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The width of a rectangle is 2/5 its length. Find the dimensions if the perimeter is 42 meters.

length = 13m, Width = 8m.
not enough information.
length = 15m, Width = 6m.
length =1 11m, Width = 10m.

Respuesta :

SkyCas
The formula to find the perimeter of a rectangle is P= 2(l+w). Substitute what we know in to the equation:

•42=2(l+(2/5l)). We can have width represented at (2/5l) because it’s 2/5 of the length. Now add them together to get a single coefficient for l.
•42= 2(7/5l). Multiply 7/5 by 2.
•42= (14/5l). Multiply both sides by the reciprocal of 14/5, which is 5/14, to get l by itself.
•l= 15.

Now that we know the value of l, we know that the length is 15m. To find the width, we multiply 15 by 2/5, which is 6. Therefore, the third option is correct because the length is 15m and the width is 6m.

Answer:

the dimensions of rectangle are length = 15 m and width = 6 m

Step-by-step explanation:

Given;  The width  of a rectangle is [tex]\frac{2}{5}[/tex] its length i.e, [tex]w=\frac{2}{5}l[/tex]

and perimeter of rectangle is 42 meters.

Perimeter(P)of a rectangle in meters is given by:

[tex]P = 2(l+w)[/tex] ;  where [tex]l[/tex] is the length of the rectangle and w is the width of the rectangle.

then, substitute the value of P =42 m and [tex]w=\frac{2}{5}l[/tex] in above formula we get;

[tex]42 = 2(l+\frac{2}{5}l) =2 \cdot (\frac{7}{5}l)[/tex]

divide both side by 2 we get;

[tex]21=\frac{7}{5}l[/tex]

On Simplify:

[tex]l = \frac{21 \cdot 5}{7} =3 \cdot 5 =15[/tex] m

Now, substitute the value of [tex]l = 15[/tex] m  in [tex]w=\frac{2}{5}l[/tex] to solve for w;

[tex]w= \frac{2}{5} \cdot 15 =2 \cdot 3 =6[/tex] m

Therefore, the dimensions length = 15 m and width = 6 m




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