Question

a wire of length 80 cm is made into a rectangle.two sides of the rectangle are 10 cm long.all of the wire is used.what is the length of one of the longer sides of the rectangle?​

Answer 1

Given -

• Length of the wire = 80cm
• Breadth of the rectangle = 10cm

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To find -

• Length of the rectangle.

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Solution -

Firstly, we have length of the wire. It is given that whole wire is used to form a rectangle, that means

• Perimeter of the rectangle = 80cm

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Now, we have perimeter and breadth of the rectangle and need to find the length of the rectangle.

$\large{\bf{\longmapsto{\boxed{\pink{Perimeter = 2(l + b)}}}}}$

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$\tt\dashrightarrow{Perimeter = 80}$

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$\tt\dashrightarrow{2(l + b) = 80}$

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$\tt\dashrightarrow{l + 10 = \dfrac{80}{2}}$

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$\tt\dashrightarrow{l + 10 = 40}$

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$\tt\dashrightarrow{l = 40 - 10}$

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$\frak\dashrightarrow{\purple{l = 30}}$

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$\underline{\sf{Thus,\: length\: of\: the\: rectangle\: is\: 30cm.}}$

Answer 2

➤ Given :-

Total measurement of wire :- 80 cm

Shorter side of a rectangle :- 10 cm

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➤ To Find :-

The longer sides of the rectangle formed...

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★ How to do :-

Here, we are given with the total length of a wire i.e, 80 cm. It's bent and made into a rectangle. We are also given with the shorter sides of the rectangle formes as 10 cm. We are asked to find the value of the longer sides of the rectangle. So, here we use the formula of perimetre of the rectangle, in which the total length of wire is the perimetre which is bent into a rectangle. The other concept used here is the transportation of numbers from one hand side to the other which changes the sign of the particular number. So, let's solve!!

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➤ Solution :-

Longer side of the rectangle :-

${\sf \leadsto \underline{\boxed{\sf P = 2 \: (Length + Breadth)}}}$

Substitute the given values.

${\tt 80 = 2 \: (L + 10)}$

Multiply the number 2 with both numbers in bracket.

${\tt \leadsto 80 = 2L + 20}$

Shift the number 80 from LHS to RHS, changing it's sign.

${\tt \leadsto 2L = 80 - 20}$

Subtract the values on RHS.

${\tt \leadsto 2L = 60}$

Shift the number 2 from LHS to RHS, changing it's sign.

${\tt \leadsto L = \dfrac{60}{2}}$

Simplify the fraction to get the value of greater side.

${\tt \leadsto \pink{\underline{\boxed{\sf Length = 30 \: \: cm}}}}$

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Verification :-

${\sf \leadsto 2 \: (Length + Breadth)}$

Substitute the values of length and breadth.

${\tt 2 \: (30 + 10) = 80}$

Add the values in the bracket.

${\tt \leadsto 2 \: (40) = 80}$

Multiply the numbers on LHS.

${\tt \leadsto 80 = 80}$

So,

${\sf \leadsto LHS = RHS}$

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Hence solved !!