# Math 6: Unit 7: Geometry Notes 2-Dimensional Figures

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1 Math 6: Unit 7: Geometry Notes -Dimensional Figures Prep for 6.G.A.1 Classifying Polygons A polygon is defined as a closed geometric figure formed by connecting line segments endpoint to endpoint. Polygons Not Polygons Polygons are named by the number of sides. We know a triangle has 3 sides. Prep for 6.G.A.1 Classifying Triangles Triangles can be classified by the measures of their angles: acute triangle 3 acute angles right triangle 1 right angle obtuse triangle 1 obtuse angle Example: Classify each triangle by their angle measure: M A 0 D G E B C F H J K Acute (Equiangular) Right Obtuse Acute L Triangles can also be classified by the lengths of their sides. You can show tick marks to show congruent sides. equilateral triangle 3 congruent sides isosceles triangle at least congruent sides scalene triangle no congruent sides Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 1 of 35

2 equilateral isosceles scalene Example: Classify the triangle. The perimeter of the triangle is 15 cm. 10 cm Using the information given regarding the perimeter:.5 cm x = 15 x x = 15 x =.5 Since sides are congruent, the triangle is isosceles. A tree diagram could also be used to show the triangle relationships. Tree Diagram for Triangles triangles acute obtuse right scalene isosceles scalene isosceles scalene isosceles equilateral Regular polygons have all sides and all angles congruent. Also in regular polygons, there is a point (its center) which is equidistant from all of its vertices. Which of the triangles we have identified so far, would be a regular triangle? Equilateral Triangles Prep for 6.G.A.1 Classifying Quadrilaterals A quadrilateral is a plane figure with four sides and four angles. They are classified based on congruent sides, parallel sides and right angles. Quadrilateral Type Definition Example >> Parallelogram Quadrilateral with both pairs of opposite sides parallel. >> Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page of 35

3 Rhombus Parallelogram with four congruent sides. Note: This polygon is a parallelogram. Rectangle Parallelogram with four right angles. Note: This polygon is a parallelogram. Square Parallelogram with four right angles and four congruent sides. Note: This polygon is a parallelogram. Trapezoid Quadrilateral with exactly one pair of parallel sides. >> >> Another way to show the relationship of the parallelograms is to complete a Venn diagram as shown below. Parallelograms Rectangles Squares Rhombi Which of the polygons we have identified so far, would be a regular polygons? Equilateral triangles and squares Example: A quadrilateral has both pairs of opposite sides parallel. One set of opposite angles are congruent and acute. The other set of angles is congruent and obtuse. All four sides are NOT congruent. Which name below best classifies this figure? A. parallelogram B. rectangle C. rhombus D. trapezoid Vocabulary becomes very important when trying to solve word problems about quadrilaterals. Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 3 of 35

4 We have both pairs of opposite sides parallel, so it cannot be the trapezoid. Since the angles are not 90 in measure, we can rule out the rectangle. We are told that the 4 sides are not congruent, so it cannot be the rhombus. Therefore, we have a parallelogram. (A) Classifying Other Polygons Polygons Pentagon Hexagon Heptagon Octagon Nonagon Decagon # of sides Regular Irregular Regular pentagon An example of an irregular pentagon Regular hexagon An example of an irregular hexagon Regular heptagon An example of an irregular heptagon Regular octagon An example of a irregular octagon Regular nonagon An example of a irregular nonagon Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 4 of 35

5 Regular decagon An example of a irregular decagon 6.G.A.1 Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems. Area of Triangles and Quadrilaterals One way to describe the size of a room is by naming its dimensions. A room that measures 1 ft. by 10 ft. would be described by saying it s a 1 by 10 foot room. That s easy enough. There is nothing wrong with that description. In geometry, rather than talking about a room, we might talk about the size of a rectangular region. For instance, let s say I have a closet with dimensions feet by 6 feet (sometimes given as 6). That s the size of the closet. ft. 6 ft. Someone else might choose to describe the closet by determining how many one foot by one foot tiles it would take to cover the floor. To demonstrate, let me divide that closet into one foot squares. ft. By simply counting the number of squares that fit inside that region, we find there are 1 squares. If I continue making rectangles of different dimensions, I would be able to describe their size by those dimensions, or I could mark off units and determine how many equally sized squares can be made. Rather than describing the rectangle by its dimensions or counting the number of squares to determine its size, we could multiply its dimensions together. Putting this into perspective, we see the number of squares that fits inside a rectangular region is referred to as the area. A shortcut to determine that number of squares is to multiply the base by the height. 6 ft. Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 5 of 35

6 The area of a rectangle is equal to the product of the length of the base and the length of a height to that base. That is A = bh. Most books refer to the longer side of a rectangle as the length (l), the shorter side as the width (w). That results in the formula A = lw. The answer in an area problem is always given in square units because we are determining how many squares fit inside the region. Example: Find the area of a rectangle with the dimensions 3 m by m. A = lw A = 3 A=6 Example: Find the area of the rectangle. The area of the rectangle is 6 m. 9 ft. yd. Be careful! Area of a rectangle is easy to find, and students may quickly multiply to get an answer of 18. This is wrong because the measurements are in different units. We must first convert feet into yards, or yards into feet. yards 1 x = feet 3 9 9= 3x 3 = x We now have a rectangle with dimensions 3 yd. by yd. A = lw A = ( 3)( ) A=6 The area of our rectangle is 6 square yards. If I were to cut one corner of a rectangle and place it on the other side, I would have the following: height height base base Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 6 of 35

7 A parallelogram! Notice, to form a parallelogram, we cut a piece of a rectangle from one side and placed it on the other side. Do you think we changed the area? The answer is no. All we did was rearrange it; the area of the new figure, the parallelogram, is the same as the original rectangle. This allows us to find a formula for the area of a parallelogram. Since the bottom length of the rectangle was not changed by cutting, it will be used as the base length (b), the height of the rectangle was not changed either, we ll call that h. h Now we arrive at the formula for the area of a parallelogram. A = bh. b Example: The height of a parallelogram is twice the base. If the base of the parallelogram is 3 meters, what is its area? First, find the height. Since the base is 3 meters, the height would be twice that or (3) or 6 m. To find the area, A = bh A = 36 A = 18 The area of the parallelogram is 18 m. We have established that the area of a parallelogram is A = bh. Let s see how that helps us to understand the area formula for a triangle and trapezoid. Remember: Once a formula for a figure has been developed, it can be used for any figure that meets its criteria. For example: The parallelogram formula can be used for rectangles, rhombi, and squares. The rectangle formula can be used for squares. The rhombus formula (derived in HS Geometry) can be used for squares. This is based on the Venn Diagram given previously (pg. 3 of these notes). The inner sets have all the same attributes and properties of the sets they are contained within. Therefore, what must be true about any element of the outer set must be true of all elements of that set. Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 7 of 35

8 Let s see how that helps us to understand the area formula for a triangle. If I were to cut the rectangle using its diagonal, I would have the following: height height base base Cutting the rectangle as shown above to create the triangle allows me to see that the triangle is half the rectangle. Therefore since the area of the rectangle is A = bh and the triangle is half of 1 that rectangle, the area of the triangle should be A = bh. We can easily see/count the rectangle contains 4 square units and the triangle contains 1 square units. Mathematically, we could show the area of the rectangle is A = 6 4 = 4units. The Area of the triangle = 1 bh or the Area of the triangle = bh 1 1. To compute the area we get A = 6 4 = 4 = 1units or bh A = = = = 1 square units. h base h base For this parallelogram, its base is 4 units and its height is 3 units. Therefore, the area is. If we draw a diagonal, it cuts the parallelogram into triangles. That means one triangle would have one-half of the area or 6 units. Note the base and height stay the same. So for a triangle, Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 8 of 35

9 h h base base For this square, let s say its base is 4 units. Then, the area is 4 4 = 16 units. If we draw a diagonal, it cuts the square into triangles. That means one triangle would have one-half of the area or 6 units. Note the base and height stay the same. So for a 1 1 triangle, A = bh, or ( 4)( 4) = 8 units h h base base For this rhombus, let s say its base is 4 units and its height is 3 units. Then, the area is 4 3 = 1 units. If we draw a diagonal, it cuts the rhombus into triangles. That means one triangle would have one-half of the area or 6 units. Note the base and height stay the same. So for a triangle, 1 1 ( )( ) A = bh, or 4 3 = 6 units We have shown that half the rectangle, half the square, half the parallelogram and half the rhombus (when cut by a diagonal) form triangles whose area is half the original polygon. To us this may seem redundant, but students need to see and experience this for themselves. NVACS calls for students to compose and decompose figures using this knowledge. At this point we do NOT teach the formula for trapezoids but instead have them decompose (divide the figure) into parts they know. Composite Figures are figures made up of multiple shapes. (linkage - Composite numbers have multiple factors) In order to find the area of these oddly-shaped figures they must be decomposed into figures we are familiar with. Let s start with this isosceles trapezoid 10 in. 4 in. Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 9 of 35 6 in.

10 10 in. We can decompose this trapezoid into a rectangle and two triangles. The area of this trapezoid would be the area of the rectangle added to the areas of the two triangles. In this 4 in. case, because the trapezoid is isosceles, the two triangles will be congruent. 6 in. Notice the b 1 length 10 is equal to the b length 6, plus 4 more inches. Those 4 inches can be split into and and would indicate the lengths of the bases of the two triangles. Now we can use the rectangle formula and triangle formula (twice) to find the total area. Rectangle A = lw A = A = 4 ( 6)( 4) Each Triangle 1 A = bh 1 A = ( )( 4) A = 4 in. 4 in. 6 in. 6 in. in. Since the two triangles are congruent, A =A ) A A = 4 + A = 3 ( )( 4) ( ) total rectangle triangle The area of this trapezoid is 3 square inches. Now that we see that it can be done, we can explore the areas of other, less common, composite shapes. These shapes must be decomposed and the area formulas of the decomposed parts can be used to find each individual area. The total area of the figure is the sum of the areas of its decomposed parts. Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 10 of 35

11 Example: Find the area of the given shape. A B C Start by decomposing 7cm-5cm = cm One way to decompose is given above. Students may find multiple ways to decompose the given figures. If the needed measurements are not available for them to complete the problem, they may want to consider trying a different combination. (Re-decompose?) From the diagram we can see that the total area of this figure will be the sum of the areas of the three rectangles that composed it. Notice that the measures of the sides of the figure can be found by counting the blocks that run along the side of that portion. (Be careful of the scale of the diagram, sometimes a block represents more than one unit.) A TOTAL =A rectangle A +A rectangle B +Arectangle C ( 7)( 3) ( )( 3) ( 6)( 3) A = + + A = A = 45 The area of this figure is 45 square cm. Students can verify this answer by counting the squares inside the figure. Another way to find the area of the above shape is to complete a rectangle. Then subtract off the section(s) not needed. A = 7( ) = 7(9) = 63 TOTAL Minus the area shown in green (and below) in orange A smallrectangle = 1(3) = 3 Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 11 of 35

12 A = 5(3) = 15 largerectangle Area of the composite figure = A = = 45 45square cm Example: Find the area of the given polygon. (DOK ) First, the figure must be decomposed Students should be able to find a rectangle and a triangle. A. 10 m B. 187 m C. 89 m D. 391 m A =A +A TOTAL remember... A A = rectangle 1 ( 11)( 17) + ( 17)( 1) A = A = 89 rectangle triangle 1 = lw and Atriangle = bh So, the area of this figure is 89 m. (C) Example: Find the area of the given polygon. (DOK ) A. 45 cm B. 90 cm C. 110 cm D. 180 cm This figure can be decomposed a few ways: Vertically down the middle, forming two congruent triangles: b = = 15 cm, h = 1 = 6 cm Therefore, the total area is twice the area of one triangle. Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 1 of 35

13 A A A triangle TOTAL TOTAL A = 90 1 = bh = = ( ) A = ( ) 1 triangle ( ) ( 15)( 6) 1 bh The total area is 90 cm. (b) Another way would be: Horizontally, along the diagonal, forming two non-congruent triangles: b = 1 cm (on both), h 1 = 4 cm (top ) & h = 11 cm (bottom ) In this case we must find the sum of both areas to find the total sum. 1 Atriangle = bh A = A + A A A TOTAL top bottom TOTAL TOTAL A = = + = ( 1)( 4) ( 1)( 11) The total area is 90 cm. (b) A final alternative would be: Horizontally and vertically, forming four right triangles, with two sets congruent: (these measurements are interchangeable based on your perspective when viewing the triangles) b 1 = 4 cm (both smaller s) & b = 11 cm (both larger s), h = 1 = 6 cm This time we must add the areas of all four triangles together, but recall that there were two sets of congruent triangles formed when we decomposed the original figure in this manner. So We can find the area of one small triangle and double it, then find the area of one larger triangle and double it, and finally add those two doubled areas together. 1 Atriangle = bh A = A + A ( ) ( ) TOTAL smaller larger 1 1 ATOTAL = + ATOTAL = A = 90 ( ) ( 4)( 6) ( ) ( 11)( 6) Taa daa The total area is 90 cm. (b) AGAIN!! Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 13 of 35

14 Example: Find the area of the given polygon. The area of this figure can be found multiple ways. It could be decomposed into rectangles this way A rectangle = lw A =A +A +A TOTAL rectangle A rectangle B rectangle C ( 1.)( 6.4) ( 3.6)( 0.6) ( 0.8)( 6.4) A = + + A = A = The total area is cm. An alternative to this method and extension on this topic is to decompose and subtract. In this case, students can picture an imaginary rectangle surrounding the entire figure then subtract the region that is not a part of the original figure. A rectangle = lw A =A -A TOTAL outer rectangle ( 5.6)( 6.4) ( 3.6)( 5.8) A = A = A = inner rectangle Again, the total area is cm. Extensions Examples: Find the areas of the shaded regions in the figures below. a) A = lw All squares are rectangles... rectangle A =A -A TOTAL A = outer square ( 4)( 4) ( )( ) inner square A = 16 4 A = 1 The area of the shaded region is 1 square cm. Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 14 of 35

15 b) A =A -A shaded region outer region inner region To find the area of the inner region, we must decompose it. A =A +A inner region rectangle A rectangle B A inner region ( 4)( ) ( )( ) = + A = 8+ 4 A = 1 Now the shaded region can be found. A shaded region =Aouter region -Ainner region A = ( 8)( 9) 1 A = 7 1 A = 60 The area of the shaded region is 60 square cm. Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 15 of 35

16 6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Apply these techniques in the context of solving real-world and mathematical problems. Example: Find the length of segment CD. (DOK 1) Simply count the units along the indicated side. Common error Remind students to start at 0 and count each swoop as 1 unit. 1 3 CD = 3 units Example: In each part below two sides of a rectangle are shown. Write the coordinates of the fourth corner of each rectangle. Then answer the questions. a) The fourth corner is at (5, -5). Can the perimeter and area of this rectangle be found? If so, what are they? If not, why not? b) Yes, they are P = 0l ) 0w A = bh P = ( 10) + ( 9) A = 10 P = A = 90 P = 38 The perimeter is 38 units. The area is 90 units. and ( )( 9) The fourth corner is at (5, 0). Can the perimeter and area of this rectangle be found? If so, what are they? If not, why not? Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 16 of 35

17 NO, because the sides are not horizontal or vertical so the measures cannot be determined. Example: Which gives the perimeter of the square? (DOK ) a) 4 c) 10 b) 8 d) 1 By counting the units we can see that l = 3 & w = 3, therefore P = l ) w P = 3 ( ) + 3 ( ) P = 6+ 6 P = 1 OR Since the side lengths of a square are always equal, it may be faster for students to find one side length and multiply by 4. P square P = 1 ( ) P = 43 = 4s Both methods yield the answer 1 units. (d) Example: Plot the sixteen points in the table below on this graph. After graphing the points, connect them to make a 16-pointed star. POINTS POINTS POINTS POINTS A(4, 0) E(-4, 0) I(0, 4) M(0, -4) B(1, ) F(1, -) J(3, 3) N(3, -3) C(, 1) G(, -1) K(-1, ) P(-1, -) D(-3, 3) H(-3, -3) L(-, 1) Q(-, -1) Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 17 of 35

18 E D L K I B C J A Q P H F M G N a) Find all horizontal lengths: DJ = 6 units KB = units LC = 4 units EA = 8 units QG = 4 units PF = units HN = 6 units b) Find all vertical lengths: DH = 6 units LQ = units KP = 4 units IM = 8 units BF = 4 units CG = units JN = 6 units c) Find the area of the triangle found by connecting the vowels ( AEI) 1 A triangle = bh 1 A = ( 8)( 4) The area of AEI is 16 square units or 16 units. A = 16 d) Discussion: I purposely skipped labeling any point with an O, what point is usually labeled with an O? Give its name and coordinates. The origin, (0, 0) Example: In each question below the coordinates of three corners of a square are given. Find the coordinates of the other corner in each case. You may find it helpful to draw a sketch. a) (, -), (, 3) and (-3, 3). The other corner of the square is at (, ). (-3, -) b) c) (, 3), (3, 4) and (1, 4). The other corner of this square is at (, ). (, 5) d) e) (, ), (4, 4) and (4, 0). The other corner of this square is at (, ). (6, ) Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 18 of 35

19 Answers: a) b) c) 5 units 5 units Example: The line marked on the coordinate grid below is one side of a square: What are the possible coordinates of the corners of the square? There are two possible places the square could be placed. Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 19 of 35

20 Students can visualize (or sketch) the imaginary right triangle with the given line as the hypotenuse. By using it as a reference, students can then find the location of the missing vertices of the square. LINKAGE: slope, rate of change The missing coordinates could be at (6, 3) and (1, 6) OR (-5, -4) and (0, -7). Example: Two corners of a square are shown in the coordinate plane below: a) If the third corner is at (7, -1), where is the fourth corner? (5, 3) b) If the third corner is at (-3, -1) where is the fourth corner? (-1, -5) It is possible to make a different square to those above by placing the third and fourth points in two new positions. c) What are the coordinates which need to be plotted? By looking at the given points as opposite vertices, we can find the other corners at (4, 0) and (0, -). Example: Identify the coordinates of vertex D after quadrilateral DEFG is translated 7 units up: (DOK ) A translation up will change the y-coordinate only and move the figure up seven units. Therefore, D(3, -) would move to a) (3, -) b) (3, 5) c) (5, 3) d) (-4, -) (3, 5). (b) Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 0 of 35

21 Example: John rode his bike on Monday and again on Friday. His path for Monday is shown on the graph below. Each unit represents 1 mile. (DOK 3) On Friday, John rode 14 miles farther than he did on Monday. How could his path have changed while remaining rectangular? a) John could have ridden 14 miles farther north. b) John could have ridden 3½ miles farther north and 7 miles farther east. c) John could have ridden 7 miles farther east. d) John could have ridden 1¾ miles farther north and 3½ miles farther east. 5 miles 6 miles Students will need to determine the perimeter of the rectangle graphed to answer this question. P = l + w Then adding 14 would result in a larger perimeter of 6. P = ( )( ) ( )( ) P = P = Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 1 of 35

22 Students must remember that each change in a dimension will result in twice as much change in the perimeter. Since the perimeter must increase by 14, the only choice that will result in that amount of change would be when John rides an additional 7 miles east. (d) Sample Questions from OnCore Example: Example: Example: Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page of 35

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28 Example: Example: 013 SBAC Questions Standard: 6.G.A.1, 6.G.A.3 DOK: Difficulty: Medium Question Type: TE Technology Enhanced Math 6 Notes Unit 7: Geometry: -Dimensional Figures Page 8 of 35

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