Question types

Areas Of Parallelograms And Triangles question types

47 questions across 5 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.

47
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5
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5
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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
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $8\ cm$ and $6\ cm$ is:
  • A rhombus of area $24 \mathrm{~cm}^2$
  • B
    A rectangle of area $24 \mathrm{~cm}^2$
  • C
    A square of area $26 \mathrm{~cm}^2$
  • D
    A trapezium of area $14 \mathrm{~cm}^2$

Answer: A.

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Q 2M.C.Q1 Mark
$ABCD$ is a trapezium with parallel sides $AB = a$ and $DC = b$. If $E$ and $F$ are mid-points of non-parallel sides $AD$ and $BC$ respectively, then the ratio of areas of quadrilaterals $ABFE$ and $EFCD$ is:
  • A
    $a : b$
  • B
    $(a + 3b) : (3a + b)$
  • $(3a + b) : (a + 3b)$
  • D
    $(2a + b) : (3a + b)$

Answer: C.

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Q 3M.C.Q1 Mark
If $AD$ is median of $\triangle\text{ABC}$ and $P$ is a point on $AC$ such that $\text{ar}(\triangle\text{ADP}):\text{ar}(\triangle\text{ABD})=2:3,$ then $\text{ar}(\triangle\text{PDC}):\text{ar}(\triangle\text{ABC})$ is:
  • A
    $1 : 5$
  • B
    $1 : 5$
  • $1 : 6$
  • D
    $3 : 5$

Answer: C.

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Q 4M.C.Q1 Mark
The median of a triangle divides it into two:
  • A
    Congruent triangle.
  • B
    Isosceles triangles.
  • C
    Right triangles.
  • Triangles of equal areas.

Answer: D.

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Q 5M.C.Q1 Mark
$A, B, C, D$ are mid-points of sides of parallelogram $PQRS$. If ar(PQRS) $=36 \mathrm{~cm}^2$, then $\operatorname{ar}(A B C D)=$
  • A
    $24 \mathrm{~cm}^2$
  • $18 \mathrm{~cm}^2$
  • C
    $30 \mathrm{~cm}^2$
  • D
    $36 \mathrm{~cm}^2$

Answer: B.

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Q 103 Marks Question3 Marks
If $P$ is any point in the interior of a parallelogram $ABCD$, then prove that area of the triangle $APB$ is less than half the area of parallelogram.
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Q 113 Marks Question3 Marks
$ABCD$ is a parallelogram whose diagonals $AC$ and $BD$ intersect at $O$. A line through $O$ intersects $AB$ at $P$ and $DC$ at $Q$. Prove that $\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$
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Q 123 Marks Question3 Marks
Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $P$. Show that: $\text{ar}(\triangle\text{APB})\times\text{ar}(\triangle\text{CPD})\\=\text{ar}(\triangle\text{APD})\times\text{ar}(\triangle\text{BPC})$
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Q 133 Marks Question3 Marks
In figure, $OCDE$ is a rectangle inscribed in a quadrant of a circle of radius $10\ cm$. If $\text{OE}=2\sqrt5\text{cm},$ find the area of the rectangle. 
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If $ABCD$ is a parallelogram, then prove that $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})\\ \ =\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}$ $(||^{gm} ABCD)$
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$\text{ABCD}$ is a parallelogram whose diagonals intersect at $O$ .If $P$ is any point on $BO,$ prove that:
$i. \text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO})$
$ii. \text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP})$
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In figure, $\text{ABCD}$ is a trapezium in which $AB \| DC$ and $DC = 40\ cm$ and $AB = 60\ cm$. If $X$ and $Y$ are, respectively, the mid$-$points of $AD$ and $BC,$ prove that:
$i. XY = 50\ cm$
$ii. \text{DCYX}$ is a trapezium
$iii. \text{ar}($trap.$\ \text{DCYX})=\Big(\frac{9}{11}\Big)\text{ar}(\text{XYBA}).$
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$\text{PQRS}$ is a rectangle inscribed in a quadrant of a circle of radius $13\ cm. A$ is any point on $PQ$. If $PS = 5\ cm$, then find $\text{ar}(\triangle\text{RAS}).$
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In figure, $ABC$ and $ABD$ are two triangles on the base $AB$. If line segment $CD$ is bisected by $AB$ at $O$, show that $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$ 
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In a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$. Prove that:
$i. \text{ar}(\triangle\text{LCM})=\text{ar}(\triangle\text{LBM})$
$ii. \text{ar}(\triangle\text{LBC})=\text{ar}(\triangle\text{MBC})$
$iii. \text{ar}(\triangle\text{ABM})=\text{ar}(\triangle\text{ACL})$
$iv. \text{ar}(\triangle\text{LOB})=\text{ar}(\triangle\text{MOC})$

 

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$\text{ABCD}$ is a parallelogram. $G$ is a point on $AB$ such that $AG = 2GB$ and $E$ is point on $DC$ such that $CE = 2DE$ and $F$ is the point of $BC$ such that $BF = 2FC$. Prove that:
$i. \text{ar}(\text{ADEG})=\text{ar}(\text{GBCE})$
$ii. \text{ar}{(\triangle\text{EGB}})=\frac{1}{6}\text{ar}(\text{ABCD})$
$iii. \text{ar}(\triangle\text{EFC})=\frac{1}{2}\text{ar}(\triangle\text{EBF})$
$iv. \text{ar}(\triangle\text{EBG})=\text{ar}(\triangle\text{EFC})$
$v.$ Find what portion of the parallelogram is the area of $\triangle\text{EFG}.$
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If $ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$, then find $\text{ar}(\triangle\text{ABC}) : \text{ar}(\triangle\text{BDE}).$
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