 5 cm and AE = 6 cm

Now,

Area of parallelogram = Base Ã— Altitude

CD Ã— AE = AD Ã— CF

10 Ã— 6 = AD Ã— 5

AD = 60/5

AD = 12 cm

**2) If E, F, G and H are the mid-points of the sides of a parallelogram ABCD, respectively, then ar (EFGH) is equal to:**

a. 1/2 ar(ABCD)

b. Â¼ ar(ABCD)

c. 2 ar(ABCD)

d. ar(ABCD)

Answer:** a**

Explanation: Join H and F as shown in the below figure:

AD || BC and AD = BC

Â½ AD = Â½ BC

AH || BF and and DH || CF

AH = BF and DH = CF (H and F are midpoints)

âˆ´, ABFH and HFCD are parallelograms.

Î”EFH and llgmABFH, both lie on a common base, FH.

âˆ´ area of EFH = Â½ area of ABFH â€” 1

area of GHF = 1/2area of HFCD â€” 2

Adding eq. 1 and 2 we get;

area of Î”EFH + area of Î”GHF = Â½ (area of ABFH + area of HFCD)

ar (EFGH) = Â½ ar(ABCD)

**3) 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

Answer:** c**

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

ar(Î”APB) = Â½ ar(parallelogram ABCD) — 1

ar(Î”BQC) = Â½ ar(parallelogram ABCD) — 2

From eq. 1 and 2:

ar(Î”APB) = ar(Î”BQC)

**4) 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

Answer:** c**

**5) A median of a triangle divides it into two**

a. Congruent triangles

b. Isosceles triangles

c. Right triangles

d. Equal area triangles

Answer:** d**

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

AD is the median of Î”ABC.

âˆ´ It will divide Î”ABC into two triangles of equal area.

âˆ´ ar(ABD) = ar(ACD) â€” (i)

also,

ED is the median of Î”ABC.

âˆ´ ar(EBD) = ar(ECD) â€” (ii)

Subtracting (ii) from (i),

ar(ABD) â€“ ar(EBD) = ar(ACD) â€“ ar(ECD)

â‡’ ar(ABE) = ar(ACE)

**6) In a triangle ABC, E is the mid-point of median AD. Then:**

a. ar(BED) = 1/4 ar(ABC)

b. ar(BED) = ar(ABC)

c. ar(BED) = 1/2 ar(ABC)

d. ar(BED) = 2 ar(ABC)

Answer:** a**

Explanation: See the figure below:

ar(BED) = Â½ BD.DE

AE = DE (E is the midpoint)

BD = DC (AD is the median on side BC)

DE = Â½ AD —- 1

BD = Â½ BC —- 2

From eq. 1 and 2, we get;

ar(BED) = (Â½ ) x (Â½) BC x (Â½) AD

ar(BED) = (Â½) x (Â½) ar(ABC)

ar(BED) = Â¼ ar (ABC)

**7) If D and E are points on sides AB and AC respectively of Î”ABC 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

Answer:** b**

Explanation: Î”DBC and Î”EBC are on the same base BC and also having equal areas.

âˆ´ They will lie between the same parallel lines.

âˆ´ DE || 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

Answer:** a**

Explanation: â–³DAC and â–³DBC lie on the same base DC and between the same parallels AB and CD.

ar(â–³DAC) = ar(â–³DBC)

â‡’ ar(â–³DAC) âˆ’ ar(â–³DOC) = ar(â–³DBC) âˆ’ ar(â–³DOC)

â‡’ ar(â–³AOD) = ar(â–³BOC)

**9) If Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(â–³AOD) = ar(â–³BOC). Then ABCD is a:**

a. Parallelogram

b. Rectangle

c. Square

d. Trapezium

Answer:** d**

Explanation: ar(â–³AOD) = ar(â–³BOC)

ar(â–³AOD) = ar(â–³BOC)

â‡’ ar(â–³AOD) + ar(â–³AOB) = ar(â–³BOC) + ar(â–³AOB)

â‡’ ar(â–³ADB) = ar(â–³ACB)

Areas of â–³ADB and â–³ACB are equal.

Therefore, they must lie between the same parallel lines.

Therefore, AB âˆ¥ CD

Hence, ABCD is a trapezium.

**10) 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

Answer:** a**