Question

a rod of 28 cm long is to be bent to make a rectangle.
●how about a rectangle of diagnol 14 cm?​

Answer 1

Answer:

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Step-by-step explanation:

length = 8 and width = 6.

Step-by-step explanation:

As per the question,

A rod of length 28 cm is bent to make a rectangle. That mean perimeter of rod is equal to perimeter of rectangle.

Perimeter of rectangle = 2(l + b)

where l = length

b = breadth

Now, the first equation we have:

2(l + b) = 28

l + b = 14

b = (14 - l)

As the diagonal is also given,d = 8 cm

Therefore,

l^{2}+ b^{2} = 8^{2}l

2

+b

2

=8

2

is the second equation,

substitute the value of b = (14 - l) in l^{2}+ b^{2} = 8^{2}l

2

+b

2

=8

2

∴ l^{2} (14-l)^{2} = 8^{2}l

2

(14−l)

2

=8

2

l^{2}+14^{2}+l^{2} -28l = 8^{2}l

2

+14

2

+l

2

−28l=8

2

l^{2} +-14l+66=0l

2

+−14l+66=0

As the discriminant is less than 0 that's why a rectangle of a diagonal 8 cm can not be made.

Now for the rectangle to be possible the sum of the length and width is equal to 14.

So one of the possible case is when length = 8 and width = 6.