Question

Frank needs to find the area enclosed by the figure. The figure is made by attaching semicircles to each side of a 54 dash m​-by-54 dash m square. Frank says the area is 1 comma 662.12 msquared. Find the area enclosed by the figure. Use 3.14 for pi. What error might Frank have​ made?

Answer 1

The error made by Frank was that he subtracted the area of the square with that of the area of the four semi-circles instead it has to be added.

Step-by-step explanation:

We are given that the figure is made by attaching semicircles to each side of a 54 dash m​-by-54 dash m square. Frank says the area is 1 comma 662.12 m squared.

We have to find the error made by Frank.

Firstly, as we know that the area of the square is given by;

Area of the square = Side of the square $\times$ Side of the square

In the question; the side of the square given is 54 m and this would also be the diameter of the semicircle attached to each side of a square.

So, the radius of the semicircle =  $\frac{\text{Diameter}}{2}$ = $\frac{54}{2}$ = 27 m

Now, the area of the square = $54 \times 54$ = 2916 $\text{m}^{2}$.

Also, the area of the semi-circle = $\frac{\pi r^{2} }{2}$ =  $\frac{3.14 \times 27^{2} }{2}$ = 1144.53 $\text{m}^{2}$.

As there are a total of 4 semi-circles attached to the square, so the area of all the 4 semi-circles = 4 $\times$ 1144.53 $\text{m}^{2}$ = 4578.12 $\text{m}^{2}$

Now, the total area of the figure = Area of the square + Area of 4 semi-circles

                 = 2916 $\text{m}^{2}$ + 4578.12 $\text{m}^{2}$

                 = 7494.12 $\text{m}^{2}$.

The error made by Frank was that he subtracted the area of the square and the area of 4 semi-circles to find the area of the whole figure as (4578.12 $\text{m}^{2}$ - 2916 $\text{m}^{2}$ = 1662.12 $\text{m}^{2}$).

Answer 2

Answer:

The error made by Frank was that he subtracted the area of the square with that of the area of the four semi-circles instead it has to be added.

Step-by-step explanation:

We are given that the figure is made by attaching semicircles to each side of a 54 dash m-by-54 dash m square. Frank says the area is 1 comma 662.12 m squared.

We have to find the error made by Frank.

Firstly, as we know that the area of the square is given by;

Area of the square = Side of the square \times× Side of the square

In the question; the side of the square given is 54 m and this would also be the diameter of the semicircle attached to each side of a square.

So, the radius of the semicircle = \frac{\text{Diameter}}{2}

2

Diameter

= \frac{54}{2}

2

54

= 27 m

Now, the area of the square = 54 \times 5454×54 = 2916 \text{m}^{2}m

2

.

Also, the area of the semi-circle = \frac{\pi r^{2} }{2}

2

πr

2

= \frac{3.14 \times 27^{2} }{2}

2

3.14×27

2

= 1144.53 \text{m}^{2}m

2

.

As there are a total of 4 semi-circles attached to the square, so the area of all the 4 semi-circles = 4 \times× 1144.53 \text{m}^{2}m

2

= 4578.12 \text{m}^{2}m

2

Now, the total area of the figure = Area of the square + Area of 4 semi-circles

= 2916 \text{m}^{2}m

2

+ 4578.12 \text{m}^{2}m

2

= 7494.12 \text{m}^{2}m

2

.

The error made by Frank was that he subtracted the area of the square and the area of 4 semi-circles to find the area of the whole figure as (4578.12 \text{m}^{2}m

2

- 2916 \text{m}^{2}m

2

= 1662.12 \text{m}^{2}m

2

).

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