Question

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

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

$[/tex] <strong>Find </strong><strong>the </strong><strong>perimeter</strong><strong> </strong><strong>of </strong><strong>the </strong><strong>rectangle</strong><strong> </strong><strong>whose </strong><strong>length</strong><strong> </strong><strong>is </strong><strong>4</strong><strong>0</strong><strong>c</strong><strong>m</strong><strong> </strong><strong>and </strong><strong>a </strong><strong>diagonal</strong><strong> </strong><strong>is </strong><strong>4</strong><strong>1</strong><strong>c</strong><strong>m</strong><strong>.</strong><strong> </strong></p><p></p><p>As given in the rectangle,</p><p><strong>By Pythagoras theorem,</strong> </p><p>[tex] {(diagonal)}^{2} = {(length)}^{2} + {(width)}^{2}$

Now as given in the question

Diagonal = 41 cm.

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 ( Length + Width )

= 2 ( 40 cm + 9 cm )

= 2 x 49cm

= 98 cm

Hence the perimeter of the rectangle is 98 cm.