Difference between revisions of "Egyptian Geometry Exploration"
From EscherMath
Jump to navigationJump to searchm (Egyptian Geometry Eploration moved to Egyptian Geometry Exploration: typo) 
(Making into a list of problems. Also testing.) 

(2 intermediate revisions by 2 users not shown)  
Line 3:  Line 3:  
{{objectiveExplore some of the geometry shown in the Rhind and Moscow papyri.}}  {{objectiveExplore some of the geometry shown in the Rhind and Moscow papyri.}}  
−  +  <ol>  
+  <li>In problem 6 of the Moscow Papyrus the sides of a rectangular region are computed, given the area. <br>  
a. What would the Egyptian solution be if the area enclosed 15 units? Write out the solution as they would have.<br>  a. What would the Egyptian solution be if the area enclosed 15 units? Write out the solution as they would have.<br>  
b. Using these two examples (i.e the one from ancient Egypt and the example you just did) explain how the lengths of the sides are computed. <br>  b. Using these two examples (i.e the one from ancient Egypt and the example you just did) explain how the lengths of the sides are computed. <br>  
Line 9:  Line 10:  
d. Are the solutions basically the same? Different?<br>  d. Are the solutions basically the same? Different?<br>  
e. Which explanation do you find easier to follow? The Egyptian version or the American version? <br>  e. Which explanation do you find easier to follow? The Egyptian version or the American version? <br>  
−  +  </li>  
−  +  <li>  
+  In problem 14 of the Moscow papyrus the volume of a frustrum (truncated pyramid) is computed.<br>  
a. Using the Egyptian method find the volume if the base is a square of side 2 cubits, the top is a square of side 3 cubits and the height of the truncated pyramid is 5 cubits. <br>  a. Using the Egyptian method find the volume if the base is a square of side 2 cubits, the top is a square of side 3 cubits and the height of the truncated pyramid is 5 cubits. <br>  
b. Using the Egyptian method find the volume if the base is a square of side x cubits, the top is a square of side y cubits and the height of the truncated pyramid is z cubits. <br>  b. Using the Egyptian method find the volume if the base is a square of side x cubits, the top is a square of side y cubits and the height of the truncated pyramid is z cubits. <br>  
+  </li>  
+  </ol>  
+  {{handinA sheet with answers to the questions.}}  
−  +  [[category:Egyptian Mathematics Explorations]] 
Latest revision as of 18:24, 11 January 2017
Objective: Explore some of the geometry shown in the Rhind and Moscow papyri.
 In problem 6 of the Moscow Papyrus the sides of a rectangular region are computed, given the area.
a. What would the Egyptian solution be if the area enclosed 15 units? Write out the solution as they would have.
b. Using these two examples (i.e the one from ancient Egypt and the example you just did) explain how the lengths of the sides are computed.
c. Write out a modern American solution of the problem using the variables x and y for the lengths of the sides. Write out a detailed solution showing all your work.
d. Are the solutions basically the same? Different?
e. Which explanation do you find easier to follow? The Egyptian version or the American version?

In problem 14 of the Moscow papyrus the volume of a frustrum (truncated pyramid) is computed.
a. Using the Egyptian method find the volume if the base is a square of side 2 cubits, the top is a square of side 3 cubits and the height of the truncated pyramid is 5 cubits.
b. Using the Egyptian method find the volume if the base is a square of side x cubits, the top is a square of side y cubits and the height of the truncated pyramid is z cubits.
Handin: A sheet with answers to the questions.