Egyptian geometry

Egyptian geometry is the way people in Ancient Egypt used math to measure shapes and spaces. They needed this math because the Nile river flooded every year and changed the shape of fields.

We learn about Egyptian geometry from two important papers: the Moscow Mathematical Papyrus and the Rhind Mathematical Papyrus. These papers show that Egyptians could find the area of shapes and the space inside cylinders and pyramids. This helped them take care of their land and build big buildings.

Area

The ancient Egyptians solved problems about shapes and sizes. They wrote their problems in steps to find answers without using letters for numbers.

They measured lengths using a unit called a cubit, which was based on the length of an arm. Special rods helped them make sure everyone used the same sizes.

They knew how to find the size of triangles by using half the base times the height. For rectangles, they could find the area using the length times the width.

They also worked with circles, using special shapes to guess the area. For a circle, they often used the area of a square instead, which gave them a close answer.

Areas

Object	Source	Formula (using modern notation)
triangle	Problem 51 in RMP and problems 4, 7 and 17 in MMP	A = 1 2 b h {\displaystyle A={\frac {1}{2}}bh} b = base, h = height
rectangles	Problem 49 in RMP and problems 6 in MMP and Lahun LV.4. problem 1	A = b h {\displaystyle A=bh} b = base, h = height
circle	Problems 51 in RMP and problems 4, 7 and 17 in MMP	A = 1 4 ( 256 81 ) d 2 {\displaystyle A={\frac {1}{4}}({\frac {256}{81}})d^{2}} d= diameter. This uses the value 256/81 = 3.16049... for π = 3.14159... {\displaystyle \pi =3.14159...}
hemisphere	Problem 10 in MMP

Volumes

Some problems help us find out how much space is inside round grain stores. Problem 60 looks at a pillar or a cone, not a pyramid. It is small and has a steep slope.

One problem in the Lahun Mathematical Papyri shows how to find the space inside a grain store with a round bottom. We see a similar way in the Rhind papyrus (problem 43). Problems in the Moscow Mathematical Papyrus (problem 14) and in the Rhind Mathematical Papyrus (numbers 44, 45, 46) find the space inside grain stores with square or rectangular sides.

Problem 14 of the Moscow Mathematical Papyrus finds the space inside a cut-off pyramid.

Volumes

Object	Source	Formula (using modern notation)
Cylindrical granaries	RMP 41	V = 256 81 r 2   h {\displaystyle V={\frac {256}{81}}r^{2}\ h} measured in cubic-cubits
Cylindrical granaries	RMP 42, Lahun IV.3	V = 32 27 d 2   h = 128 27 r 2   h {\displaystyle V={\frac {32}{27}}d^{2}\ h={\frac {128}{27}}r^{2}\ h} (measured in khar).
Rectangular granaries	RMP 44-46 and MMP 14	V = w   l   h {\displaystyle V=w\ l\ h} w = width, l = length, h = height
Truncated pyramid (frustum)	MMP 14	V = 1 3 ( a 2 + a b + b 2 ) h {\displaystyle V={\frac {1}{3}}(a^{2}+ab+b^{2})h}

Seked

Problem 56 of the RMP shows that ancient Egyptians knew about geometric similarity. This problem talks about the ratio of run to rise, called the seked. This was important for building pyramids. In Problem 57, the height of a pyramid is found using the base length and the seked. Problem 58 uses the base length and height to find the seked.

In Problem 59, part 1 finds the seked. The second part checks the answer: If you build a pyramid with a base side of 12 cubits and a seked of 5 palms and 1 finger, what is its height?