Tag: perimeter and area of rectilinear figures

Questions Related to perimeter and area of rectilinear figures

In a parallelogram ABCD, if $ \angle B = 135^{\circ},$ determine the measures of its other angles.

  1. $ \angle A = 45^{\circ}, \angle C = 45^{\circ},\angle D = 135^{\circ} $

  2. $ \angle A = 135^{\circ}, \angle C = 45^{\circ},\angle D = 45^{\circ} $

  3. $ \angle A = 40^{\circ}, \angle C = 50^{\circ},\angle D = 135^{\circ} $

  4. None of these


Correct Option: A
Explanation:

We have, $\angle B = 135^{\circ} $ 


Since ABCD is a parallelogram.

$ \therefore \angle A = \angle C, \angle B = \angle D $ and $ \angle A + \angle B = 180^{\circ} $

$ \Rightarrow \angle A = \angle C = 45^{\circ} $ and $ \angle B = \angle D = 135^{\circ} $

A flooring tile has the shape of a parallelogram whose base is $24: cm$ and the corresponding height is $10: cm$. How many such tiles are required to cover a floor of area $1080$ $m^2$? 

  1. $20000$

  2. $35000$

  3. $45000$

  4. $65000$


Correct Option: C
Explanation:

Area of the parallelogram $=$ Base$\times $Height


So area of each tiles $=24\times 10=240 :cm^2$

Area of the floor $=1080 : m^2=(1080\times 100\times 100)  : cm^2$

$\therefore$ Required number of tiles $=\dfrac{\text{Area of the floor}}{\text{Area of each tiles}}=\dfrac{10800000}{240}=45000$

$ABCD$ is a parallelogram of area $162\ sq. cm$. $P$ is a point on $AB$ such that $AP : PB = 1 : 2$. 
Calculate the ratio $PA : DC$.
  1. $1 : 3$

  2. $3:1$

  3. $3:2$

  4. $2:3$


Correct Option: A
Explanation:

Given,

$ABCD$ is a parallelogram, Area of parallelogram $ABCD=162cm^2$

$P$ is such a point on $AB$ such that $AP:PB=1:2$

Now, $\frac{AP}{PB}=\frac{1}{2}=k$[Let]

Thus, $AP=k$ and $PB=2k$

And, $AB=AP+PB$

   $=>AB=k+2k$

$=>AB=3k$

$\therefore \frac{AP}{AB}=\frac{1}{3}$

Now, $AB=CD$ [Opposite sides of a parallelogram are equal]

$\therefore \frac{AP}{CD}=\frac{1}{3}$


$ABCD$ is a parallelogram of area $162\: cm^2$. $P$ is a point on $AB$ such that $AP : PB = 1 : 2$. Calculate the area of $\Delta APD$
  1. $20\; cm^2$

  2. $27\ cm^2$

  3. $24\ cm^2$

  4. $25\ cm^2$


Correct Option: B
Explanation:

If $DB$ is the diagonal,
Area of parallelogram $\displaystyle \frac{1}{2}ABCD =$ area $ADB = $ area $BDC$
Area ADB$=\displaystyle \frac{1}{2} (162)=81$
$P$ is mid point of $AB$ in the ratio $1:2$ $ \left (\dfrac{1}{3}+\dfrac{2}{3}\right)$ 

Area APD $=\displaystyle \frac{1}{3}\times $ Area of $ADB=\displaystyle \dfrac{1}{3}(81)=27 $ sq. cm

The area of a parallelogram is $y$ $cm^{2}$ and its height is $h\ cm$. The base of another parallelogram is $x\ cm$ more than the base of the first parallelogram and its area is twice the area of the first. Find, in terms of $y, h$ and $x$, the expression for the height of the second parallelogram.

  1. $\displaystyle \frac{hy}{yh-x}$

  2. $\displaystyle \frac{y}{y-xh}$

  3. $\displaystyle \frac{2hy}{y+xh}$

  4. None of these


Correct Option: C
Explanation:

area of 1st parallelogram = y cm square, height = h cm

Area= Base\times Height
$y=b\times h$ ----eq 1
Base =$b=y/h$
area of 2nd parallelogram = 2y cm square, height = H cm
Base of 2nd parallelogram $\dfrac { y }{ h } +x$
$2y=(\dfrac { y }{ h }+x)\times H$
$H=\dfrac { 2yh }{ (y+hx) } $

A field is in the form of a parallelogram whose base is $420\ m$ and altitude is $3.6\ dam$. Find the cost of watering at $10$ paise per sq. m.

  1. Rs. $15.120$

  2. Rs. $1512$

  3. Rs. $151.20$

  4. None


Correct Option: B
Explanation:

Base  $= 420\ m$
Height $=  36\ m$
Area $= b \times h = 420 \times 36$
$= 15,120 \displaystyle \ m^{2}$
The cost of watering per sq m
$= 10$ paise
Cost  of watering the field $= 15120 \times 0.1$
Rs. $1512$

The base of a parallelogram is three times its height. If the area of the parallelogram is $75$ sq cm, then its height is

  1. $5 cm$

  2. $\displaystyle 5\sqrt{2}cm$

  3. $\displaystyle 3\sqrt{2}cm$

  4. $15 cm$


Correct Option: A
Explanation:

let height of ||g be 'b', then base of ||g be 3b.
 area of ||g$=$base$\times $height
$75c{ m }^{ 2 }=3b\times b\ { b }^{ 2 }=25$
 $b=5$cm, therefore height $=5$cm

A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is $100 m $, then the altitude of the triangle is:

  1. $100 m$

  2. $200 m$

  3. $100\sqrt{2}$ m

  4. $10\sqrt{2}$ m


Correct Option: B
Explanation:

Let the altitude of the $ \Delta =h _{1}$ and altitude of the parallelogram $  =h _{2}$ 

Let base of both $\Delta $ and parallelogram$ = b$
Then, $ b\times h _{2}=\cfrac{1}{2}\times b\times h _{1}$ 
$ \Rightarrow h _{1}=2h _{2}$
$\Rightarrow h _1 = 2\times 100m=200m$

The ratio of two adjacent sides of a parallelogram is $3 : 4$ Its perimeter is $105$ cm Find its area if altitude corresponding to the larger is $15$ cm

  1. $900$ $\displaystyle cm^{2}$

  2. $600$ $\displaystyle cm^{2}$

  3. $300$ $\displaystyle cm^{2}$

  4. $450$ $\displaystyle cm^{2}$


Correct Option: D
Explanation:

Let the two adjacent sides of the parallelogram be $3x$ and $4x$ Then
$2(3x + 4x) = 105$
$\displaystyle \Rightarrow 14x=105.$
$\displaystyle \Rightarrow x=7.5$
$\displaystyle \therefore $ The two sides are $3 \times 7.5$ cm and $4 \times 7.5$ cm 
i.e, $22.5$ cm and $30$ cm
Area of the parallelogram = base x altitude 
                                         = $30$ cm $\times$ $15$cm = $450$
$\displaystyle cm^{2}$

ABCD is a parallelogram P and R are two points on AB such that the area of parallelogram ABCD is 8 times the are of $\displaystyle \Delta DPR$ If PR = 5cm then CD is equal to 

  1. $10$ cm

  2. $5$ cm

  3. $20$ cm

  4. $12$ cm


Correct Option: C
Explanation:

Area of parallelogram ABCD 
= $8$ X Area of $\displaystyle \Delta $DPR
$\displaystyle \Rightarrow AB\times \ height=8\times \left ( \frac{1}{2}\times PR\times height \right )$
(Note: Height will be same for both the $\displaystyle \Delta $ and parallelogram)
$\displaystyle \Rightarrow  $ $AB = 4\times  PR = 4$ $\displaystyle \times  5$ cm = $20$ cm
$\displaystyle \therefore $ $\displaystyle CD = 20 $ cm