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Geometry Session 6: Solutions
 
Session 6 Part A Part B Part C Homework
 
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Solutions for Session 6, Part B

See solutions for Problems: B1 | B2 | B3 | B4


Problem B1

By construction, all four triangles are right triangles. The longer leg of each triangle has length b (since it equals the length of the side of the larger inscribed square), and the shorter leg of each triangle has length a (since it equals the length of the side of the smaller inscribed square). So all four triangles have two congruent sides as well as the angle between the two sides. So, by SAS congruence, the four triangles are indeed congruent to the original.

<< back to Problem B1


 

Problem B2

The sides of the piece in the center are all equal to c, since they are the longest sides of four triangles that are congruent to the original triangle (see the argument of Problem B1). Now we have to show that the angles are each 90°. Notice that three angles line up to make a straight line, or 180°: the small and the middle angles from the original right triangle (marked with one and two lines, respectively), and the one from the quadrilateral that lies between the two (see picture). Since the three angles of a triangle also add up to 180°, the angle in the quadrilateral must be the same as the largest angle in the triangle -- a right angle. This is true for all four angles in the shape, and so we know that we have a quadrilateral with four congruent sides and four right angles, so it must be a square.

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Problem B3

Here is an interactive illustration that describes the reasoning behind "Behold!"

This interactive illustration requires the Flash plug-in, which you can download for free from Macromedia's Web site. See below for a non-interactive solution.

The larger (shaded) square has side b, so its area is b2 (see picture below). The smaller square has side a, so its area is a2. The pieces that make up all of this are four right triangles with sides a and b, and a small square with side b - a. (It is the difference between the side b square and the smaller side of the triangle.) The same five pieces make up the large square with area c2 in the third picture. So a2 + b2 = c2.

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Problem B4

The picture shows a trapezoid made up of three triangles. The areas of the triangles are ab / 2 (two of them), and c2 / 2. On the other hand, using the formula for the area of the trapezoid, we get that the overall area is (a + b)(a + b) / 2. So:

ab / 2 + ab / 2 + c2 / 2 = (a + b)2 / 2

Multiplying both sides by 2 and expanding the right-hand side, we get the following:

2ab + c2 = a2 + 2ab + b2
or
a2 + b2 = c2

<< back to Problem B4


 

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