Math, asked by sumukatoch, 9 months ago

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

Answers

Answered by Gr8AYU
1

Step-by-step explanation:

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Answered by SarcasticL0ve
4

GivEn:

  • Length of Rectangle = 40 cm

  • Diagonal of Rectangle = 41 cm

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To find:

  • Perimeter of Rectangle

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SoluTion:

\bf We\; have \begin{cases} & \text{Length = 40 cm }  \\ & \text{Diagonal = 41 cm} \end{cases}

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As we know that,

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Diagonal of Rectangle is,

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\star\;{\boxed{\sf{\purple{D = \sqrt{l^2 + b^2}}}}}

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\;\;\;\;\;\small\sf \underline{Put\;the\;given\;values\;:}

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:\implies\sf 41 = \sqrt{(40)^2 + (b)^2}

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\;\;\;\;\;\small\sf \underline{Squaring\;both\;sides\;:}

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:\implies\sf (41)^2 = \sqrt{(40^2 + b^2)^2}

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:\implies\sf 41^2 = 40^2 + b^2

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:\implies\sf b^2 = 41^2 - 40^2

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:\implies\sf b^2 = 1681 - 1600

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:\implies\sf b^2 = 81

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\;\;\;\;\;\small\sf \underline{Taking\;sqrt.\;both\;side\;:}

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:\implies\sf \sqrt{b^2} = \sqrt{81}

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:\implies\sf{\underline{\boxed{\sf{\pink{b = 9\;cm}}}}}\;\bigstar

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\therefore Breadth of Rectangle is 9 cm.

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DIAGRAM:

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\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\tt\large{A}}\put(7.7,1){ \tt\large{B}}\put(9.5,0.7){\sf{\large{40 cm cm}}}\put(11.5,1){ \tt\large{C}}\put(8,1){\line(1,0){3.5}}\put(8,1){\line(0,2){2}}\put(11.5,1){\line(0,3){2}}\put(8,3){\line(3,0){3.5}}\put(11.6,2){\sf{\large{9 cm}}}\qbezier(8,1)(8,1)(11.5,3)\put(11.5,3){ \tt\large{D}}\put(11.3,1){\line(0,2){0.2}}\put(11.3,1.2){\line(2,0){0.2}}\end{picture}

━━━━━━━━━━━━━━━

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\bf We\; have \begin{cases} & \text{Length = 40 cm }  \\ & \text{Breadth = 9 cm}  \end{cases}

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{\underline{\sf{\bigstar\;Now,\;we\;have\;to\;find\; Perimeter\;of\; Rectangle\;:}}}

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As we know that,

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\star\;{\boxed{\sf{\purple{Perimeter\;of\; Rectangle = 2(l + b)}}}}

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\;\;\;\;\;\small\sf \underline{Put\;the\;given\;values\;:}

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:\implies\sf P = 2(40 + 9)

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:\implies\sf P = 2(49)

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:\implies\sf{\underline{\boxed{\sf{\pink{P = 98\;cm}}}}}\;\bigstar

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\therefore Hence, Perimeter of Rectangle is 98 cm.

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