Answer:
Length:8 m
Width:3 m
Step-by-step explanation:
The complete question is
If the perimeter of a rectangle is 22 meters, and the perimeter of a right triangle is 12 meters (the sides of the triangle are half the length of the rectangle, the width of the rectangle, and the hypotenuse is 5 meters). How do you solve for L and W, the dimensions of the rectangle.
step 1
Perimeter of rectangle
we know that
The perimeter of rectangle is equal to
[tex]P=2(L+W)[/tex]
we have
[tex]P=22\ m[/tex]
so
[tex]22=2(L+W)[/tex]
Simplify
[tex]11=L+W[/tex] -----> equation A
step 2
Perimeter of triangle
The perimeter of triangle is equal to
[tex]P=\frac{L}{2}+W+5[/tex]
[tex]P=12\ m[/tex]
so
[tex]12=\frac{L}{2}+W+5[/tex]
Multiply by 2 both sides
[tex]24=L+2W+10[/tex]
[tex]L+2W=14[/tex] ----> equation B
Solve the system of equations by graphing
Remember that the solution is the intersection point both graphs
using a graphing tool
The solution is the point (8,3)
see the attached figure
therefore
The dimensions of the rectangle are
Length:8 m
Width:3 m
