Question types

Areas of Parallelograms and Triangles question types

28 questions across 3 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.

28
Questions
3
Question groups
5
Question types
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.

Q 1M.C.Q1 Mark
Write the correct answer of the following: The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $8\ cm$ and $6\ cm$, is:
  • A
    A rectangle of area $24 \mathrm{~cm}^2$.
  • B
    A square of area $25 \mathrm{~cm}^2$.
  • C
    A trapezium of area $24 \mathrm{~cm}^2$.
  • A rhombus of area $24 \mathrm{~cm}^2$.

Answer: D.

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Q 2M.C.Q1 Mark
Write the correct answer of the following: In Fig. if parallelogram $ABCD$ and rectangle $ABEF$ are of equal area, then: 
  • A
    Perimeter of $ABCD$ = Perimeter of $ABEM$.
  • B
    Perimeter of $ABCD$ < Perimeter of $ABEM$.
  • Perimeter of $ABCD$ > Perimeter of $ABEM$.
  • D
    Perimeter of $ABCD$ $=\frac{1}{2}$ (perimeter of $ABEM$).

Answer: C.

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Q 3M.C.Q1 Mark
Write the correct answer of the following: Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is:
  • A
    $1 : 2$
  • $1 : 1$
  • C
    $2 : 1$
  • D
    $3 : 1$

Answer: B.

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Q 4M.C.Q1 Mark
Write the correct answer of the following: The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to:
  • $\frac{1}{2}\ \text{ar}\ (\text{ABC})$
  • B
    $\frac{1}{3}\ \text{ar}\ (\text{ABC})$
  • C
    $\frac{1}{4}\ \text{ar}$
  • D
    $\text{ar}\ (\text{ABC})$

Answer: A.

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Q 5M.C.Q1 Mark
Write the correct answer of the following: The median of a triangle divides it into two:
  • Triangles of equal area.
  • B
    Congruent triangles.
  • C
    Right triangles.
  • D
    Isosceles triangles.

Answer: A.

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Write True or False and justify your answer: In the figure, $ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then, $\text{ar}(\triangle\text{DPC})=\frac{1}{2}\text{ar}(\text{EFGD}).$ 
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Write True or False and justify your answer: $PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13\ cm$ and $A$ is any point on $PQ.$ If $PS = 5\ cm,$ then ar $(\triangle\text{PAS})= 30\ cm^2$.
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Write True or False and justify your answer: $ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC.$ Then $\text{ar}(\triangle\text{BDE})=\frac{1}{4}\text{ar}(\triangle\text{ABC}).$
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Write True or False and justify your answer: $ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If ar $(AXCD) = 24\ cm^2,$ then ar $(ABC) = 24\ cm^2.$
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In Fig. $X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively, $QP || BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $ar\ (ABP) = ar\ (ACQ).$
 
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In $\triangle\text{ABC}, D$ is the mid-point of $AB$ and $P$ is any point on $BC.$ If $CQ || PD$ meets $AB$ in $Q ($shown in figure$),$ then prove that $\text{ar}(\triangle\text{BPQ})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$ 
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In Fig.$9\ PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA || QB || RC.$ Prove that $ar\ (PQE) = ar\ (CFD). $
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