Question

a wire of 160m in the shape of a square was rebent into a rectangle of width 60m. which has got more area and by how much​

Answer 1

Answer :

• A wire of 160m in the shape of a square.
• The wire is rebent into a rectangle of width 60 m.

We have to find which has got more area and by how much.

◗ According to the Question Now :

⇒Perimeter of square = Perimeter of rectangle

⇒4 × side = 2(Length + Breadth)

⇒4 × 160 = 2(L + 60)

⇒640 = 2L + 120

⇒640 - 120 = 2L

⇒520 = 2L

⇒L = 520 ÷ 2

⇒L = 260 m

⛬ The Length of the rectangle wire is 260 m.

◦ VERIFICATION ◦

⇢ 4 × 160 = 2(L + 60)

⇢ 640 = 2(260 + 60)

⇢ 640 = 2(320)

⇢ 640 = 640

▵ HENCE VERIFIED ▵

◈ Finding area of square :

⇸ Area of square = (Side)²

⇸ Area of square = (160)²

⇸ Area of square = 25,600 m²

◈ Finding area of rectangle :

⇸ Area of rectangle = Length × Breadth

⇸ Area of rectangle = 260 × 60

⇸ Area of rectangle = 15,600 m²

◈ DIFFERENCE :

⇸Area of square - Area of rectangle

⇸25,600 - 15,600

⇸10,000 m²

• Hence,the area of square is greater than area of rectangle by 10,000 m².

Answer 2

Question :-

• a wire of 160m in the shape of a square was rebent into a rectangle of width 60m. which has got more area and by how much

Given :-

• A wire of 160m and its shape is Square.
• When it was rebent into a rectangle it's, width was 60m

To Find :-

• Which has got more area and by how much ?

Solution :-

According to the Question :-

${:} \longrightarrow \sf\boxed{\bf{Perimeter\: of \: Square \: = \: Perimeter\: of \: Rectangle }}$

${:} \longrightarrow \sf{\sf{ 4 \times 160 \: = \: 2(length \: + \: 60) }}$

${:} \longrightarrow \sf{\sf{ 640 \: = \: 2Length \: + \: 120 }}$

${:} \longrightarrow \sf{\sf{ 640 \:- \: 120 \: = \: 2Length \: }}$

${:} \longrightarrow \sf{\sf{ 520 \: = \: 2Length \: }}$

${:} \longrightarrow \sf{\sf{ \cancel\dfrac{520}{2} \: = \: Length \: }}$

${:} \longrightarrow \sf{\bf{ 260m \: = \: Length \: }}$

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$\longmapsto \sf\boxed{\bf{Area \: of \: Square \: = \: {(Side)}^{2} }}$

$\longmapsto \sf{\sf{Area \: of \: Square \: = \: {160}^{2} }}$

${:}\longrightarrow \sf{\tt{Area \: of \: Square \: = \: 25600\: m^{2} }}$

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$\longmapsto \sf\boxed{\bf{Area \: of \: Rectangle \: = \: Length \: \times \: Breadth }}$

$\longmapsto \sf{\sf{Area \: of \: Rectangle \: = \: 260 \: \times \: 60 }}$

${:}\longrightarrow \sf{\tt{Area \: of \: Rectangle \: = \: 15600\: m^{2} }}$

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If can be seen that Area of Square is more then the Area of Rectangle . Now , let's minus the Area of Rectangle from Area of Square to find how much longer is the area.

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${:} \longrightarrow \sf\boxed{\bf{Area \: of \: Square\: - \: Area \: of \: Rectangle }}$

${:} \longrightarrow \sf{\sf{25600\: - \: 15600 }}$

$\longmapsto \sf{\tt{ 10000\: m^{2}}}$

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$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ Area \: of \: Square\: is \: larger \: by \: \underline {\underline{ 10000\: m^{2}}}}\\\end{gathered}\end{gathered}$

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