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# Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

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As we can see in the rectangle,

By Pythagoras theorem,

$(Diagonal)^2=(Length)^2+(Width)^2$

Now as given in the question,

Diagonal = 41 cm.

Length = 40 cm.

So, Putting these value we get,

$(41)^2=(40)^2+(Width)^2$

$(Width)^2=(41)^2-(40)^2$

$(Width)^2=1681-1600$

$(Width)^2=81$

$Width=9cm$

Hence the width of the rectangle is 9 cm.

So

The perimeter of the rectangle = 2 ( Length + Width )

= 2 ( 40 cm + 9 cm )

= 2 x 49 cm

= 98 cm

Hence the perimeter of the rectangle is 98 cm.

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