Corridor of Comprehensive Knowledge

Corridor of Comprehensive Knowledge

Geometry 1

Geometry 1 (Pythagoras theorem, Circumference and area of circles, and Volume)

This scenario is a fun way for school age students to learn more about Pythagoras theorem, Circumference and area of circles, and cylinder and spheroid volumes. The curriculum included in this document is suitable for an hour workshop with grade 4-7 participants. The students receive a square board and four of the same right triangles with legs a and b and hypotenuse c. The length of each edge of the board is a+b.

They make the following two arrangements.

Because the blue area is the same for both arrangements, the area of the red surfaces are also equal. In the first arrangement this area is equal to c^{2}. In the second arrangement the red area is equal to a^{2}+b^{2}. Hence:
c^{2}=a^{2}+b^{2}
The above is a very simple proof for the Pythagoras theorem. Based on the Pythagoras theorem we may develop various problems.
Problem 1: In the following figure, the length of the sheep leash is equal to 3.2 m. Does the sheep have access to the grass?

In the next activities of this scenario we focus on the circumference and area of a circle and volume of a spheroid. Each pair of students receives four cylinders on a board as shown in the figure, thread and scissors.

عكس 2 سناريو هندسه

Here we have


\frac{R_{1}}{R_{2}}=\frac{4}{3}

\frac{R_{1}}{R_{3}}=\frac{3}{2}

\frac{R_{1}}{R_{4}}=2

The teacher asks the students to cut four pieces of the thread in such a way that each piece completely fits the circumference of its corresponding base circle.

عكس 3 سناريو هندسه

If L_{i} shows the length of the ith thread (i=1,2,3,4), which is related to the cylinder i,

then, it can be seen that

L_{2}=\frac{4}{3}L_{1}

L_{3}=\frac{3}{2}L_{1}

L_{4}=2L_{1}

The above experiment justifies that:
The ratio of the circumference to radius of each circle is a constant number. It has been shown that the constant ratio is 2\pi, where 3.14<\pi<3.15. The monitor shows an approximate value of π. How many numbers do you see after the decimal point?!!

Hence, the circumference of a circle with radius R is equal to 2\pi R. Based on this relation, let us consider the following problem.
Problem 2: You have a very long yellow thread that can completely roll around the earth! (See the following figure.) How longer the thread must be so that when role this longer thread (the red one in the figure) around the earth, there exists one meter space between it and the yellow thread?

While it seems that the answer is a very large number, the correct answer is 2\pi \approx 6.28 m. In the next activity of this scenario, we consider the concept of the area of circle. Each student will receive a wooden disk, which looks like inside a watermelon and two other similar disks, which one of them is partitioned into eight and the other into sixteen equal parts. The aim of this activity is to find the relation between the area and circumference of the disk.

The following alignments of the disks parts (has been done by each student) illustrate why the area of a circle (and therefore our disk) is given by multiplication of its radius and half of its perimeter (R*(\pi R))

Now each student receives a sheet of paper. The student, then will be asked to cut a square out of the sheet in such a way that the watermelon disk will be completely covered when the square is cut into different pieces with no leaved part of it.

The goal of this activity is to teach the students the concept of spherical volume. Each pair of students receives a spherical container with radius R=5 cm, a cylindrical container of radius 5 cm and height 5 cm and a dropper.

With each drop, one cc of water comes out of the dropper.
The question that could be asked is: How many times in dropping is needed to fill the spherical container?
What is the height of water of cylindrical container if the water in the spherical container is poured in it? Based on the result what is the volume of a spherical as a function of its radius?

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