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the rectangle shown has a perimeter of 56 cm and the given area. it's length is 8 more than three times it's width. write and solve a system of equations to find the dimensions of the rectangle.​

The Rectangle Shown Has A Perimeter Of 56 Cm And The Given Area Its Length Is 8 More Than Three Times Its Width Write And Solve A System Of Equations To Find Th class=

Sagot :

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

Let's solve ~

Assume width of rectangle be " x ", length = 3×width + 8 = 3x + 8 ~

Now, Perimeter of rectangle is :

[tex]\qquad \sf  \dashrightarrow \:2(l + w) = 56[/tex]

[tex]\qquad \sf  \dashrightarrow \:2(3x + 8 + x) = 56[/tex]

[tex]\qquad \sf  \dashrightarrow \:2(4x + 8) = 56[/tex]

[tex]\qquad \sf  \dashrightarrow \:4x + 8 = 56 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:4x + 8 = 28[/tex]

[tex]\qquad \sf  \dashrightarrow \:4x = 28 - 8[/tex]

[tex]\qquad \sf  \dashrightarrow \:x = 20\div 4[/tex]

[tex]\qquad \sf  \dashrightarrow \:x =5 \: cm[/tex]

Hence, width = x = 5 cm

[tex]\qquad \sf  \dashrightarrow \:l = 3w + 8[/tex]

[tex]\qquad \sf  \dashrightarrow \:l = 3(5)+ 8[/tex]

[tex]\qquad \sf  \dashrightarrow \:l = 15+ 8[/tex]

[tex]\qquad \sf  \dashrightarrow \:l =23 \:cm[/tex]

And, length = 26 cm

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