Parallelogram and rectangle - class-IX

Area of parallelogram $=base\times height$

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 2sqrt(5).

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

The base of parallelogram is $36cm$

$2 : 1$
Diagonal resolves a parallelogram into two equal triangles.

Base $= 17\ cm$
Height $= 0.8\ m =0.8 \times 100 = 80\ cm$
Area of parallelogram 
$= b \times h$
$= 17\times 80$
$= 1360\ cm^2$

Area of the rectangle $= l \times b = 10 m \times 12 m$
                                     $= 120 m^{2}$
Area of parallelogram $= Base \times Altitude= 120 m^{2}$
$ \Rightarrow$ Altitude $= \cfrac{Area }{Base}=\cfrac{120m^{2}}{20m}=6m$

Area of the rectangle $=$ Area of parallelogram
                                     $= 20 cm \times 6 m = 120 \displaystyle cm^{2}$
Now,                 
Ratio $= 7 : 5$             Ratio $= 40 : 3$
Product $= 35$            Product $= 120$
-------------------------------------------------------------
Ratio $= 15 :2$           Ratio $= 30 : 1$
Product $= 30$            Product $= 30$
$\displaystyle \therefore $ Ratio $= 7 : 5$ does not match with the condition as $120$ is not divisible by $35$.

Area of parallelogram 
= $Base \times height = 120 cm^{2}$.
$\therefore $  Base = $12 cm$

As we know,

Given One side of parallelogram is 8 cm and altitude is 6 cm 

Given the area of parallelogram is 350 sq cm and base is 25 cm

Given the two adjacent sides of a parallelogram is x and y and angle is $\alpha $

Area = Base $\times$ Altitude
         = 6 cm $\times$ 3.5 cm
         = 21 square cm

Base = 16 cm
Height = 0.4 m
            = 0.4 $\times$ 100 = 40 cm
Area of parallelogram = b $\times$ h
                                      = 16 $\times$ 40
                                      = 640 $cm^2$

Area = base * height = 360. Given base:height = 5:2, let base = 5x and height = 2x. Then 10x^2 = 360, so x^2 = 36, x = 6. Base = 30m, height = 12m.

Area of parallelogram $=$ length $\times$ height
Here altitude is nothing but the height
$\therefore $ Area of parallelogram $= 8 \times 6$
$= 48: cm^2$

Area of parallelogram $S _1=bh$

The diagonals of parallelogram is $5cm,6cm$

In a parallelogram ABCD, the diagonals bisect each other. The midpoint of AC must equal the midpoint of BD. Using this property with C on the line 3x-2y=1, we derive the locus of D.

We have,

$ \overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k} $

$ \overrightarrow{b}=3\widehat{i}-2\widehat{j}+\widehat{k} $

We know that,

Area of parallelogram $=\left| \begin{matrix} i & j & k \\ 1 & 2 & 3 \\ 3 & -2 & 1 \end{matrix} \right| $

$ =\left( -6-2 \right)\widehat{i}-\widehat{j}\left( 9-1 \right)+\widehat{k}\left( 6+2 \right) $

$ =-8\widehat{i}-8\widehat{j}+8\widehat{k} $

$ Now, $

$ \left| \overrightarrow{a}\times \overrightarrow{b} \right|=\sqrt{{{\left( -8 \right)}^{2}}+{{\left( -8 \right)}^{2}}+{{8}^{2}}} $

$ =\sqrt{64+64+64} $

$ =\sqrt{3\times 64} $

$ =8\sqrt{3}\,\,sq.\,unit $

Hence, this is the answer.

A quadrilateral with three equal angles is not necessarily a parallelogram. For example, a kite can have three equal angles but is not a parallelogram.

$\Rightarrow$  We know that, in parallelogram opposite sides are parallel and equal in length.

Since opposite angles of a parallelogram are equal. Therefore,
$3x-2=50-x\Rightarrow x=13$

Let the smallest angle be of $x^{\circ}$. 

We have, $\angle B = 135^{\circ} $ 

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

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]

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

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

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. $1512$

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

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

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}$

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

Area of the parallelogram $=$ base $\times$ altitude 
$ \Rightarrow \left ( x+4 \right )\times\left ( x-3 \right )=x^{2}+4x-3x-12$
$ \displaystyle=x^{2}+x-12$
Given, $ \displaystyle x^{2}+x-12=x^{2}-4$

Given that, a parallelogram whose sides$l=10cm$ and$b=5cm$ has one diagonal${{d} _{1}}=8cm$.

Let, length of other diagonal $={{d} _{2}}$.

Now we know that,

  $ {{d} _{1}}^{2}+{{d} _{2}}^{2}=2\left( {{l}^{2}}+{{b}^{2}} \right) $

 $ {{8}^{2}}+{{d} _{2}}^{2}=2\left( {{10}^{2}}+{{5}^{2}} \right) $

 $ {{d} _{2}}^{2}=250-64 $

 $ {{d} _{2}}^{2}=186 $

 $ {{d} _{2}}=\sqrt{186}cm $

Hence,
this is the answer. 

Area of a parallelogram = base $\times$ height
$60 = 12 \times height$
height = $60 \div 12$
height = $5$ cm

Area of a parallelogram = base $\times$ height
= $34 \times 8$
= $272$ $m^{2}$

Area of a parallelogram$=base\times height$. Let the base be 'b'.
Hence
$125m^{2}=5b$ or $b=25m$
Therefore $base=25m$.

Area of parallelogram $\displaystyle =b\times a$
$\displaystyle 80=b\times 10$
$\displaystyle b=8cm$

Area of a parallelogram = base $\times$ height
= $200 \times 2.5$
= $500$ $cm^{2}$

Area of a parallelogram = base $\times$ height
= $12 \times 5$
= $60$ $m^{2}$

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

Area of parallelogram $=Base \times height $