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The area of the logo is the sum of the areas of the involved figures. Then, start by recalling the formulas for finding the area of the corresponding figures.

Figure	Area's Formula
Rectangle	$A=\ell \cdot w$
Trapezoid	$A=\frac{(b_1+b_2)h}{2}$
Semicircle	$A=\frac{\pi r^2}{2}$

Next, find the area of each of the pieces that make up the logo.

Figure	Dimensions (cm)	Area $({\text{cm}}^2)$
Green rectangle	$\ell =40$ $w=20$	$A_1=40\cdot 20=800$
Purple trapezoid	$b_1=100$ $b_2=20$ $h=30$	$A_2=\frac{(100+20)30}{2}=1800$
Gray trapezoid	$b_1=180$ $b_2=100$ $h=30$	$A_3=\frac{(180+100)30}{2}=4200$
Gray rectangle	$\ell =100$ $w=10$	$A_4=100\cdot 10=1000$
Red semicircle	$r=20$	$A_5=\frac{\pi (20{)}^2}{2}\approx 628.32$

Since all the figures but the green rectangle appear twice in the logo, the corresponding areas must be multiplied by $2.$ Finally, choose the appropriate conversion factor to convert the area from square centimeters to square meters. Recall that $1$ square meter is the same as $10000$ square centimeters. The gym's logo has an area of about $1.6$ square meters. Notice that all the slant segments have the same length because each of them is the hypotenuse of a right triangle with legs measuring $40$ and $30$ centimeters. Therefore, their lengths can be found by using the Pythagorean Theorem. The length of each slant segment is $50$ centimeters. The length of each semicircle is $\pi$ times the radius. Since the radius is $20$ centimeters, the length of each semicircle is $20\pi$ centimeters. Finally, all the side lengths are added. There are six segments of $40$ centimeters, eight segments of $50$ centimeters, four segments of $10$ centimeters, four segments of $30$ centimeters, and two semicircles of $20\pi$ centimeters. The perimeter of the logo, rounded to the nearest integer, is $926$ centimeters.

Notice that the three rooms together form a composite figure that is shaped as two overlapping rectangles.

The area of the entire floor is then the sum of the areas of the big rectangles minus the area of the locker room. The area of the locker room must be subtracted because it is common for both rectangles.

Room	Figure	Dimensions (ft)	Area $({\text{ft}}^2)$
Spinning	Upper Rectangle	$\ell =50$ $w=45$	$$\begin{array}{c}{A}_1=50\cdot 45=2250\end{array}$$
Weight	Lower Rectangle	$\ell =55$ $w=40$	$$\begin{array}{c}{A}_2=55\cdot 40=2200\end{array}$$
Locker	Small Rectangle	$\ell =55-35=20$ $w=40-25=15$	$$\begin{array}{c}{A}_3=20\cdot 15=300\end{array}$$

Finally, find the area of the entire floor. The gym owner has to buy $4150$ square feet of rubber matting to cover the entire floor.

The front view of the kettlebell can be separated into two parts — the body and the handle. The body is a regular octagon and the handle looks like half a ring. The kettlebell's front view is then a composite figure whose area is the sum of these two areas.

Area of the Octagon

The area of a regular polygon is half the perimeter times the apothem.

The perimeter of a regular polygon is equal to the side length times the number of sides. In this case, the number of sides is $8$ and all of them have a length of $13.5$ centimeters. The apothem of a regular polygon is the distance from the center of the polygon to any of its sides. From the given diagram, the apothem of the kettlebell's front view is $16.3$ centimeters. The area of the regular octagon is $880.2$ square centimeters.

Area of the Handle

The handle has the shape of half a ring which is what is left when a small semicircle is removed from a large semicircle. The inner semicircle has a diameter of $12$ centimeters so its radius is $r_1=6$ centimeters. The outer semicircle has a diameter of $13.5$ centimeters so its radius is $r_2=6.75$ centimeters. The area of a semicircle is half the product of $\pi$ and the radius squared. Use this formula to find the area of the semicircles.

Semicircle	Radius (cm)	$A=\frac{1}{2}\pi r^2$	Area $({\text{cm}}^2)$
Outer	$6.75$	$A_1=\frac{1}{2}\pi (6.75{)}^2$	$A_1=71.57$
Inner	$6$	$A_2=\frac{1}{2}\pi (6{)}^2$	$A_2=56.55$

The area of the handle is the difference between the area of the outer semicircle and the area of the inner semicircle. The area of the front view of the handle is $15.02$ square centimeters.

Area of the Kettlebell's Front View

Finally, add the areas of the regular octagon and the area of the handle. The area of the front view of the kettlebell is about $895$ square centimeters.

The first step involves finding the perimeter of the entire floor. After that, the area of the floor will be calculated.

Perimeter of The Floor

Start by identifying the coordinates of the vertices of the blueprint and the lengths of the horizontal and vertical sides. Let $m$ be the length of the slant side of the entrance.

The value of $m$ can be calculated by substituting $(\text{-}4,1)$ and $(0,4)$ into the Distance Formula. The slant side is $5$ units long. The length of the curved wall at the back of the stage is the length of a semicircle whose radius is $3$ units. The length of a semicircle is $\pi$ times the radius. The wall at the back of the stage is about $9.42$ units long. By adding the lengths of the exterior sides of the blueprint, the perimeter of the entire floor can be calculated. The perimeter of the entire floor in the blueprint is $43.42$ units. Convert this to meters by using the fact that every unit in the blueprint is the same as $4$ meters. The entire floor has a perimeter of about $174$ meters.

Area of The Floor

Notice that the entire floor can be broken down into some known figures.

The area of the floor is the sum of the areas of the involved figures.

Place	Figure	Dimensions	Area
Entrance and locker room	Trapezoid	$b_1=6$ $b_2=3$ $h=4$	$A_1=\frac{(6+3)4}{2}=18$
Seating area, hall, and boxing room	Rectangle	$\ell =8$ $w=6$	$A_2=8\cdot 6=48$
Multipurpose room without the stage	Rectangle	$\ell =6$ $w=3$	$A_3=6\cdot 3=18$
Stage	Semicircle	$r=3$	$A_4=\frac{1}{2}\pi (3{)}^2\approx 14.14$

Next, add these four areas. The area of the entire floor in the blueprint is about $98.14$ square units. Finally, convert this to square meters. The entire floor has an area of about $1570$ square meters.