Diagram : a 40cm wire bent to make a closed figure that consists a rectangle and a semicircle. (Kinda like a door with a curved top)
Total perimeter of diagram / length of bent wire (rectangle + semicircle) = $40cm$, Total area inside of diagram / bent wire (rectangle + semicircle) = $105cm^2$.
Width of rectangle = $2x$, Height of rectangle = $y$, Area of rectangle = $(2x)(y)$ = $2xy$.
Find the radius of the semicircle.
note : no pic of diagram because can't post pictures yet
Let "x" be the width of the rectangle and let "y" be the height of the rectangle, in cm. The semicircle at the top has diameter x so radius x/2.
The circumference of the semicircle is $\pi d/2= \pi x/2$ and that is the length of wire needed for that. The rectangle requires x+ 2y cm of wire (since there no top to the rectangle so we have $x+ 2y+ \pi x/2= 40$ cm. The area is $xy+ \pi x^2/4= 105$ square cm. From the first equation $2y= 40- x- \pi x/2$ so $y= 20- x/2- \pi x/4$. Putting that into the equation $x(20- x/2- \pi x/4)+ \pi x^2/4= 20x- x^2/2= 105$. Solve the quadratic equation $x^2- 40x+ 210= 0$.