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Parallelogram and Triangle are a section of the basic geometric figures. Parallelogram is a type of a polygon, which has 4 sides and it is also called a Quadrilateral. In a parallelogram, the opposite sides are exactly parallel to the one another. Triangles on the other side are a geometric shape with the three sides. A triangle has a vertex and an interior angle.
Question 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:
- 10cm
- 6cm
- 12cm
- 15cm
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Ans. (c)
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
Question 2. If parallelogram ABCD and rectangle ABEM are of equal area for the given figure, then
- Perimeter of ABCD = Perimeter of ABEM
- Perimeter of ABCD = (½) Perimeter of ABEM
- Perimeter of ABCD < Perimeter of ABEM
- Perimeter of ABCD > Perimeter of ABEM
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Ans. (d)
Explanation: The Perimeter of parallelogram ABCD is greater than the Perimeter of rectangle ABEM.
Question 3. If E, F, G and H are the mid-points of the sides of a parallelogram ABCD, respectively, then ar (EFGH) is equal to:
- 1/2 ar (ABCD)
- ¼ ar (ABCD)
- 2 ar (ABCD)
- ar (ABCD)
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Ans. (a)
Explanation: Join H and F as shown in the below figure:
AD || BC and AD = BC
½ AD = ½ BC
AH || BF and DH || CF
AH = BF and DH = CF (H and F are midpoints)
∴, ABFH and HFCD are parallelograms.
ΔEFH and parallelogram ABFH, 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)
Question 4. If P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD, then:
- ar(APB) > ar(BQC)
- ar(APB) < ar(BQC)
- ar(APB) = ar(BQC)
- None of the above
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Ans. (c)
Explanation: ΔAPB and parallelogram ABCD lie on the same base AB and in-between the 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)
Question 5. A median of a triangle divides it into two
- Congruent triangles
- Isosceles triangles
- Right triangles
- Equal area triangles
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Ans.(d)
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)
Question 6. In a triangle ABC, E is the midpoint of median AD. Then:
- ar(BED) = 1/4 ar(ABC)
- ar(BED) = ar(ABC)\
- ar(BED) = 1/2 ar(ABC)
- ar(BED) = 2 ar(ABC)
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Ans. (a)
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)
Question 7. If D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC). Then:
- DE is equal to BC
- DE is parallel to BC
- DE is not equal to BC
- DE is perpendicular to BC
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Ans. (b)
Explanation: ΔDBC and ΔEBC are on the same base BC and also have equal areas.
∴ they will lie between the same parallel lines.
∴ DE || BC.
Question 8. ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD
- is always a rhombus
- is a rectangle
- is a parallelogram
- need not be any of (A), (B) or (C)
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Ans. (d)
Explanation: A quadrilateral ABCD need not be any rhombus, parallelogram or rectangle. If ABCD is a square, its diagonal AC divides it into two parts and are equal in area.
Question 9. If Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Then,
- ar (AOD) = ar (BOC)
- ar (AOD) > ar (BOC)
- ar (AOD) < ar (BOC)
- None of the above
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Ans. (a)
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)
Question 10. 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:
- Parallelogram
- Rectangle
- Square
- Trapezium
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Ans. (d)
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.
Question 11. The area of a parallelogram with base “b” and height “h” is
- b×h square units
- b2 square units
- h2 square units
- b+h square units
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Ans. (a)
Explanation: The area of a parallelogram with base “b” and height “h” is b×h.(i.e) A = base × height square units.
Question 12. Parallelogram is a
- Pentagon
- Quadrilateral
- Heptagon
- Octagon
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Ans. (b)
Explanation: Parallelogram is a quadrilateral, as it has four sides.
Question 13. Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
- 1: 1
- 1: 2
- 2: 1
- 3: 1
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Ans. (a)
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.
Question 14. The midpoint of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to
- ar (ABC)
- (1/2) ar(ABC)
- (1/3) ar(ABC)
- (1/4) ar(ABC)
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Ans. (b)
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
=(½) A.
Hence, option (B) is the correct Ans.
Question 15. Triangles on the same base and between the same parallels are
- Greater in area
- Equal in area
- smaller in area
- None of the above
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Ans. (b)
Explanation: Triangles on the same base and between the same parallels are equal in area.
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