Question

find the perimeter of rectangle whose length is 20cm and digonal is 21cm​

Answer 1

I hope this will help you ✌️

Answer 2

Given:-

Length of Rectangle = 20 cm

Diagonal of Rectangle = 21 cm

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To find:

Perimeter of Rectangle

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Solution:

$\bf We\; have \begin{cases} & \text{Length = 20 cm } \\ & \text{Diagonal = 21 cm} \end{cases}$

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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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$:\implies\sf 21 = \sqrt{(20)^2 + (b)^2}$

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

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$:\implies\sf (21)^2 = \sqrt{(20^2 + b^2)^2}$

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$:\implies\sf 21^2 = 20^2 + b^2$

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$:\implies\sf b^2 = 21^2 - 20^2$

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$:\implies\sf b^2 = 441 - 400$

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$:\implies\sf b^2 = 41$

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

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$:\implies\sf \sqrt{b^2} = \sqrt{41}$

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$:\implies\sf{\underline{\boxed{\sf{\pink{b = 6.40\;cm}}}}}\;\bigstar$

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$\therefore$ Breadth of Rectangle is 6.40 cm.

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$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\tt\large{A}}\put(7.7,1){ \tt\large{B}}\put(9.5,0.7){\sf{\large{20 cm cm}}}\put(11.5,1){ \tt\large{C}}\put(8,1){\line(1,0){3.5}}\put(8,1){\line(0,2){2}}\put(11.5,1){\line(0,3){2}}\put(8,3){\line(3,0){3.5}}\put(11.6,2){\sf{\large{6.40 cm}}}\qbezier(8,1)(8,1)(11.5,3)\put(11.5,3){ \tt\large{D}}\put(11.3,1){\line(0,2){0.2}}\put(11.3,1.2){\line(2,0){0.2}}\end{picture}$

━━━━━━━━━━━━━━━

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$\bf We\; have \begin{cases} & \text{Length = 20 cm } \\ & \text{Breadth = 6.40 cm} \end{cases}$

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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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$:\implies\sf P = 2(20 + 6.40)$

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$:\implies\sf P = 2(26.40)$

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$:\implies\sf{\underline{\boxed{\sf{\pink{P = 52.80\;cm}}}}}\;\bigstar$

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$\therefore$ Hence, Perimeter of Rectangle is 52.80 cm.