Lesson 6 Plane Geometry Practice Test Answer Explanations

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1 Lesson 6 Plane Geometry Practice Test Answer Explanations Question 1 One revolution is equal to one circumference: C = r = 6 = 1, which is approximately inches. Multiply that by 100 to get 3,768 inches. Then divide by 1 to get 314 feet. The bike rider goes 314 feet. Question Draw the flagpole and its shadow on your scratch paper. 40 x 30 The flagpole is a vertical line measuring 40 feet. Its shadow is perpendicular to the base of the vertical line and measures 30 feet. The rope forms the hypotenuse of the triangle. Its length is the unknown value. To find the value of x, you could use the Pythagorean theorem, but it would be faster to notice that the triangle is a right triangle. The legs are 10 times 3 and 4. Therefore, the length of the hypotenuse (rope) is 10 times 5 or 50 feet. Question 3 The best way to start working on this question is to draw a cylinder on your page. Mark the height as 10, and draw in a line representing the radius of the circle on the top of the cylinder. The question is asking you to figure out the diameter of the can (which we know is just times the radius of the can) knowing that the volume of the can is 50. The formula for the volume of a cylinder is Volume = r h, where r is the radius of the cylinder and h is its height. Let's start by plugging in what we know:

2 = 50 r (10) Because we have only one variable remaining (r), we can solve the equation. Start by dividing both sides by 10: 5 = r Then divide both sides by pi: 5 r = Now take the square root of both sides: 5 = r Because the square root of 5 is 5, we can reduce the left side to: 5 r = So the radius of the can is 5 divided by the square root of pi. To figure out the diameter, simply multiply this radius by : 5 ( ) d = Because times 5 is 10, the left side becomes: 10 d = So the diameter of the cylinder is 10 divided by the square root of pi, which cannot be reduced further. Question 4 This question is asking you to imagine that the bottom of the can you drew for Question 3 has a big hemispheric indentation in the bottom. It may help for you to draw this indentation so you can picture what's being described. (Your picture should now have a half-circular "bubble" coming up from the bottom and into the cylinder.) The question tells you that the indentation is the largest hemispheric indentation possible on a can of this size. We now know (from Question 3) that the radius of the can is 5, so the largest half-

3 sphere we can make with a radius of that size is a sphere with 5 half-sphere's radius. We're on our way. as that The question is asking you to figure out the volume of the cylinder with the indentation. We already know that the volume of the cylinder without the indentation is 50. With the indentation, the cylinder will lose volume. How much volume will it lose? Well, it will lose the volume inside the hemispheric indentation we've drawn. In formula form: Volume of Indented Cylinder = Volume of Regular Cylinder - Volume of Hemisphere or, with numbers: V = 50 - Hemisphere Volume Let's see if we can figure out the hemispheric volume. We know that the 4 3 volume of a full sphere is r. Because a hemisphere is just half of a sphere, the volume of the hemisphere will just be half of the volume of the full sphere. So let's start by figuring out the volume of the full sphere. Once we do that, we'll just divide our result by. Okay, so the volume of a sphere is: 4 3 Sphere Volume = r We know that the radius of our sphere is 5, so the first thing we need to do is cube that number: 5 3 = ( ) (5)(5)(5) = ( )( )( ) = 15 So r 3 in this question is 15 divided by pi times the square root of pi. Let's plug that back into our formula for the volume of a sphere: Sphere Volume = 4 15 ( )

4 The pi in the numerator and the pi in the denominator cancel each other out, so we're left with: Sphere Volume = 4 15 ( ) Multiplying the 4 and 15 in the numerator gives you: Sphere Volume = 500 This fraction cannot be reduced further because 3 does not go evenly into 500. So we know the volume of the full sphere would be 500 divided by 3 times the square root of pi. But we need to know the volume of the half sphere. So let's divide our result by, which is the same as multiplying the fraction by 1/: = 500 ( 1 ) Or: = 500 ()(3)( ) Dividing 500 by reduces the fraction to: = 50 The fraction cannot be reduced further. So the volume of the half sphere (or hemisphere) is 50 divided by 3 times the square root of pi. Plugging this back in for "Hemisphere Volume" in our very first equation above gives us: 50 V = 50

5 Because 50 and 50 do not have a common denominator, they cannot be subtracted as is. So we can either leave our answer in the form above or we can multiply 50 by to give both pieces a common denominator of. If we do the latter, we get: V = (50)(3)( ) 50 Because 50 times 3 is 750, our final answer is: V = (750 ) 50 It's messy, but that's the answer! Because we're dealing with volume, don't 3 forget to add the units at the end, which are cm. Question 5 Fortunately Question 5 is a little easier than Question 4. All we need to do here is figure out how much aluminum is needed to make the indented bottom of the can (which we drew in Question 4) as opposed to the flat bottom of the can (which we drew in Question 3). This question is asking us to think about surface area! Specifically, it's asking us to compare the surface area of the hemisphere we drew in Question 4 to the area of the circle at the bottom of the can we drew in Question 3. No problem. Let's just figure out both areas and then subtract our results. Let's start with the hemisphere. The surface area of the hemisphere will be half of the surface area of the full sphere. We know that the surface area of a sphere is: Surface Area = 4 r Plugging in our radius which is 5 divided by the square root of pi we have: 5 = 4 ( ) Because 5 squared is 5 and the square root of pi squared is just pi, we get: 5 = 4 ( )

6 The pi in the numerator cancels with the pi in the denominator, so we're left with: = 4(5) 4 times 5 is 100, so the surface area of the full sphere would be 100. Since we have only a hemisphere, the surface area is 1/ of 100, or 50. Great! That's a nice, clean number (thank goodness!). Now we just need to figure out the area of the regular circular can bottom we had in Question 3. The area of a circle is: Area = r Again, the radius of our circle is 5 divided by the square root of pi, so let's plug that in: 5 = ( ) Because 5 squared is 5 and the square root of pi squared is just pi, we get: 5 = ( ) The pi in the numerator cancels with the pi in the denominator, so we're left with: = 5 Great! The area of the circular bottom is 5. The question asks us to find the difference between the surface area of the hemispheric bottom and the circular bottom. So we simply subtract: or or Difference = Hemisphere Surface Area - Circle Area Difference = 50-5 Difference = 5 So the answer to Question 5 is 5. Don't forget to add back in the units to make the answer 5cm.

7 Question 6 The question asks for the area of a trapezoid. The formula for the area of a trapezoid is the average of the two bases times the height. One of the bases measures 5 units. The measurement of the other base is the sum of the segments. Because AB and CD are parallel and AC and BD are parallel, CD must equal AB. So segment CD is 5 units. Therefore, the measure of the other base is = 7.5 units. Substitute the values into the formula: A= x 1.5 A= x A= 6.5 x square units The area of ABFE is 6.5x square units.

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