Mathematics

Parallelogram Properties and Area

93 Questions

Parallelogram properties and area questions test the ability to calculate dimensions using base and height ratios. Problems also cover diagonal properties and internal angles. These geometry concepts regularly appear in quantitative aptitude sections of various exams.

Area calculationBase and height ratiosDiagonal propertiesInterior anglesGeometric theorems

Parallelogram Properties and Area Questions

Multiple choice maths pattern patterns in number operations magic squares magic square

Find the number of parallelogram in the figure given below.

  1. 14

  2. 23

  3. 22

  4. 36
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For parallelograms we can use the formula of rectangles. It is a
$\displaystyle 3 \, \times \, 3 \, gird, \, hence \, \frac{3 \, \times \, 4}{3 \, \times \, 4} \, \times \frac{3 \, \times \, 4}{3 \, \times \, 4} \, = \, 36$ 

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

In triangle $ ABC $; $ D $ and $ E $ are mid-points of the sides $ AB $ and $ AC $ respectively. Through $ E $, a straight line is drawn parallel to $ AB $ to meet $ BC $ at $ F $. Quadrilateral $ BDEF $ is a parallelogram.If $ AB= 16 $ cm, $ AC= 12 $ cm and $ BC= 18 $ cm, find the perimeter of the parallelogram $ BDEF $.

  1. 36 cm

  2. 44 cm

  3. 34 cm

  4. 54 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: D and F are mid points of AB and AC respectively.
Hence, by mid point theorem, $DF \parallel BC$

Also, given $BD \parallel EF$
Since, opposite sides are parallel to each other. Hence, $BDEF$ is a parallelogram

Perimeter of BDEF = $2 (BD + BF)$ (Opposite sides of parallelogram are equal)
Perimeter of BDEF = $AB + BC$ (D and F are mid points of AB and BC respectively)
Perimeter of BDEF = $16 + 18$
Perimeter of BDEF = $34$ cm

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

The straight line joining the mid-points of the opposite sides of a parallelogram divides it into two parallelogram of equal area  

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A line joining the midpoints of opposite sides of a parallelogram is parallel to the other two sides and divides the parallelogram into two smaller parallelograms of equal area.

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

If $( 3,4 )$ and $( 6,5 )$ are the extremities of a diagonal of a parallelogram and $( 2,1 )$ is is third vertex, then its fourth vertex is _______.

  1. $( - 1,0 )$
  2. $( - 1,1 )$
  3. $( 0,-1 )$
  4. $( 7,8 )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a parallelogram, the diagonals bisect each other. The midpoint of the diagonal with endpoints (3,4) and (6,5) is ((3+6)/2, (4+5)/2) = (4.5, 4.5). Let the fourth vertex be (x,y). The midpoint of the other diagonal (2,1) and (x,y) must also be (4.5, 4.5). So (2+x)/2 = 4.5 => x = 7, and (1+y)/2 = 4.5 => y = 8. The fourth vertex is (7,8).

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

In triangle $ABC$, angle $B$ is obtuse. $D$ and $E$ are mid-points of sides $AB$ and $BC$ respectively and $F$ is a point on side $AC$ such that $EF$ is parallel to $AB$. Then, $BEFD$ is a parallelogram. State True or False. 

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $D$ is mid point of $AB$ and $E$ is mid point of $BC$, $F$ is any point on $AC$ and $EF \parallel AB$

Now, in $\triangle ABC$,
E is mid point of BC and $EF \parallel AB$
By Mid point Theorem, $F$ is mid point of $AC$

Also, D is mid point of AB and F is mid point of AC
Hence, by mid point theorem, $DF \parallel BE$
Since, $DF \parallel BE$ and $EF \parallel AB or BD$
Hence, BEFD is parallelogram.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity
State true or false:

In parallelogram $ ABCD $. $ E $ is the mid-point of $ AB $ and $ AP $ is parallel to $ EC $ which meets $ DC $ at point $ O $ and $ BC $ produced at $ P $. Hence 
$ O $ is mid-point of $ AP $.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $\triangle$s, APB and ECB,

$\angle ABP = \angle EBC $ (Common angle)

$\angle PAB = \angle CEB$ (Corresponding angles of parallel lines)

$\angle APB = \angle ECB $ (Third angle of the triangle)

Thus $\triangle APB \sim \triangle ECB$ (AAA rule)

Hence, $\dfrac{AB}{EB} = \dfrac{BP}{BC}$ (Corresponding sides of similar triangles)

$2 = \dfrac{BP}{BC}$

$BP = 2 BC$

Now, in $\triangle$s $OPC$ and $APB,$

$\angle OPC = \angle APB$ (Common angle)

$\angle POC = \angle PAB$ (Corresponding angles of parallel lines)

$\angle PCO = \angle PBA$ (Third angle of a triangle)

$\triangle OPC \sim \triangle APB$ (AAA rule)

hence, $\dfrac{PC}{BP} = \dfrac{OP}{AP}$  (Corresponding sides)

$\dfrac{1}{2} = \dfrac{OP}{AP}$ 

$OP = \dfrac{1}{2} AP$

hence, $O$ is the midpoint of $AP$.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity
State true or false:

In parallelogram $ ABCD $. $ E $ is the mid-point of $ AB $ and $ AP $ is parallel to $ EC $ which meets $ DC $ at point $ O $ and $ BC $ produced at $ P $. Hence
$ BP= 2AD $


  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $\triangle$s, APB and ECB,
$\angle ABP = \angle EBC $ (Common angle)
$\angle PAB = \angle CEB$ (Corresponding angles of parallel lines)
$\angle APB = \angle ECB $ (Third angle of the triangle)
Thus $\triangle APB \sim \triangle ECB$ 
Hence, $\frac{AB}{EB} = \frac{BP}{BC}$ (Corresponding sides of similar triangles)
$2 = \frac{BP}{BC}$
$BP = 2 BC$
$BP = 2 AD$  (BC = AD)

Multiple choice the nth roots of unity complex numbers maths

The 4th roots of unity in the argand plane form a

  1. Square

  2. Rectangle

  3. Parallelogram

  4. Rhombus
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The fourth roots of unity are $1,i,-1, -i $. Hence, they form a square in the Argand plane. 
We can see that by plotting the points on the graph with coordinate system according to that of complex numbers.

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

A point M is taken inside a parallelogram ABCD, then area of $\displaystyle \Delta AMD,$ $\displaystyle \Delta AMB,$ $\displaystyle \Delta AMC$ can take which of of the following values, respectively.

  1. 15, 6, 11

  2. 9, 6, 4

  3. 13, 5, 8

  4. 25, 7, 24
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The area of triangles formed by a point inside a parallelogram with vertices A, B, C, D follows specific geometric constraints. The sum of areas of opposite triangles (AMD and BMC) equals half the area of the parallelogram, and the sum of the other two (AMB and DMC) also equals half the area.

Multiple choice maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

If the area of a parallelogram is $144 \operatorname { cm } ^ { 2 }$ and its base is $9 cm$. then its height is 

  1. $8 cm$
  2. $12 cm$
  3. $24 cm$
  4. $16 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Area of parallelogram $=base\times height$


$\Rightarrow$ $144{cm}^{2}=9cm\times 10cm$


$\Rightarrow$ $h$ in $cm=\cfrac{144{cm}^{2}}{9cm}$

$\therefore$ $h=16cm$

Hence height $=16cm$

Multiple choice maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

Area of the parallelogram formed by the pairs of lines $x^{2}+xy-^{2}=0$ and $x^{2}+xy-y^{2}-3x-4y+1=0$ is

  1. $\sqrt {5}$
  2. $\dfrac {1}{\sqrt {5}}$
  3. $2\sqrt {5}$
  4. $\dfrac {2}{\sqrt {5}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The area of a parallelogram formed by two pairs of parallel lines is given by the formula |c1-c2|*|d1-d2| / |ab'-a'b|. Calculating this for the given equations yields 2*sqrt(5).

Multiple choice maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

A parallelogram has sides 12 cm and 9 cm. If the distance between its shorter sides is 8 cm, find the distance between its longer side.

  1. $3 \ cm$
  2. $6 \ cm$
  3. $9 \ cm$
  4. $12 \ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Adjacent sides of parallelogram $= 12\ cm$ and $9\ cm$

Distance between shorter sides $= 8\ cm$

Area of parallelogram = $b \times h =9 \times 8 =72 \ cm^2$

Again, area of parallelogram = $b \times h$
$72 =12 \times h$
$h= 6 \ cm$

Therefore, the distance between its longer side $= 6\ cm.$