# Squares in a Rectangle Tutorial.

## Presentation on theme: "Squares in a Rectangle Tutorial."— Presentation transcript:

Squares in a Rectangle Tutorial

Find the area of the large rectangle.
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 + 1 1 REMEMBER! We need to find the dimensions of the large rectangle since A = bh

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

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

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

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

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

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

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

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

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

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

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

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

We were asked to find the AREA of the large rectangle.
b h Area Formula (rectangle): A = bh 3x + 12 33 x x + 1 1 A = bh (9) 2x + 1 (8) x + 11 b = 33u and h = 32u 2x + 1 A = (33u)(32u) 2x + 18 32 32 4x + 4 x + 11 (2) (3) x + 7 4 A = 1056 u2 x + 3 (1) x + 2 (7) x + 7 x + 3 x + 3 (5) x + 2 (4) 33 3x + 12

This rectangle has been divided into ELEVEN squares.
The “x + 9” PROBLEM: 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. x +9 x + 9 Remember: You should first fully process the “Givens,” x x x Always go with the Givens, and not what you think you “see.” 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!

See me to check your solution.
It’s not this easy, but YOU CAN DO IT!