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 BYJU’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 ⊥ DC and CF ⊥ AD. If AB = 10 cm, AE = 6 cm and CF = 5 cm, then AD is equal to:
AB = CD = 10 cm (Opposite sides of a parallelogram)
CF = 5 cm and AE = 6 cm
Area of parallelogram = Base × Altitude
CD × AE = AD × CF
10 × 6 = AD × 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. ¼ ar(ABCD)
c. 2 ar(ABCD)
Explanation: Join H and F as shown in the below figure:
AD || BC and AD = BC
½ AD = ½ BC
AH || BF and and DH || CF
AH = BF and DH = CF (H and F are midpoints)
∴, ABFH and HFCD are parallelograms.
ΔEFH and llgmABFH, both lie on a common base, FH.
∴ area of EFH = ½ area of ABFH — 1
area of GHF = 1/2area of HFCD — 2
Adding eq. 1 and 2 we get;
area of ΔEFH + area of ΔGHF = ½ (area of ABFH + area of HFCD)
ar (EFGH) = ½ 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
Explanation: ΔAPB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.
ar(ΔAPB) = ½ ar(parallelogram ABCD) — 1
ar(ΔBQC) = ½ ar(parallelogram ABCD) — 2
From eq. 1 and 2:
ar(ΔAPB) = ar(ΔBQC)
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
5) A median of a triangle divides it into two
a. Congruent triangles
b. Isosceles triangles
c. Right triangles
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)
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)
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)
Explanation: See the figure below:
ar(BED) = ½ BD.DE
AE = DE (E is the midpoint)
BD = DC (AD is the median on side BC)
DE = ½ AD —- 1
BD = ½ BC —- 2
From eq. 1 and 2, we get;
ar(BED) = (½ ) x (½) BC x (½) AD
ar(BED) = (½) x (½) ar(ABC)
ar(BED) = ¼ ar (ABC)
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
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) 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
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) 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:
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) 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: