Question

1. A wire length 40 cm is bent so as to form a rectangle. Find the maximum area that can
be enclosed by the wire.
April. 2015 [DCE]; Oct. 2011 [ECE], 2010 [EEE]​

Answer 1

Answer:

Solution:

Length= 40cm

Breadth= 22cm

Perimeter of the rectangle= Length of the wire

=2(l+b)= 2(40cm + 22cm)

=2\times622×62cm = 124cm

Now, the wire is rebent into a square.

Perimeter= 124cm

4\times side=1244×side=124

Therefore side= \frac{124}{4}  

4

124

​

cm= 31cm

So, the measure of each side= 31cm

Area of rectangular shape= l\times bl×b

=40cm\times22cm40cm×22cm

=880cm^2880cm  

2

 

Area of a square shape= \left(Side\right)^2(Side)  

2

 

=\left(31\right)^2=961cm^2(31)  

2

=961cm  

2

 

Since 961cm^2\ >\ 880cm^2961cm  

2

 > 880cm  

2

 

Hence, the square encloses more area.

Step-by-step explanation:

Answer 2

Answer:

The maximum area of the rectangle = 99cm²

Step-by-step explanation:

Given,

Length of the wire = 40cm

To find,

The maximum area of the rectangle can be enclosed in the wire.

Recall the formula

Area of the rectangle = length × breadth

Perimeter of the rectangle = 2(length +breadth)

Solution:

Since, the  length of the wire is 40cm,

The perimeter of the rectangle which can be enclosed in the wire = 40cm

2(length +breadth) = 40cm

length +breadth = 20cm

The possible values of length and breadth are

(1,19)(2,18),(3,17),(4,16),(5,15),(6,14),(7,13),(8,12),(9,11),(10,10)

Out of these dimensions, the maximum area is obtained, when length = 10 and breadth = 10

If length = 10 and breadth = 10, then it becomes a square.

Eliminating (10,10) from the above options we get the maximum area when the dimensions are (9,11)

∴ The maximum area of the rectangle = 99cm²

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