MCQs for Areas Of Parallelogram Triangles: Chapter 9

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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.

Read Also:

Perimeter of Parallelogram	Laws of Vector Addition	Vector Formula
Remainder Theorem	Areas of Parallelograms and Triangles	Congruence of Triangles
Types of Quadrilaterals	Quadrilaterals	Types of Triangles
Difference Between Variance and Standard Deviation	Logarithm Formula	Geometry Formula

MCQs for Areas Of Parallelogram Triangles: Chapter 9

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MCQs for Areas Of Parallelogram Triangles: Chapter 9

CBSE X Related Questions

2.A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is $10\sqrt{3}$ m away from the base of the tree, then angle of depression of the snake from the eye of the peacock is

3.Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p + 1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.

4.In \(\triangle ABC, \angle B = 90^\circ\). If \(\frac{AB}{AC} = \frac{1}{2}\), then \(\cos C\) is equal to

5.If \(\alpha, \beta\) are zeroes of the polynomial \(8x^2 - 5x - 1\), then form a quadratic polynomial in x whose zeroes are \(\frac{2}{\alpha}\) and \(\frac{2}{\beta}\).

6.Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)

Similar Mathematics Concepts

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SUBSCRIBE TO OUR NEWS LETTER

2.
A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is $10\sqrt{3}$ m away from the base of the tree, then angle of depression of the snake from the eye of the peacock is

3.
Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p + 1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.

4.
In \(\triangle ABC, \angle B = 90^\circ\). If \(\frac{AB}{AC} = \frac{1}{2}\), then \(\cos C\) is equal to

5.
If \(\alpha, \beta\) are zeroes of the polynomial \(8x^2 - 5x - 1\), then form a quadratic polynomial in x whose zeroes are \(\frac{2}{\alpha}\) and \(\frac{2}{\beta}\).

6.
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