Question

Find the perimeter of a rectangle where the
|length is 40 cm and diagonal is 41 cm.​

Answer 1

Answer:

Perimeter of the rectangle = 98 cm

Step-by-step explanation:

Given :

Length of the rectangle = 40 cm

Diagonal of the rectangle = 41 cm

Diagram :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(2,3){\sf\large 40 cm}\put(-0.5,-0.4){\bf D}\put(-0.5,3.2){\bf A}\put(5.3,-0.4){\bf C}\put(5.3,3.2){\bf B}\qbezier(0,0)(0,0)(5,3)\put(1.65,2){\sf\large 41 cm}\put(-1.6,-1){\qbezier(1.6,4)(1.6,4)(6.6,1)}\end{picture}$

To find :

The perimeter of the rectangle

Concept :

Let AB of the diagram is length . AC is the diagonal of the rectangle . And the AD is the Breadth .

So , we have to first find the Breadth of the rectangle . Use the Pythagoras theorem to find the Breadth .

$\boxed{\rm{{AB}^{2}+{AD}^{2}={AC}^{2}}}$

After that to find the perimeter of the rectangle use this formula -:

$\boxed{\tt{P=2(AB+AD)}}$

Where,

P = Perimeter of the rectangle

Solution :

Breadth -:

$\leadsto\rm{{40cm}^{2}+{AD}^{2}={41cm}^{2}}$

$\leadsto\rm{AD=\sqrt{{41cm}^{2}-{40cm}^{2}}}$

$\leadsto\rm{AD=\sqrt{(41\times41)-(40\times40)}cm}$

$\leadsto\rm{AD=\sqrt{1681-1600}cm}$

$\leadsto\rm{AD=\sqrt{81}cm}$

$\leadsto\rm{AD=9cm}$

So , the Breadth of the rectangle is 9 cm .

Perimeter of the rectangle -:

$\leadsto\tt{P=2(40cm+9cm)}$

$\leadsto\tt{P=2(49cm)}$

$\leadsto\tt{P=98cm}$

So , the perimeter of the rectangle is 98 cm.

Answer 2

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$\large\underline{\red{\sf \blue{\bigstar} .}}$In Δ ACD $\large\underline{\blue{\sf \red{\bigstar} .}}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• AB = CD = 40 CM

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• AD=41 CM

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$\large\underline{\red{\sf \green{\bigstar} .}}$Diagram$\large\underline{\red{\sf \green {\bigstar} .}}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(2,3){\sf\large 40 cm}\put(-0.5,-0.4){\bf D}\put(-0.5,3.2){\bf A}\put(5.3,-0.4){\bf C}\put(5.3,3.2){\bf B}\qbezier(0,0)(0,0)(5,3)\put(1.65,2){\sf\large 41 cm}\put(-1.6,-1){\qbezier(1.6,4)(1.6,4)(6.6,1)}\end{picture}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

_____________________

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• ∠ACD = 90° ( RECTANGLE HAS 90° ON ALL THE BOTH SIDES)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\red{\sf \purple{\bigstar} .}}$By Pythagoras theorem $\large\underline{\red{\sf \blue{\bigstar} .}}$

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• AD² = AC² + CD²

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• AD² - CD² = AC²

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• (41)² - (40)² = AC²

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• 1681 - 1600 = 81 = AC²

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• √81 = AC

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• $\large\underline{\red{\sf \green{\bigstar} .}}$AC = 9 CM$\large\underline{\red{\sf \blue{\bigstar} .}}$

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• Perimeter of Rectangle ABCD = 2 (l+b)

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• = 2(40 + 9)

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• 2(49) = 98 CM

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$\pink{\sf{\star\;98cm\; is \;the \;perimeter\; of\; rectangle }}$

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$\red{\sf{\star\;Note }}$

• kindly view the answer from brainly.in⠀⠀⠀⠀⠀⠀⠀