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The figure below shows a rectangle that has been divided into nine squares, all of different dimensions. The smallest square has sides 1 unit long, and.

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Presentation on theme: "The figure below shows a rectangle that has been divided into nine squares, all of different dimensions. The smallest square has sides 1 unit long, and."— Presentation transcript:

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2 The figure below shows a rectangle that has been divided into nine squares, all of different dimensions. The smallest square has sides 1 unit long, and the sides of two other squares have been labeled. Find the area of the large rectangle. x x x x x + 1 1 REMEMBER! We need to find the dimensions of the large rectangle since A = bh

3 First, let’s number our squares. Now do you see where the side measure of square number 3 (x + 1) came from? x x x x x + 1 1 The highlighted segment is the SUM of “x” plus “1” (see below) (2)(3) (4) (5) (6) (7) (8) (9) Next, we need to examine the “ Givens ” (1) Square 1: 1 by 1Square 2: x by xSquare 3: (x + 1) by (x + 1) TIP #1: If a polygon is a square, then all sides are congruent. TIP #3: Always look at both sides of a shared segment to find the next measure. TIP #2: If sides are congruent, then they have the same measure

4 Next, we move into square #4. Now you should see that each side measure of square number 4 must be “x + 2” x x x x x + 1 1 The highlighted segment is the SUM of “x + 1” plus “1” (see below) (2)(3) (4) (5) (6) (7) (8) (9) (1) x + 2

5 Next, we move into square # 5. Now you should see that each side measure of square number 5 must be “x + 3” x x x x x + 1 1 The highlighted segment is the SUM of “x + 2” plus “1” (see below) (2)(3) (4) (5) (6) (7) (8) (9) (1) x + 2 x + 3

6 Next, we move into square # 6. From beneath, the measure of this segment is the sum of (x + 3) + 1 x x x x x + 1 1 This is the square that is the “deal maker,” or “deal breaker” !!! (2)(3) (4) (5) (6) (7) (8) (9) (1) x + 2 x + 3 (x + 3)(1)+(x + 4) From above, we know part of the measure of this segment is x. So, (x + 4) – x = 4 The measure of each side of square # 6 is 4! (x) + 4

7 Next, we move into square # 6. x x x x x + 1 1 This is the square that is the “deal maker,” or “deal breaker” !!! (2)(3) (4) (5) 44 (7) (8) (9) (1) x + 2 x + 3 (x + 4) The measure of each side of square # 6 is 4! (x) + 4

8 Next, we move into square # 7. x x x x x + 1 1 Remember to examine the highlighted segment from both sides! (2)(3) (4) (5) 44 (7) (8) (9) (1) x + 2 x + 3 x + 7 Now you should see that each side measure of square number 7 must be (x + 3) + 4 x + 7 For square number 7, each side measures (x + 3 + 4) = x + 7

9 Next, we move into square # 8. x x x x x + 1 1 Remember to examine the highlighted segment from both sides! (2)(3) (4) (5) 44 (7) (8) (9) (1) x + 2 x + 3 x + 7 Now you should see that each side measure of square number 8 must be (x + 7) + 4 x + 7 For square number 8, each side measures (x + 7 + 4) = x + 11

10 x x x x x + 1 1 (2)(3) (4) (5) 44 (7) (8) (9) (1) x + 2 x + 3 x + 7 The sides have measures of x and x + 1, so: (x) + (x + 1) x + 7 x + 11 2x + 1 For square # 9, we simply find the sum of the sides of squares 2 and 3. For square number 9, each side measures (x + x + 1) = 2x + 1

11 x x x x x + 1 1 (2)(3) (4) (5) 44 (7) (8) (9) (1) x + 2 x + 3 x + 7 x + 11 2x + 1 We were asked to find the AREA of the large rectangle. ALSO: Remember an important property of all rectangles is that both pairs of opposite sides are CONGRUENT! Area Formula (rectangle): A = bh Bottom ≅ Top (x + 7) + (x + 3) + (x + 2) Equals (x + 11) + (2x + 1) 3x + 12 = 3x + 12 √ Left ≅ Right (x + 7) + (x + 11) Equals (2x + 1) + (x + 1) + (x + 2) 2x + 18 = 4x + 4 Solve for x! b h 3x + 12 This property of rectangles allows us to write the following equations... Congruent sides  Equal measures

12 x x x x x + 1 1 (2)(3) (4) (5) 44 (7) (8) (9) (1) x + 2 x + 3 x + 7 2x + 18 x + 7 3x + 12 x + 11 2x + 1 4x + 4 2x + 1 3x + 12 We were asked to find the AREA of the large rectangle. ALSO: Remember an important property of all rectangles is that both pairs of opposite sides are CONGRUENT! Area Formula (rectangle): A = bh 2x + 18 = 4x + 4 Solve for x! b h 4x + 4 = 2x + 18 -2x -2x 2x + 4 = 18 - 4 = - 4 2x = 14 x = 7

13 x x x x x + 1 1 (2)(3) (4) (5) 44 (7) (8) (9) (1) x + 2 x + 3 x + 7 2x + 18 x + 7 3x + 12 x + 11 2x + 1 4x + 4 2x + 1 3x + 12 We were asked to find the AREA of the large rectangle. ALSO: Remember an important property of all rectangles is that both pairs of opposite sides are CONGRUENT! Area Formula (rectangle): A = bh b h x = 7 base: 3x + 12 base: 3(7) + 12 base: 21 + 12 base: 33 units 33 33

14 x x x x x + 1 1 (2)(3) (4) (5) 44 (7) (8) (9) (1) x + 2 x + 3 x + 7 2x + 18 x + 7 3x + 12 x + 11 2x + 1 4x + 4 2x + 1 3x + 12 We were asked to find the AREA of the large rectangle. ALSO: Remember an important property of all rectangles is that both pairs of opposite sides are CONGRUENT! Area Formula (rectangle): A = bh b h x = 7 height: 2x + 18 height: 2(7) + 18 height: 14 + 18 height: 32 units 33 height: 4x + 4 height: 4(7) + 4 height: 28 + 4 height: 32 units 3 2 3 2

15 x x x x x + 1 1 (2)(3) (4) (5) 44 (7) (8) (9) (1) x + 2 x + 3 x + 7 2x + 18 x + 7 3x + 12 x + 11 2x + 1 4x + 4 2x + 1 3x + 12 We were asked to find the AREA of the large rectangle. Area Formula (rectangle): A = bh b h 33 32 A = bh b = 33u and h = 32u A = (33u)(32u) A = 1056 u 2

16 x +9 x This rectangle has been divided into ELEVEN squares. 1. Write an equation that will yield the value of x. 2. Solve your equation and write the solution for x. The “x + 9” PROBLEM: Remember: 1)You should first fully process the “Givens,” 2)Always go with the Givens, and not what you think you “see.” 3)Never over-determine the meaning of the Givens. 4) Always look at each shared segment from BOTH sides! 5) Know that you’ll encounter a deal-maker or deal-breaker segment!

17 It’s not this easy, but YOU CAN DO IT!


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