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 perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

The base and the corresponding altitude of a parallelogram are $10: cm$ and $3.5: cm$, respectively. The area of the parallelogram is

  1. $30\: cm^2$
  2. $35\: cm^2$
  3. $70\: cm^2$
  4. $ 17.5\:cm^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The area of the parallelogram is base $\times$ height $cm^2$

Area of the parallelogram$=(10)(3.5)=35:cm^2$.

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

A parallelogram has sides $30 m, 70 m$ and one of its diagonals is $80 m$ long. Its area will be

  1. $600\displaystyle m^{2}$
  2. $\displaystyle 1200\sqrt{3}m^{2}$
  3. $1200\displaystyle m^{2}$
  4. $\displaystyle 600\sqrt{3} m^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The diagonal of parallelogram divides it into two congruent triangles. 
$\therefore $ Area (parallelogram $ABCD) = 2 \times$ Area $ \left (\Delta ABC  \right ) $
In $ \Delta ABC$,
$ s=\cfrac{80m+ 30m+70m}{2}=\cfrac{180m}{2}=90m$
$ \therefore Area=\sqrt{90\left ( 90-80 \right )(90-30)(90-70)}m^{2}$
$ =\sqrt{90\times10\times60\times20m^{2} }$
$= 600\sqrt{3} m^{2}$

$ \therefore $ Area of parallelogram $ABCD =$$ 2\times 600\sqrt{3}m^{2}$ $ =1200\sqrt{3}m^{2}$
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

Let $A= \left ( 1,2,3 \right )B= \left ( -1,-2,-1 \right )C= \left ( 2,3,2 \right )$ and $ D= \left ( 4,7,6 \right )$. Then $ABCD$ is a

  1. rectangle

  2. square

  3. parallelogram

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

AB${=}$ $\sqrt{{(-1-1)}^{2}+{(-2-2)}^{2}+{(-1-3)}^{2}}$
AB${=}$ $\sqrt{{(-2)}^{2}+{(-4)}^{2}+{(-4)}^{2}}$
AB${=}$ $\sqrt{36}$
AB${=}$ 6
Similarly you find that BC${=}$ $\sqrt{43}$  CD${=}$ 6  and DA${=}$ $\sqrt{43}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(2-1)}^{2}+{(3-2)}^{2}+{(2-3)}^{2}}$
AC${=}$ $\sqrt{3}$
similarly BD${=}$ $\sqrt{155}$ 
Diagonals are not equal
direction ratio of line passing through AB is (-2,-4,-4)
direction ratio of line passing through  CD is (2,4,4), As the dr of AB and CD are proportional which means AB is parallel to CD,
Similarly check for BC and DA then you will find that they are also parallel
hence it is parallelogram

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If $A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C= \left ( \sqrt{2},\sqrt{2},0 \right )$ and $D= \left ( \displaystyle \frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17} \right )$, then $ABCD$ is a

  1. rhombus

  2. square

  3. parallelogram

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given points are, $A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C=

\left ( \sqrt{2},\sqrt{2},0 \right )$ and $D= \left ( \displaystyle

\frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17}

\right )$

Length $AB = \sqrt{\sqrt{2}^2 + \sqrt{2}^2 + (2-2)^2} = 2$
Length $BC = \sqrt{(\sqrt{2}-\sqrt{2})^2 + (\sqrt{2}-\sqrt{2})^2 + 2^2} = 2$
Length $CD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}-\sqrt{2}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}-\sqrt{2}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17}\right)^2} = 2$
Length $AD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17} - 2\right)^2} = 2$

Angle between two vectors $\overline{AB} = p _{1}\hat{i}+q _{1}\hat{j}+r _{1}\hat{k} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + 0\hat{k}$ and $\overline{BC} = p _{2}\hat{i}+q _{2}\hat{j}+r _{2}\hat{k} = 0\hat{i} + 0\hat{j} + 2 \hat{k}$ is $cos\theta = 0 \Rightarrow \theta = 90^o$

All sides are equal and angle between $AB$ and $BC$ is $90^o$, Hence, its a square.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

A rectangular parallelopiped is formed by drawing planes through the points $(-1,2,5)$ and $(1,-1,-1)$ and parallel to the coordinate planes. the length of the diagonal of the parallelopiped is

  1. $2$
  2. $3$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The plane forming the parallelipiped are

$x=-1,x=1;y=2,y=-1$ and $z=5,z=-1$
Hence, the lengths of the edges of the parallelopiped are
$1-\left( -1 \right) =2,\left| -1-2 \right| =3$ and $\left| -1-5 \right| =6$
$($ length of an edge of the parallelopiped is the distance between the parallel plane sperpendicular to the edge$)$
$\therefore$ Length of diagonal of the parallelopiped
$=\sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 49 } =7$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The arrangement of the following conics in the descending order of their lengths of semi latus rectum is
A) $ 6= r (1 + 3\cos \theta )$
B) $10= r (1 + 3\cos \theta )$
C) $8= r (1 + 3\cos \theta )$
D) $12= r (1 + 3\cos \theta )$

  1. $D, A, B, C$
  2. $B, C, D, A$
  3. $D, B, C, A$
  4. $A, C, B, D$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Comparing given equation with standard equation $r(1+e\cos\theta)=l$ where $l$ is semi latus rectum
Hence order is $D,B,C,A$

Multiple choice maths constructions mid-point formula midpoints division of a line segment

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

  1. $(-1, 0)$
  2. $(-1, 1)$
  3. $(0, -1)$
  4. $(-5, 0)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a parallelogram, the diagonals bisect each other. Let the vertices be A(3, -4), B(x, y), C(-6, 5), and D(2, 1). The midpoint of diagonal AC is ((3-6)/2, (-4+5)/2) = (-1.5, 0.5). The midpoint of diagonal BD is ((x+2)/2, (y+1)/2). Setting these equal gives x+2 = -3 (x=-5) and y+1 = 1 (y=0).

Multiple choice maths constructions mid-point formula midpoints division of a line segment

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

  1. $(-1,0)$
  2. $(0,-1)$
  3. $(-1,1)$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$P(-6,-4),Q(3,5),R(-2,1),S(\alpha ,\beta )$

Let $P$ and $Q$ are the extremities of diagonals of a parallelogram, and 

$R$ and $S$ will be the extremities of diagonals of a parallelogram

Now,

midpoint of $PQ=\dfrac{3-6}{2},\dfrac{5-4}{2}=\dfrac{-3}{2},\dfrac{1}{2}$

midpoint of $RS\Rightarrow \dfrac{-2+\alpha }{2}=-\dfrac{3}{2}$

$\Rightarrow \alpha =-3+2=-1$

Now,

$\dfrac{\alpha +\beta }{2}=\dfrac{1}{2}$

$\Rightarrow \beta =0$

Therefore, coordinates of 4th vertex is $(-1,0)$
Multiple choice maths constructions mid-point formula midpoints division of a line segment

Three consecutive vertices of a parallelogram are $(1, -2)$, $(3,6)$ and $(5,10)$. The coordinates of the fourth vertex are:

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

Let the fourth vertex D $ = (x,y) $
We know that the diagonals of a parallelogram bisect each other. So,the

midpoint of AC is same as the mid point of BD.

Mid point of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y

} _{ 2 }) $ is  calculated by the formula $ \left( \frac { { x } _{ 1 }+{

x } _{ 2 } }{ 2 } ,\frac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $

So, midpoint of $ AC = $ Mid point of $ BD $

$ => \left( \frac { 1+5 }{ 2 } ,\frac { -2+10 }{ 2 }  \right) \quad =

\left( \frac { 3+x }{ 2 } ,\frac { 6 +y }{ 2 }  \right) \quad $

$ => \left( \frac { 6 }{ 2 } ,\frac { 8 }{ 2 }   \right) \quad =

\left( \frac { 3+x }{ 2 } ,\frac { 6 +y }{ 2 }  \right) \quad $

$ => 3+x=6 ; 6 + y = 8 $

$ => x = 3 ; y = 2 $

Hence, $ D = (3,2) $

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The vertices of a parallelogram are $(3, -2)$, $(4,0)$, $(6, -3)$ and $(5, -5)$. The diagonals intersect at the point M. The coordinates of the point M are:

  1. $\begin{pmatrix} \frac { 9 }{ 2 },-\frac { 5 }{ 2 } \end{pmatrix}$
  2. $\begin{pmatrix} \frac { 7 }{ 2 },-\frac { 5 }{ 2 }\end{pmatrix}$
  3. $\begin{pmatrix} \frac { 7 }{ 2 },-\frac { 3 }{ 2 }\end{pmatrix}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The diagonals of a parallelogram bisect each other. Hence M is the mid point of the vertices $ (3,-2) ; (6,-3) $ or of the vertices  $ (4,0) ; (5,-5) $

Mid point of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{

2 }) $ is  calculated by the formula $ \left( \frac { { x } _{ 1 }+{x} _{ 2 } }{ 2 } ,\frac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $
Using this formula, mid point of $ (3, -2) , (6, -3) = \left( \frac { 3\quad +6 }{ 2 } ,\frac { -2-3 }{ 2 }  \right) =\left( \frac { 9 }{ 2 } ,\frac { -5 }{ 2 }  \right) $

Multiple choice maths constructions mid-point formula midpoints division of a line segment

Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points $(-2, -1)$, $(1,0)$, $(4,3)$ and $(1,2)$ meet.

  1. $(5,1)$
  2. $(1,1)$
  3. $(1,5)$
  4. $(1,1-)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the vertices of the parallelogram be $A (-2,-1), B(1,0),  C(4,3),  D(1,2) $

The diagonals AC and BD would meet at the midpoint of AC and BD.

Midpoint of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{2 }) $ is  calculated by the formula $ \left( \cfrac { { x } _{ 1 }+{ x} _{ 2 } }{ 2 } ,\cfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $

Hence, mid point of AC $= \left( \cfrac { -2+4 }{ 2 } ,\cfrac { -1+3}{ 2 }  \right) = (1,1) $