Tag: perimeter and area of rectilinear figures

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 $