Sustiosocharmingidn
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3. The perimeter of a rectangle is 140 cm. If the length is longer than width by 10 cm, find its dimensions.​

Sagot :

TheBlueStaridn

Given :

  • Perimeter of rectangle = 140cm
  • Length is 10cm long than width.

[tex] \\ \\ [/tex]

To find :-

  • Length of rectangle
  • Width of rectangle

[tex] \\ \\ [/tex]

Solution:-

So let take width as x.

{why width as x? , cause it's clearly visible that width is smaller than Length, so we will take width as x}

[tex] \\ [/tex]

and Length as x + 10.

{why Length is x + 10 and not 10x? , cause they have told 10 cm more not 10 times, If it would be 10 times then surely it would be 10x.

[tex] \\ [/tex]

Now Let's further exceed.

[tex] \\ [/tex]

We know:-

[tex] \bigstar \boxed{ \rm perimeter \: of \: rectangle = 2(length + width)}[/tex]

[tex] \\ \\ [/tex]

So:-

[tex] \\ [/tex]

[tex] \dashrightarrow \small\sf perimeter \: of \: rectangle = 2(length + width) \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf 140= 2(x + 10 + x) \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf 140= 2( \underbrace{x +x }+ 10 ) \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf 140= 2( 2x+ 10 ) \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf 140 \div 2= ( 2x+ 10 ) \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf \frac{70 \times 2}{2} = ( 2x+ 10 ) \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf \frac{70 \times \cancel2}{\cancel2} = ( 2x+ 10 ) \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf 70= ( 2x+ 10 ) \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf ( 2x+ 10 ) = 7 0 \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf 2x+ 10 = 7 0 \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf 2x= 7 0 - 10\\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf 2x= 60\\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf x= 60 \div 2\\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf x= \dfrac{60}{2} \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf x= \dfrac{30 \times 2}{2} \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\sf x= \dfrac{30 \times \cancel2}{\cancel2} \\ [/tex]

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \small\bold{ x= 30}[/tex]

[tex] \\ \\ [/tex]

  • Width = x
  • Width = 30 cm

[tex] \\ [/tex]

  • Length = 10 + x
  • Length = 10 + 30
  • Length = 40 cm
Hanimeyaidn

DIMENSIONS

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[tex]\large\sf\underline{Problem:} [/tex]

  • The perimeter of a rectangle is 140 cm. If the length is longer than width by 10 cm, find its dimensions.

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[tex]\large\sf\underline{Answer:}[/tex]

[tex] \qquad \qquad \qquad \huge \rm \: l = 10 \\ \qquad \qquad \qquad\huge \rm \: w = 60[/tex]

==================================

[tex]\large\sf\underline{Solution:}[/tex]

Determine the dimensions of the rectangle using the formula of perimeter.

  • [tex]\bold{Formula\:||\:P = 2L + 2W}[/tex]

suppose the perimeter is 140 square centimeter and the length is 10 centimeter.Fill in the perimeter equation. It will look like this:

  • [tex]\rm 140 = 2(10) + 2(W).[/tex]

Multiply the "2 by 10" on the right side of the equation, and you will now have

  • [tex]\rm 140 = 20 + 2(W). [/tex]
  • [tex]\rm 120 = 2(W) [/tex]
  • [tex]\rm W = 60[/tex]

Therefore,the final answer is Width = 60. So the dimensions of the rectangle are 10 cm for each of the lengths and 60 cm for each of the widths.

To check if the dimensions are true;

  • [tex]\rm P = 2(L) + 2(W) [/tex]
  • [tex]\rm P = 2(10) + 2(60) [/tex]
  • [tex]\rm P = 20 + 120[/tex]
  • [tex]\rm P = 140 [/tex]

Hence, the equation is correct.

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