Tag: maths

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.