Area and Volume of Complex Objects In this lesson we will look at finding the area and volume of objects that are created either by putting together simple objects or by cutting out one simple object from another. It is assumed that you are already familiar with the geometry formulas associated with rectangles, circles, and triangles. If you need some review of this, go to the lesson on basic shapes. The strategy that we will take to find the area of these complex objects is the following. Strategy to Find the Area of Complex Objects
Example 1 Find the area of the figure shown below
Solution
Now try one by yourself. If you want to see the answer, put your mouse on the yellow rectangle and the answer will appear. Exercise 1 Find the area of the region shown below. (Hint: The area of a right triangle with base b and height h is A = 1/2 bh).
Answer
Example 2 A square with side length 7 is inside in a circle of diameter 10. Find the area of the part of the circle that does not contain the square.
Solution
Now try one by yourself. If you want to see the answer, put your mouse on the yellow rectangle and the answer will appear. Exercise 2 A circle of radius 2.8 is removed from the triangle shown below. Find the remaining area. (Round your answer to one decimal place.
Answer
Volume We will use a similar strategy to find volumes of three dimensional shapes. Here are the steps to find such volumes. Steps for finding the volume of complex solids
Example 3 A building is constructed by topping a cylindrical can with height 20 feet and base radius 10 feet with a hemispherical solid as shown in the figure below. Find the approximate volume of this building. Recall that the volume of a hemisphere is A = 2/3 p r^{3} and the volume of a cylinder is A = p r^{2}h. Use 3.14 for p and 0.67 for 2/3. Round your answer to the nearest whole number.
Solution
≈ (0.67)(3.14) (10^{3}) = (0.67)(3.14)(1000) ≈ 2104
Now try one by yourself. If you want to see the answer, put your mouse on the yellow rectangle and the answer will appear. Exercise 3 A tower is formed by attaching a cone of height
30 feet onto a cylinder that has radius
40 feet and height 50
feet as shown below. Find the volume of the tower. The volume of a
cone is Use 3.14 for p and round your answer to the nearest whole number.
Answer
Surface Area There are many different types of surface area problems. We will focus on finding the surface area when the solid is composed of cubes that all have the same side length. To solve such a surface area problem, find the area of the top and bottom, left and right, and front and back. Then add the areas to get the total area. Example 4 One inch cubes are stacked as shown in the drawing below. What is the total surface area?
Solution We will first find the area of the top and the bottom. Notice that these areas are the same. Looking at the top from above (or the bottom from below), we can see that this is just a rectangle.
The rectangle has base 5 and height 4. The area is Area of Top = (5)(4) = 20 The area of the bottom is also 20. Now let's find the area of the left and right side. Looking from the right, we also see a rectangle. The rectangle's base is 4 and height is 2.
Its area is Area of Left Side = (4)(2) = 8 The area of the right side is also 8. Next we find the area of the front. To find this area, we can just count the squares. There are 7 squares.
Area of the Front = 7 the area of the back is the same as the area of the front. That is, the area of the back is also 7. We could have also arrive at an area of 7 by cutting the figure into two rectangles and adding the areas of these rectangles. To find the total area, we add up the areas of all of the sides. Total Surface Area = 20 + 20 + 8 + 8 + 7 + 7 = 70 The total surface area is 70 square inches. Exercise 4 One inch cubes are stacked as shown in the drawing below. What is the total surface area?
Answer

