A piece of wire 42 cm long is bent into a rectangle whose length is twice its width. Find the dimensions of the rectangle. A. The rectangle is 12 cm long and 9 cm wide
B. The rectangle is 10 cm long and 11 cm wide.
C. The rectangle is 15 cm long and 6 cm wide.
D. The rectangle is 17 cm long and 3 cm wide.
E. The rectangle is 16 cm long and 5 cm wide.
F. The rectangle is 14 cm long and 7 cm wide.

If the rectangle has a length that is 2x (double) the width, then the answer is F.

Reasoning:
The question states that the total length of wire is 42cm, which is the perimeter of the rectangle. The wire is bent into a rectangle, so we know the total length around the rectangle is 42cm. Therefore, we can use the perimeter formula of a rectangle:
• 1) P=2L+2W [Formula]
• 2) P=2(2W)+2W [Given information]
*In step 2, we simply substituted the length of the rectangle, which is 2x the width.*
• 3) P=4W+2W [Distributive Property of Equality]
• 4) P=6W [Addition of Like Terms]
*This means that the perimeter of the rectangle (42cm) will equal 6 times the width of the rectangle. This simply means that the length of one side is 2x the width, so two sides would be a total of 4x the width, and then we have to add on the both side widths.*

• 5) P=42cm. This is because we know that the total length around the rectangle is 42cm.
• 6) 42=6W [Substitution Property of Equality]

Now, we must solve for W (width) because if we know the width, then we know that the length is just 2x the width and the perimeter is 6x the width.

•7) 42/6=6W/6 [Division Property of Equality]
• 8) 7=W [Defined variables].

So, we know the width of the rectangle is 7cm, and we know that the length is double the width, so:
7(2)=14.
Therefore, the side lengths are 14cm since 7+7=7(2)=14.

And, 6(7)=42. So the answer of F is correct.

Equatorial triangle and obtuce angle