Question

If one rectangle is formed by 100 cm long wird
then what can be maximum area of rectangle?​

Answer 1

Answer:

length of wire is 100 cm

length of wire = perimeter of rectangle

\begin{gathered}100 = 2(l + b) \\ l + b = \frac{100}{2} \\ l + b = 50\end{gathered}

100=2(l+b)

l+b=

2

100

l+b=50

we have to find maximum area . So, let

l = b .

Then,

\begin{gathered}l + b = 50 \\ l + l = 50 \\ 2l = 50 \\ l = \frac{50}{2} \\ l = 25 \\ l = b = 25 \: cm\end{gathered}

l+b=50

l+l=50

2l=50

l=

2

50

l=25

l=b=25cm

Area of ∆

\begin{gathered} = l \times b \\ = 25 \times 25 \\ = 625 \: { cm}^{2} \end{gathered}

=l×b

=25×25

=625cm

2

Maximum area of ∆ is 625 cm² .

Step-by-step explanation:

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