Question

Two plots of land one a square with side 40m and another a rectangle with length 55m have the same perimeter which one has the greater area

Answer 1

➲Given that, Two plots of land one a square with side 40m and another a rectangle with length 55m have the same perimeter.

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To Find:

★which has a greater area★

➺Understanding the question :

so here we find the area of the square and rectangle and compare them,but as the breadth of the rectangle is not given so we have to find its breadth on basing the perimeter of the square as they are equal.

Solution:

⇨ given side of the square is 40m

$: \longrightarrow \tt \: perimeter = 4 \times side \\ \\ \\ : \longrightarrow \tt \: primeter = 4 \times 40m \: \: \: \\ \\ \\ : \longrightarrow \tt \: perimeter = 160m \: \: \: \: \: \: \:$

$\sf \underline \orange{ \bigstar \: perimeter \: of \: the \: square = 160m} \:$

now, we have to find the breadth of the rectangle.

⇨ the rectangle has the length 55m

$: \longrightarrow \tt \: perimeter = 2(l + b) \: \: \: \\ \\ \\ : \longrightarrow \tt \: 160m = 2(55 + b) \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \longrightarrow \tt \: 160m = (110 + 2b) \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \longrightarrow \tt 2b = 160 - 110 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \longrightarrow \tt \:2b = 50m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \longrightarrow \tt \: b = \cancel \frac{50}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \longrightarrow \tt \: b = 25m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \underline \pink{ \bigstar \: breadth \: of \: the \: rectangle = 55m}$

➥ now let's find their areas

$\sf \underbrace \red{area \: of \: a \: square = {side}^{2} } \:$

$: \implies \sf \: area = {side}^{2} \: \: \: \: \: \\ \\ \\ : \implies \sf \: area = 1600 {m}^{2}$

$\blue{ \underline{ \boxed{ \pink{ \mathfrak{ \therefore \: area \: of \: the \: square = 1600 {m}^{2} }}}}}$

$\sf \underbrace \purple{area \: of \: a \:rectangle= lenght \times breadth }$

${: \implies \sf \: area \: of \: recrangle = lenght \times breadth} \\ \\ \\ : \implies \sf \: area \: of \: recrangle =1375 {m}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\blue{ \underline{ \boxed{ \pink{ \mathfrak{ \therefore \: area \: of \: recrangle = 1375 {m}^{2} }}}}}$

$\sf \underline \green{hence \: the \: area \: of \: square \: is \: greater \: ^ \bigstar}$