Question

the length of an rectangle is twice its width if the perimeter of the rectangle is 66 cm find its length
A 11cm
B. 22cm
C. 33cm
D. 20cm​

Answer 1

Given :

• Length of rectangle is twice its width.
• Perimeter of rectangle = 66 cm

To find :

• The length of the rectangle.

Solution :

Let's assume the width of the rectangle as x

Then Length will be 2(x) = 2x

Now, We know that,

★ Perimeter (rectangle) = 2(length + breadth)

[ Width can also be said as breadth ]

Thus,

⇝ 66 = 2(2x + x)

⇝ 66 = 2(3x)

⇝ 66 = 6x

⇝ 66/6 = x

⇝ 11 = x

Therefore,

• The width of the rectangle = 11 cm
• Then, the length will be (2x) = 2×11 = 22 cm

Check —

Perimeter of rectangle = 2(l + b)

• Putting the value :

↦ 66 = 2(11 + 22)

↦ 66 = 2 × 33

↦ 66 = 66

Thus, Both sides are equal/same.

Hence , Solution is checked !

Answer 2

Correct Question-:

• The length of an rectangle is twice its width . If the Perimeter of the rectangle is 66 cm. Find its length.

AnswEr-:

• $\underline{\boxed{\star{\sf{\blue{ Length \:of\:Rectangle\:is\: \: =\frak{22cm}\:.}}}}}$

EXPLANATION-:

• $\frak{Given \:\: -:} \begin{cases} \sf{The\:Perimeter \:of\:Rectangle\:\:is\:= \frak{66cm}} & \\\\ \sf{The\: length\: of \:a\: rectangle\: is\: twice\: its\: width . }\end{cases}$

• $\frak{Given \:\: -:} \begin{cases} \sf{The\:Length \:of\:Rectangle. }\end{cases}$

Solution,

• $\frak{Let's \:Assume \: -:} \begin{cases} \sf{The\:width\:of\:Rectangle\:\:be\:= \frak{x\:cm}} & \\\\ \sf{Then\:, }& \\\\ \sf{Length \:of\:Rectangle\:is\:twice\:to\:it's\:Breadth= 2 × x = \frak{2x\:cm} }\end{cases}$

Then ,

• $\underline{\boxed{\star{\sf{\blue{ Perimeter _{(Rectangle)}\: = 2×(Length \:+ Breadth)\:.}}}}}$

• $\frak{Here \:\: -:} \begin{cases} \sf{The\:Perimeter \:of\:Rectangle\:\:is\:= \frak{66cm}}& \\\\ \sf{The\:width\:of\:Rectangle\:\:is\:= \frak{x\:cm}} & \\\\ \sf{Length \:of\:Rectangle\:is\: \frak{2x\:cm} } \end{cases}$

Now ,

• $\implies{\sf{\large {66cm \: = \: 2( x + 2x ) }}}$
• $\implies{\sf{\large {66cm \: = \: 2( 3x ) }}}$
• $\implies{\sf{\large {66cm \: = \: 6x }}}$
• or,
• $\implies{\sf{\large {6x \: = \: 66cm }}}$
• $\implies{\sf{\large {x \: = \: \frac{66}{6} }}}$
• $\implies{\sf{\large {x \: = \: 11 }}}$

Therefore,

• $\underline{\boxed{\star{\sf{\blue{ x\: = \: 11\:.}}}}}$

Now ,

• $\frak{Putting \:x=11\: -:} \begin{cases} \sf{The\:width\:of\:Rectangle\:\:is\:= x\:= 11cm} & \\\\ \sf{Length \:of\:Rectangle\:is\: 2x\:= 2×11 \: =22cm} \end{cases}$

Hence ,

• $\underline{\boxed{\star{\sf{\blue{ Length \:of\:Rectangle\:is\: \: =\frak{22cm}\:.}}}}}$

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♤ Verification ♤

• $\underline{\boxed{\star{\sf{\blue{ Perimeter _{(Rectangle)}\: = 2×(Length \:+ Breadth)\:.}}}}}$

• $\frak{Here \:\: -:} \begin{cases} \sf{The\:Perimeter \:of\:Rectangle\:\:is\:= \frak{66cm}}& \\\\ \sf{The\:width\:of\:Rectangle\:\:is\:= \frak{11\:cm}} & \\\\ \sf{Length \:of\:Rectangle\:is\: \frak{22\:cm} } \end{cases}$

Now,

• $\implies{\sf{\large {66cm \: = \: 2( 11+22 ) }}}$
• $\implies{\sf{\large {66cm \: = \: 2( 33 ) }}}$
• $\implies{\sf{\large {66cm \: = \: 66cm }}}$

Therefore,

• $\underline{\boxed{\star{\sf{\blue{ LHS\: = \: RHS\:.}}}}}$
• $\underline{\boxed{\star{\sf{\blue{ Hence\:, \: Verified \:.}}}}}$

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