Answer:
[tex]L = 8 + 2w[/tex]
[tex]58=2(L+w)[/tex]
The length of the rectangle is 22 cm and width is 7 cm.
Step-by-step explanation:
Let the width of the rectangle be 'w'.
Given:
Perimeter (P) = 58 cm
Area of the rectangle (A) = 154 cm²
As per question:
Length = 8 more than twice its width
[tex]L = 8 + 2w----- 1[/tex]
Now, perimeter is given as:
[tex]P=2(L+w)\\\\58=2(L+w)\\\\29=L+w----- 2[/tex]
Now, we have to solve the given system of equations.
For that, we plug in 'L' from equation 1 to equation 2. This gives,
[tex]29=8+2w+w\\\\29=8+3w\\\\29-8=3w\\\\21=3w\\\\3w=21\\\\w=\frac{21}{3}=7\ cm[/tex]
[tex]L=8+2w\\\\L=8+2(7)=8+14=22\ cm[/tex]
Therefore, the length of the rectangle is 22 cm and width is 7 cm.
Area = Length × Width = 22 × 7 = 154 cm².
