Finding total area and perimeter

[Gr8fu13]

Find the area and perimeter of the following figure. Your final answer may include the pi.

Here is what I had:


First I needed to find the missing measurements for the triangle so I subtracted 12 from 18 to give us our base of the triangle (6yd). Next I used the Pythagorean Theorem to find the hypotenuse of the triangle.
8^2 + 6^2 = c^2
64 + 36 = c^2
100 = c^2
10 = c

Now I had the measurements for the triangle. Next was to find the radius of the semicircle. Since the measurement of 8yds was going to be the diameter. I divided 8 by 2 to find our radius of 4. We have all our missing measurements to solve now.
12 x 8 = 96yd <--area of rectangle
2(12 + 8) = 40yd <--perimeter of rectangle

1/2 x 6 x 8 = 24yd <--area of triangle
10 + 8 + 6 = 24yd <-- perimeter of triangle

pi x 4^2 = 16piyd^2 <--area of semicircle
pi x 4 = 4pi <-- perimeter of semicircle


We take all our measurements and add them together for the area of the figure:
96 + 24 + 16pi = 136piyd² is the total area of the figure

For the total perimeter I believe it would be a little different than adding them all together because we do not want to include the lines inside the figure. So we would just add the exterior measurements together.

For the total perimeter I believe it would be a little different than adding them all together because we do not want to include the lines inside the figure. So we would just add the exterior measurements together.
10 = 18 = 12 = 4pi = 44piyd^2 total perimeter of the figure


I have total area as 136piyd^2 and total perimeter of 44piyd^2.

Am I totally off on this? I thought it would be asiest to break the shape down and go from there.

 

[Deleted member 4993]

Find the area and perimeter of the following figure. Your final answer may include the pi.

Here is what I had:


First I needed to find the missing measurements for the triangle so I subtracted 12 from 18 to give us our base of the triangle (6yd). Next I used the Pythagorean Theorem to find the hypotenuse of the triangle.
8^2 + 6^2 = c^2
64 + 36 = c^2
100 = c^2
10 = c

Now I had the measurements for the triangle. Next was to find the radius of the semicircle. Since the measurement of 8yds was going to be the diameter. I divided 8 by 2 to find our radius of 4. We have all our missing measurements to solve now.
12 x 8 = 96 sq.yd <--area of rectangle
2(12 + 8) = 40yd <--perimeter of rectangle

1/2 x 6 x 8 = 24 sq.yd <--area of triangle
10 + 8 + 6 = 24yd <-- perimeter of triangle

pi x 4^2 = 16pi yd^2 <--area of semicircle
pi x 4 = 4pi <-- perimeter of semicircle


We take all our measurements and add them together for the area of the figure:
96 + 24 + 16pi = 136piyd² is the total area of the figure <<<< Incorrect - it should be = 120 + 16*π yd²
For the total perimeter I believe it would be a little different than adding them all together because we do not want to include the lines inside the figure. So we would just add the exterior measurements together.

For the total perimeter I believe it would be a little different than adding them all together because we do not want to include the lines inside the figure. So we would just add the exterior measurements together.
10 = 18 = 12 = 4pi = 44piyd^2 total perimeter of the figure<<<< Incorrect - it should be = 40 + 4*π yd


I have total area as 136piyd^2 and total perimeter of 44piyd^2.

Am I totally off on this? I thought it would be asiest to break the shape down and go from there.

Correct your answers as indicated above.

 

[Gr8fu13]

Thank you so much for your feedback I worked the problem over and figured out that I tried to add things together that shouldn't have been. It gave me a completely different answer. I do have one question though. You stated that the total area should have been 120 + 16piyd^2. Wouldn't this include the entire circle and not just the semicircle? Shouldn't it be 120 + 8piyd^2? Just thought I would double check.