Question

calculating the area of an office a new business requires an office space of at least 110,000 square meters. is there enough space in the floor plan shown here? the radius of the semicircle is blank. the area of the office shown in the floor plan blank.

Explanation:

Answer:

The radius of the semicircle is \( 200 \) meters (since the diameter of the semicircle is equal to the side length of the rectangle, which is \( 400 \) meters, so radius \( = \frac{400}{2} = 200 \) meters).

To find the area of the office, we calculate the area of the square, the rectangle, and the semicircle:

1. Area of the square: \( 200 \times 150 = 30,000 \) square meters.
2. Area of the rectangle: \( 400 \times (200 + 150) = 400 \times 350 = 140,000 \) square meters. Wait, no, actually, the height of the rectangle is \( 200 + 150 = 350 \)? Wait, no, looking at the diagram: the square is \( 200 \)m (width) and \( 150 \)m (height). The rectangle next to it has a width of \( 400 - 200 = 200 \)m? Wait, no, the total width is \( 400 \)m, and the square is \( 200 \)m wide, so the rectangle is \( 400 - 200 = 200 \)m wide? Wait, no, maybe I misread. Wait, the diagram shows: the square is \( 200 \)m (width) and \( 150 \)m (height). The tall rectangle has a height of \( 200 + 150 = 350 \)m? No, the \( 200 \)m is labeled next to the tall rectangle, maybe the height of the tall rectangle is \( 200 + 150 = 350 \)? Wait, no, let's re-express:

Wait, the floor plan consists of three parts? No, the square (200m x 150m), the rectangle (let's say width \( w \), height \( h \)), and the semicircle (radius \( r \)). Wait, the total width is \( 400 \)m, and the square is \( 200 \)m wide, so the rectangle's width is \( 400 - 200 = 200 \)m? No, maybe the tall rectangle has a width of \( 200 \)m (since the semicircle's diameter is \( 200 \)m? Wait, no, the semicircle is on top of the tall rectangle, so the diameter of the semicircle is equal to the width of the tall rectangle. Looking at the diagram, the square is \( 200 \)m (width) and \( 150 \)m (height). The tall rectangle is adjacent to the square, with a total width of \( 400 \)m, so the tall rectangle's width is \( 400 - 200 = 200 \)m? No, that can't be. Wait, maybe the tall rectangle has a width of \( 200 \)m (since the semicircle's radius is \( 100 \)m? Wait, no, the problem says "the radius of the semicircle is [ ]". Let's look at the diagram again: the square is \( 200 \)m (width) and \( 150 \)m (height). The tall rectangle is \( 200 \)m (height) and, let's see, the total width is \( 400 \)m, so the tall rectangle's width is \( 400 - 200 = 200 \)m? No, maybe the tall rectangle's width is \( 200 \)m, so the semicircle on top has a diameter of \( 200 \)m, so radius \( 100 \)m? Wait, this is confusing. Wait, maybe the tall rectangle has a height of \( 200 + 150 = 350 \)m? No, the \( 200 \)m is labeled next to the tall rectangle, so maybe the height of the tall rectangle is \( 200 \)m, and the square is \( 150 \)m tall. Wait, let's start over:

The floor plan has three components:

1. A square: \( 200 \)m (width) × \( 150 \)m (height). Area: \( 200 \times 150 = 30,000 \) m².
2. A rectangle: Let's say the width is \( 400 - 200 = 200 \)m (since total width is \( 400 \)m, and square is \( 200 \)m wide), and the height is \( 200 + 150 = 350 \)m? No, the \( 200 \)m is labeled next to the tall rectangle, so maybe the height of the tall rectangle is \( 200 \)m, and the square is \( 150 \)m tall. Wait, the diagram shows: the square is \( 150 \)m (height) and \( 200 \)m (width). The tall rectangle is \( 200 \)m (height) and, let's see, the total width is \( 400 \)m, so the tall rectangle's width is \( 400 - 200 = 200 \)m? No, that would make the tall rectangle \( 200 \)m (width) × \( 200 \)m (height), area \( 40,000 \) m². Then the semicircle: if the tall rectangle's width is \( 200 \)m, then the diameter of the semicircle is \( 200 \)m, so radius \( 100 \)m. Area of semicircle: \( \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (100)^2 = 5000\pi \approx 15,708 \) m². Then total area: \( 30,000 + 40,000 + 15,708 = 85,708 \) m², which is less than \( 110,000 \). But that can't be right. Wait, maybe the tall rectangle's width is \( 200 \)m, and height is \( 200 + 150 = 350 \)m? Then area of rectangle: \( 200 \times 350 = 70,000 \) m². Square: \( 200 \times 150 = 30,000 \) m². Semicircle: diameter \( 200 \)m, radius \( 100 \)m, area \( 5000\pi \approx 15,708 \) m². Total: \( 30,000 + 70,000 + 15,708 = 115,708 \) m², which is more than \( 110,000 \). Ah, that makes sense. So the height of the tall rectangle is \( 200 + 150 = 350 \)m (since the square is \( 150 \)m tall, and the tall rectangle is \( 200 \)m above that? Wait, the diagram has "200 m" labeled next to the tall rectangle, maybe the height of the tall rectangle is \( 200 + 150 = 350 \)m. So:

• Square: \( 200 \times 150 = 30,000 \) m².
• Rectangle: \( 200 \times (200 + 150) = 200 \times 350 = 70,000 \) m².
• Semicircle: diameter \( 200 \)m, so radius \( 100 \)m. Area: \( \frac{1}{2} \pi (100)^2 = 5000\pi \approx 15,708 \) m².

Total area: \( 30,000 + 70,000 + 15,708 = 115,708 \) m², which is more than \( 110,000 \) m². So there is enough space.

Wait, but the first blank is "The radius of the semicircle is [ ]". If the diameter of the semicircle is equal to the width of the tall rectangle, which is \( 200 \)m (since the square is \( 200 \)m wide, and the total width is \( 400 \)m, so the tall rectangle is \( 400 - 200 = 200 \)m wide? No, that would make the diameter \( 200 \)m, radius \( 100 \)m. But maybe the tall rectangle is \( 400 \)m wide? Wait, the diagram shows "400 m" at the bottom, so total width is \( 400 \)m. The square is \( 200 \)m wide, so the tall rectangle is \( 400 \)m wide? No, the square is adjacent to the tall rectangle, so the tall rectangle's width is \( 400 - 200 = 200 \)m? No, that can't be. Wait, maybe the tall rectangle is \( 400 \)m wide, and the square is \( 200 \)m wide, but the square is attached to the tall rectangle, so the total width is \( 400 \)m, meaning the tall rectangle is \( 400 \)m wide. Then the semicircle on top of the tall rectangle has a diameter of \( 400 \)m, so radius \( 200 \)m. Ah, that makes sense! So:

• Square: \( 200 \)m (width) × \( 150 \)m (height) = \( 30,000 \) m².
• Rectangle: \( 400 \)m (width) × \( 200 \)m (height) = \( 80,000 \) m².
• Semicircle: diameter \( 400 \)m, so radius \( 200 \)m. Area: \( \frac{1}{2} \pi (200)^2 = 20,000\pi \approx 62,832 \) m².

Wait, but then the total area would be \( 30,000 + 80,000 + 62,832 = 172,832 \) m², which is way more than \( 110,000 \). But the problem says "the radius of the semicircle is [ ]". Let's look at the diagram again: the tall rectangle has a height of \( 200 \)m (labeled), and the square has a height of \( 150 \)m. The total width is \( 400 \)m. The semicircle is on top of the tall rectangle, so the diameter of the semicircle is equal to the width of the tall rectangle. If the tall rectangle is \( 400 \)m wide, then diameter is \( 400 \)m, radius \( 200 \)m. That must be it. So:

Radius of semicircle: \( 200 \) meters (since diameter is \( 400 \)m, radius \( = 400/2 = 200 \)m).

Area of office:

• Square: \( 200 \times 150 = 30,000 \) m².
• Rectangle: \( 400 \times 200 = 80,000 \) m².
• Semicircle: \( \frac{1}{2} \pi (200)^2 = 20,000\pi \approx 62,832 \) m².

Total area: \( 30,000 + 80,000 + 62,832 = 172,832 \) m², which is more than \( 110,000 \) m². So there is enough space.

So the radius of the semicircle is \( 200 \) meters, and the area of the office is approximately \( 172,832 \) square meters (or exactly \( 110,000 + 20,000\pi \) m², but \( 20,000\pi \approx 62,832 \), so total \( 110,000 + 62,832 = 172,832 \)).

Wait, but maybe the rectangle's height is \( 200 + 150 = 350 \)m? No, the \( 200 \)m is labeled next to the tall rectangle, so the height of the tall rectangle is \( 200 \)m, and the square is \( 150 \)m tall, so the total height of the tall rectangle is \( 200 \)m, and the square is \( 150 \)m tall, attached to the bottom. So the tall rectangle is \( 400 \)m wide (total width) and \( 200 \)m tall, and the square is \( 200 \)m wide and \( 150 \)m tall, attached to the left of the tall rectangle. Then the semicircle is on top of the tall rectangle, with diameter \( 400 \)m, so radius \( 200 \)m. That makes the radius \( 200 \)m.

So the first blank: \( 200 \) meters.

The area of the office: area of square + area of rectangle + area of semicircle.

Square: \( 200 \times 150 = 30,000 \) m².

Rectangle: \( 400 \times 200 = 80,000 \) m².

Semicircle: \( \frac{1}{2} \pi (200)^2 = 20,000\pi \approx 62,832 \) m².

Total area: \( 30,000 + 80,000 + 62,832 = 172,832 \) m² (or \( 110,000 + 20,000\pi \) m², but \( 20,000\pi \approx 62,832 \), so total is \( 172,832 \) m², which is more than \( 110,000 \) m²).

So the answers are:

The radius of the semicircle is \( \boldsymbol{200} \) meters.

The area of the office shown in the floor plan is \( \boldsymbol{172832} \) square meters (or \( 110000 + 20000\pi \) square meters, but approximately \( 172832 \) m²), which is more than \( 110,000 \) m², so there is enough space.