At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Ask your questions and receive precise answers from experienced professionals across different disciplines. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Given:
The perimeter of a rectangle is 44 meters.
The length is 10 meters more than twice the width.
Required:
To find the dimensions.
Explanation:
The perimetero of rectangle formula is
[tex]P=2(L+W)[/tex]Let x be the width.
The length is 10 meters more than twice the width.
So
[tex]L=2x+10[/tex]Now
[tex]\begin{gathered} P=2(2x+10+x) \\ \\ 44=4x+20+2x \\ \\ 44=6x+20 \\ \\ 6x=44-20 \\ \\ 6x=24 \\ \\ x=\frac{24}{6} \\ \\ x=4m \end{gathered}[/tex]Here the width is 4m.
Now the length is
[tex]\begin{gathered} L=2(4)+10 \\ =8+10 \\ =18m \end{gathered}[/tex]Final Answer:
18 meters and 4 meters.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
