Composite Shapes
A composite shape is a complex shape that is made up of multiple other basic shapes. Once we simplify the composite shape into the basic shapes, it becomes easy to calculate different geometric quantities.
Perimeter and Area
We can determine the Perimeter by adding all the sides together. Some sides are not labelled and we will need to figure them out with the available information. Remember that the single tick means that all lengths are the same.
For example, the left side is not labelled. However, we can see that the right side is \(3.1 + 3.1 = 6.2\). The left and right sides are equal.
The Perimeter is (left, top, right, bottom):
\(P = 6.2 + 12 + 3.1 + 3.1 + 3.1 + 15.1 \)
\(P = 42.60 \; [\text{cm}]\)
Therefore, we can determine that the Perimeter is \(42.60 \; [\text{cm}]\).
The total Area of this Composite Shape is the sum of the rectangle's Area plus the square's Area:
First, we can determine the Area of the rectangle:
\(A_{\text{rectangle}} = (12)(6.2) = 74.4 \; \text{cm}^2\)
\(A_{\text{rectangle}} = 74.4 \; [\text{cm}^2]\)
Next, we can determine the Area of the square:
\(A_{\text{square}} = (3.1)(3.1)\)
\(A_{\text{square}} = 9.61 \; [\text{cm}^2]\)
Finally. we can determine the total Area by calculating the sum of the individual Areas:
\(A = A_{rectangle} + \text{A}_{square}\)
\(A = 74.4 + 9.61\)
\(A = 84.01 \; [\text{cm}^2]\)
Therefore, we can determine that the Area is \(84.01 \; [\text{cm}^2]\).
Note that there are different ways to solve this problem. For exmaple, you could calculate the Area as the difference between the Area of the big outer rectangle and the Area of the square:
\(A = A_{\text{rectangle}} - \text{A}_{\text{square}}\)
\(A = (15.1)(6.2) - (3.1)(3.1)\)
\(A = 84.01 \; [\text{cm}^2]\)
We can determine the perimeter by adding all side lengths together.
First, we can calculate the circumference (or length) of the semi-circle as such:
\(C = \cfrac{1}{2} (2\pi r)\)
\(C = \cfrac{1}{2}(2\pi \cdot 10)\)
\(C = 31.42 \; [\text{cm}]\)
Now, we can determine the Perimeter by calculating the sum of all the side lengths:
\(P = \text{left} + \text{top} + \text{right} + \text{bottom}\)
\(P = 20 + 40 + 31.40 + 40\)
\(P = 131.42 \; [\text{cm}]\)
Therefore, we can determine that the Perimeter is \(131.42 \; [\text{cm}]\).
The Area of this Composite Shape is the difference between the individual Areas of the large rectangle and the semi-circle.
First, we can determine the Area of the rectangle:
\(A_{\text{rectangle}} = lw\)
\(A_{\text{rectangle}} = (20) (40)\)
\(A_{\text{rectangle}} = 800\;[\text{cm}^2]\)
Next, we can determine the Area of the semi-circle:
\(\text{A}_{\text{semi-circle}} = \cfrac{1}{2} \pi r^2\)
\(\text{A}_{\text{semi-circle}} = \cfrac{1}{2} \pi(10)^2\)
\(\text{A}_{\text{semi-circle}} = 157.08 \; [\text{cm}^2]\)
Finally, we can determine the total Area of the Composite Shape:
\(A = \text{A}_{\text{rectangle}} - \text{A}_{\text{semi-circle}}\)
\(A = 800 - 157.08\)
\(A = 642.92 \; [\text{cm}^2]\)
Therefore, we can determine that the Area of the Composite Shape is \(642.92 \; [\text{cm}^2]\).
Surface Area and Volume
We can determine the Surface Area of the giant podium by combining the Surface Area of the Big Cuboid with the Surface Area of the Small Cuboid. But remember, we have to subtract the base of the Small Cuboid that is overlapped with the top of the Big Cuboid.
First, we can determine the Surface Area of the Big Cuboid:
\(SA_{\text{big}} = 2 ((18)(12) + (18)(4) + (12)(4)) \)
\(SA_{\text{big}} = 672 \; [\text{yd}^2]\)
Next, we can determine the Surface Area of the Small Cuboid::
\(SA_{\text{small}} = 2 ((5)(5) + (5)(12) + (12)(5)) \)
\(SA_{\text{small}} = 290 \; [\text{yd}^2]\)
Then, we can find the Area of the Small Cuboid's base:
\(A = lw\)
\(A_{base} = (5)(12)\)
\(A_{base} = 60\;[\text{yd}^2]\)
Finally, we can determine the total Surface Area. We can do so by finding the sum of the Cuboids' Surface Areas and subtracting the Area of the overlapping space (the base of the Small Cuboid):
\(\text{SA} = SA_{\text{big}} + SA_{\text{small}} - A_{\text{base}}\)
\(SA = 672 + 290 - 60\)
\(SA = 902 \; [\text{yd}^2]\)
Therefore, we can determine that the Surface Area of the podium is \(902 \; [\text{yd}^2]\).
In order to determine the total Volume of the podium, we need to find the sum of the Volumes for the \(2\) Cuboids.
First, we can determine the Volume of the Big Cuboid:
\(V_{\text{big}} = (18)(12)(4) \)
\(V_{\text{big}} = 864 \; [\text{yd}^3]\)
Next, we can determine the Volume of the Small Cuboid:
\(V = l w h\)
\(V_{\text{small}} = (5) (5) (12)\)
\(V_{\text{small}} = 300 \; [\text{yd}^3]\)
Finally, we can determine the Volume of the entire podium:
\(V = V_{\text{big}} + V_{\text{small}}\)
\(V = 864 + 300\)
\(V = 1164 \; [\text{yd}^3]\)
Therefore, we can determine that the Volume of the podium is \(1164 \; [\text{yd}^3]\).
We can determine the Surface Area of the ice cream cone shape by finding the sum of the Surface Areas for the Cone and Hemisphere.
First, we can determine the slanted length of the Cone by using the Pythagorean Theorem. The slant is the hypoteneuse, \(l\) and the other sides of the triangle are the radius, \(r\) and height, \(h\).
As the radius represents half of the diameter, we can calculate it as \(8\).
\( l^2 = h^2 + r^2\)
\(l = \sqrt{(14)^2+(8)^2} \)
\(l = 16.12 \; [\text{in}]\)
Next, we can determine the Surface Area of the Cone:
\(SA_{\text{cone}} = \pi r (l + r)\)
\(SA_{\text{cone}} = \pi \cdot 8 (16.12 + 8)\)
\(SA_{\text{cone}} = \pi \cdot 8(24.12)\)
\(SA_{\text{cone}} = 606.20 \; [\text{in}^2]\)
Then, we can determine the Surface Area of the Hemisphere, which is half that of a regular Sphere:
\(SA_{\text{hemisphere}} = \cfrac{1}{2} 4 \pi r^2 \)
\(SA_{\text{hemisphere}} = 2 \pi (8)^2\)
\(SA_{\text{hemisphere}} = 402.12 \; [\text{in}^2] \)
Finally. we can determine the total Surface Area by combining the individual Surface Areas:
\(SA_{\text{total}} = SA_{\text{cone}} + SA_{\text{hemisphere}} \)
\(SA_{\text{total}} = 606.2 + 402.12\)
\(SA_{\text{total}} = 1008.32 \; [\text{in}^2]\)
Therefore, we can determine that the total Surface Area is \(1008.32 \; [\text{yd}^2]\).
We can determine the Volume of the ice cream cone shape by adding the Volume of the Cone and Hemisphere:
First, we can determine the Volume of the Cone:
\(V_{text{cone}} = \cfrac{1}{3} \pi r^2 h\)
\(V_{text{cone}} = \cfrac{1}{3} \pi (8)^2 (14)\)
\(V_{text{cone}} = 938.29 \; [\text{yd}^3]\)
Next, we can determine the Volume of the Hemisphere, which is half that of a regular sphere:
\(V_{\text{hemisphere}} = \cfrac{1}{2} \cfrac{4}{3} \pi r^3\)
\(V_{\text{hemisphere}} = \cfrac{2}{3} \pi (8)^3\)
\(V_{\text{hemisphere}} = 1072.33 \; [\text{yd}^3]\)
Finally, we can determine the total Volume by combining the individual Volumes:
\(V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}}\)
\(V_{\text{total}} = 938.29 + 1072.33\)
\(V_{\text{total}} = 2010.62 \; [\text{yd}^3]\)
Therefore, we can determine that the total Volume is \(1164 \; [\text{yd}^3]\).