Two parallelograms are on the same base and between the same parallels. The ratio of their areas is:
- 1 : 2
- 2 : 1
- 1 : 1
- 3 : 1
Question types
Areas Of Parallelograms And Triangles question types
81 questions across 5 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
81
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5
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5
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01
M.C.Q
20 Q→022 Marks Questions
6 Q→033 Marks Question
23 Q→044 Marks Questions
29 Q→055 Marks Questions
3 Q→Sample Questions
Areas Of Parallelograms And Triangles questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
View full solution →
The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to:
View full solution →- $\text{ar}(\triangle\text{ABC})$
- $\frac{1}{2}\text{ar}(\triangle\text{ABC})$
- $\frac{1}{3}\text{ar}(\triangle\text{ABC})$
- $\frac{1}{4}\text{ar}(\triangle\text{ABC})$
The median of a triangle divides it into two:
- Congruent triangle.
- Isosceles triangles.
- Right triangles.
- Triangles of equal areas.
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8cm and 6cm is:
- A rhombus of area 24cm2
- A rectangle of area 24cm2
- A square of area 26cm2
- A trapezium of area 14cm2
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16cm and 12cm is:
- 28cm2
- 48cm2
- 96cm2
- 24cm2
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
Let ABCD be a parallelogram of area 124cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.
In figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of $\triangle\text{OTS}$ if PQ = 8cm.
View full solution →In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10cm. If $\text{OE}=2\sqrt5\text{cm},$ find the area of the rectangle.
View full solution →In figure, D and E are two points on BC such that BD = DE = EC. Show that $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{AEC}).$
View full solution →In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $\text{AX}\perp\text{DE}$ meets BC at Y. Show that:
$\triangle\text{FCB}\cong\triangle\text{ACE}$
View full solution →$\triangle\text{FCB}\cong\triangle\text{ACE}$
PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of $\triangle\text{PBQ}$ is 17cm2, find the area of $\triangle\text{ASR}.$
View full solution →PQRS is a rectangle inscribed in a quadrant of a circle of radius 13cm. A is any point on PQ. If PS = 5cm, then find $\text{ar}(\triangle\text{RAS}).$
View full solution →P is any point on base BC of $\triangle\text{ABC}$ and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If $\text{ar}(\triangle\text{ABC})=12\text{cm}^2,$ then find area of $\triangle\text{EPC}.$
View full solution →In the given figure, ABCD is a rectangle with sides AB = 10cm and AD = 5cm. Find the area of $\triangle\text{EFG}.$
View full solution →In the given figure, CD || AE and CY || BA.
View full solution →- Name a triangle equal in area of $\triangle\text{CBX}$
- Prove that $\text{ar}(\triangle\text{ZDE})=\text{ar}(\triangle\text{CZA})$
- Prove that $\text{ar}(\text{BCZY})=\text{ar}(\triangle\text{EDZ}).$
In figure, ABCD is a trapezium in which AB || DC and DC = 40cm and AB = 60cm. If X and Y are, respectively, the mid-points of AD and BC, prove that:
View full solution →- XY = 50cm
- DCYX is a trapezium
- $\text{ar}(\text{trap}.\ \text{DCYX})=\Big(\frac{9}{11}\Big)\text{ar}(\text{XYBA}).$
If ABCD is a parallelogram, then prove that
$\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})\\ \ =\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}$ (||gm ABCD)
View full solution →$\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})\\ \ =\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}$ (||gm ABCD)
ABCD is a parallelogram whose diagonals intersect at O .If P is any point on BO, prove that:
View full solution →- $\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO})$
- $\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP})$
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