Section B Section B Checkpoint

Problem 1

  1. Find the exact value of xx.

    A right triangle with hypotenuse x and legs 7 and square root 32.

  2. An envelope measures 3123\frac12 inches tall by 5 inches wide. Kiran wants to use it to mail a really cool pencil to a friend that measures 6 inches long. Will the pencil fit in the envelope? Explain your reasoning.
Show Solution
Solution
  1. x=9x=9
  2. Yes, the pencil should fit in the envelope if it is put in diagonally. Sample reasoning: Since 3.52+52=37.253.5^2 + 5^2=37.25 , the length of the diagonal of the envelope is 37.25≈6.1\sqrt{37.25}\approx6.1, which is a little bit longer than the pencil.
Show Sample Response
Sample Response
  1. x=9x=9
  2. Yes, the pencil should fit in the envelope if it is put in diagonally. Sample reasoning: Since 3.52+52=37.253.5^2 + 5^2=37.25 , the length of the diagonal of the envelope is 37.25≈6.1\sqrt{37.25}\approx6.1, which is a little bit longer than the pencil.

Problem 2

Find the distance between the two points.

Show Solution
Solution
10 units
Show Sample Response
Sample Response
10 units

Problem 3

F
First of two squares of the same area.
First of two squares of the same area. This square is divided into the following: A square with side lengths “a”. Two rectangles with side lengths “a” and “b”. A square with side lengths “b”.

G
Second of two squares of the same area.
Second of two squares of the same area. This square is divided into the following: Four identical triangles on each corner of the square with sides labeled “a” and “b”. A square in the center with unlabeled side lengths.

Complete the explanation for each step of this proof that a2+b2=c2a^2+b^2=c^2, where aa and bb are pieces of the sides of the two identical squares in Figures F and G, and cc is the length of a side of the smaller square in Figure G.

  • Step 1: a2+b2+2aba^2+b^2+2ab represents . . .
  • Step 2: 4⋅12ab+c24\boldcdot \frac12 a b + c^2 represents . . .
  • Step 3: a2+b2+2ab=4⋅12ab+c2a^2+b^2+2ab=4 \boldcdot \frac12 a b + c^2 because . . .
  • Step 4: a2+b2+2ab=2ab+c2a^2+b^2+2ab=2ab+c^2 because . . .
  • Step 5: a2+b2=c2a^2+b^2=c^2 because . . .
Show Solution
Solution

Sample response:

  • Step 1: a2+b2+2aba^2+b^2+2ab represents the total area of the 4 quadrilaterals that make up the larger square in Figure F.
  • Step 2:  4⋅12ab+c24\boldcdot \frac12 a b + c^2 represents the total area of the 4 triangles and the smaller square that make up the larger square in Figure G.
  • Step 3: a2+b2+2ab=4⋅12ab+c2a^2+b^2+2ab=4 \boldcdot \frac12 a b + c^2 because the larger squares in each Figure both have the same total area since they are both squares with side length a+ba+b. 
  • Step 4: a2+b2+2ab=2ab+c2a^2+b^2+2ab=2ab+c^2 because the area of the 4 triangles in Figure G (4⋅12ab4\boldcdot\frac12ab) is equal to the area of the 2 rectangles in Figure F (2ab2ab).
  • Step 5: a2+b2=c2a^2+b^2=c^2 because subtracting an equivalent area from each figure results in an equivalent area remaining. 
Show Sample Response
Sample Response

Sample response:

  • Step 1: a2+b2+2aba^2+b^2+2ab represents the total area of the 4 quadrilaterals that make up the larger square in Figure F.
  • Step 2:  4⋅12ab+c24\boldcdot \frac12 a b + c^2 represents the total area of the 4 triangles and the smaller square that make up the larger square in Figure G.
  • Step 3: a2+b2+2ab=4⋅12ab+c2a^2+b^2+2ab=4 \boldcdot \frac12 a b + c^2 because the larger squares in each Figure both have the same total area since they are both squares with side length a+ba+b. 
  • Step 4: a2+b2+2ab=2ab+c2a^2+b^2+2ab=2ab+c^2 because the area of the 4 triangles in Figure G (4⋅12ab4\boldcdot\frac12ab) is equal to the area of the 2 rectangles in Figure F (2ab2ab).
  • Step 5: a2+b2=c2a^2+b^2=c^2 because subtracting an equivalent area from each figure results in an equivalent area remaining.