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1.12 area w

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The document discusses the concept of area. It defines area as the amount of surface enclosed by a perimeter or border. Area is quantified by measuring the number of square units within the enclosed surface. The area of basic shapes like squares, rectangles, and rhombuses is defined. Formulas are provided to calculate the area of these shapes given the measurements of their sides. Examples are included to demonstrate calculating areas by dividing shapes into rectangles and adding the individual areas.

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Area http://www.lahc.edu/math/frankma.htm
Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop.
Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane.
Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below.
Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. 1 in 1 in 1m 1m 1 mi 1 mi
Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. If each side of a square is 1 unit, then we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit. 1 in 1 in 1m 1m 1 mi 1 mi
Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. If each side of a square is 1 unit, then we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit. Hence the areas of the following squares are: 1 in 1m 1 mi 1 in 1m 1 mi
Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. If each side of a square is 1 unit, then we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit. Hence the areas of the following squares are: 1 in 1m 1 mi 1 in 1 in2 1 square-inch 1m 1 mi
Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. If each side of a square is 1 unit, then we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit. Hence the areas of the following squares are: 1 in 1m 1 mi 1 in 1 in2 1 square-inch 1m 1 m2 1 square-meter 1 mi 1 mi2 1 square-mile
Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi2 (square miles). 3 mi 2 mi
Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi2 (square miles). 3 mi 2 mi 2x3 = 6 mi2
Area A 2 mi x 3 mi rectangle may be cut into 2 mi six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h 2). then its area A = h x w (unit * For our discussion, the “width” is the horizontal length. 3 mi 2x3 = 6 mi2 w
Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). * For our discussion, the “width” is the horizontal length.
Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s s A square * For our discussion, the “width” is the horizontal length.
Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s The perimeter of a square is s + s + s + s = 4s. The area of a square is s*s = s2. s A square * For our discussion, the “width” is the horizontal length.
Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s The perimeter of a square is s + s + s + s = 4s. The area of a square is s*s = s2. s So the perimeter of a 10 m x 10 m square is 40 m. A square 2 = 100 m2. and its area is 10 * For our discussion, the “width” is the horizontal length.
Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s The perimeter of a square is s + s + s + s = 4s. The area of a square is s*s = s2. s So the perimeter of a 10 m x 10 m square is 40 m. A square 2 = 100 m2. and its area is 10 s An area with four equal sides in general is called a rhombus (diamond shapes). s A rhombus * For our discussion, the “width” is the horizontal length.
Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s The perimeter of a square is s + s + s + s = 4s. The area of a square is s*s = s2. s So the perimeter of a 10 m x 10 m square is 40 m. A square 2 = 100 m2. and its area is 10 s An area with four equal sides in general is called a rhombus (diamond shapes). The perimeter of s a rhombus is 4s, but its area depends on its shape. A rhombus * For our discussion, the “width” is the horizontal length.
Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 4 R 12 12
Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 4 R 12 12
Area Example A. Find the area of the following shape R. Assume the unit is meter. There are two basic approaches. 4 4 R 12 12
Area Example A. Find the area of the following shape R. Assume the unit is meter. There are two basic approaches. I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2 corner removed. 8 4 R 12 12 4 4 4 R 12 12
Area Example A. Find the area of the following shape R. Assume the unit is meter. There are two basic approaches. I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2 corner removed. Hence the area of R is 144 – 32 = 112 m2 8 4 R 12 12 4 4 4 R 12 12
Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 R There are two basic approaches. 12 12 I. We may view R as a 12 x 12 = 144 m2 square Il. We may dissect R into two 2 with a 4 x 8 = 32 m rectangles I and II as shown. corner removed. Hence the area of R is 144 – 32 = 112 m2 8 8 4 R 12 4 4 I 12 12 4 II 12 4
Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 R There are two basic approaches. 12 12 I. We may view R as a 12 x 12 = 144 m2 square Il. We may dissect R into two 2 with a 4 x 8 = 32 m rectangles I and II as shown. corner removed. Area of I is 12 x 8 = 96, Hence the area of R is area of II is 4 x 4 = 16. 144 – 32 = 112 m2 8 8 4 R 12 4 4 I 12 12 4 II 12 4
Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 R 4 There are two basic approaches. 12 12 I. We may view R as a 12 x 12 = 144 m2 square Il. We may dissect R into two 2 with a 4 x 8 = 32 m rectangles I and II as shown. corner removed. Area of I is 12 x 8 = 96, Hence the area of R is area of II is 4 x 4 = 16. 144 – 32 The area of R is the sum of the = 112 m2 two or 96 + 16 = 112 m2. 8 8 4 R 12 4 4 I 12 12 4 II 12
Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 4 R There are two basic approaches. 12 12 I. We may view R as a 12 x 12 = 144 m2 square Il. We may dissect R into two 2 with a 4 x 8 = 32 m rectangles I and II as shown. corner removed. Area of I is 12 x 8 = 96, Hence the area of R is area of II is 4 x 4 = 16. 144 – 32 The area of R is the sum of the = 112 m2 two or 96 + 16 = 112 m2. 8 4 R 12 4 4 I 12 12 8 8 iii 4 II 12 12 4 4 iv (We may also cut R into iii and iv as shown here.) 12
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. 2 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. I II III 2 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, I II III 2 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, I II III 2 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. I II III 2 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. I II III 2 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? I 2 ft II III 20 ft 4 ft 25 ft 6 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? The area of the larger strip is 25 x 6 = 150 ft2 I 2 ft II III 20 ft 4 ft 25 ft 6 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. I 2 ft II III 20 ft c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? 25 ft 2 The area of the larger strip is 25 x 6 = 150 ft and the area of the smaller strip is 20 x 4 = 80 ft2. 4 ft 6 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. I 2 ft II III 20 ft c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? 25 ft 2 The area of the larger strip is 25 x 6 = 150 ft and the area of the smaller strip is 20 x 4 = 80 ft2. Their sum 150 + 80 = 230 includes the 4 ft x 6 ft (= 24 ft2) rectangular overlap twice. 4 ft 6 ft
Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. I 2 ft II III 20 ft c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? 25 ft 2 The area of the larger strip is 25 x 6 = 150 ft and the area of the smaller strip is 20 x 4 = 80 ft2. Their sum 150 + 80 = 230 includes the 4 ft x 6 ft (= 24 ft2) rectangular overlap twice. Hence the total area covered is 230 – 24 = 206 ft2. 4 ft 6 ft
Area A parallelogram is a shape enclosed by two sets of parallel lines. h b
Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h b
Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h b
Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b
Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height.
Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. 8 ft 12 ft
Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. 8 ft 12 ft 8 ft 12 ft
Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. 8 ft 12 ft 8 ft 8 ft 12 ft 12 ft
Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. 8 ft 12 ft 8 ft 8 ft 12 ft 12 ft 8 ft 12 ft
Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. For example, the area of all the parallelograms shown 8 ft 8 ft 12 ft here is 8 x 12 = 96 ft2, 12 ft so they are the same size. 8 ft 12 ft 8 ft 12 ft
Area A triangle is half of a parallelogram.
Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, h b
Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, h b h b
Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. h b h b
Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. h b h b
Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. 8 ft 12 ft h b h b
Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. 8 ft 8 ft 12 ft 12 ft h b h b
Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. 8 ft 8 ft 12 ft 12 ft 8 ft 12 ft h b h b
Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. h b h b 8 ft 8 ft 12 ft 12 ft 8 ft 8 ft 12 ft 12 ft
Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. 8 ft 8 ft 12 ft 12 ft h b h b For example, the area of all the triangles shown here is (8 x 12) ÷ 2 = 48 ft2, i.e. they are the same size. 8 ft 8 ft 12 ft 12 ft
Area A trapezoid is a 4-sided figure with one set of opposite sides parallel.
Area A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 Example B. Find the area of the following trapezoid R. Assume the unit is meter. 8 12
Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle.
Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5,
Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2.
Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 4 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2. The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m2.
Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 4 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2. The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m2. Therefore the area of the trapezoid is 40 + 10 = 50 m2.
Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 4 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2. The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m2. Therefore the area of the trapezoid is 40 + 10 = 50 m2. We may find the area of any trapezoid by slicing it one parallelogram and one triangle. A direct formula for the area of a trapezoid may be obtained by pasting two copies together as shown on the next slide.
Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. a T h b
Area a T Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. h b a h b
Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) a T h b a b h b a a+b
Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) so its area is (a + b)h. a T h b a b h b a a+b
Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) so its area is (a + b)h. Therefore the area of a trapezoid T is half of (a + b)h or that T = (a + b)h ÷ 2 a T h b a b h b a a+b
Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) so its area is (a + b)h. Therefore the area of a trapezoid T is half of (a + b)h or that T = (a + b)h ÷ 2 a T h b a b h b a a+b For example, applying this formula in the last example, we have the same answer: 8 5 12
Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) so its area is (a + b)h. Therefore the area of a trapezoid T is half of (a + b)h or that T = (a + b)h ÷ 2 a T h b a b h b a a+b For example, applying this formula in the last example, we have the same answer: 8 (12 + 8) 5÷2 = 100÷2 = 50. 5 12

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1.12 area w

  • 2. Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop.
  • 3. Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane.
  • 4. Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below.
  • 5. Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. 1 in 1 in 1m 1m 1 mi 1 mi
  • 6. Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. If each side of a square is 1 unit, then we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit. 1 in 1 in 1m 1m 1 mi 1 mi
  • 7. Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. If each side of a square is 1 unit, then we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit. Hence the areas of the following squares are: 1 in 1m 1 mi 1 in 1m 1 mi
  • 8. Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. If each side of a square is 1 unit, then we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit. Hence the areas of the following squares are: 1 in 1m 1 mi 1 in 1 in2 1 square-inch 1m 1 mi
  • 9. Area If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses an area, i.e. a surface in the plane. The word “area” also denotes the amount of surface enclosed which is defined below. If each side of a square is 1 unit, then we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit. Hence the areas of the following squares are: 1 in 1m 1 mi 1 in 1 in2 1 square-inch 1m 1 m2 1 square-meter 1 mi 1 mi2 1 square-mile
  • 10. Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi2 (square miles). 3 mi 2 mi
  • 11. Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi2 (square miles). 3 mi 2 mi 2x3 = 6 mi2
  • 12. Area A 2 mi x 3 mi rectangle may be cut into 2 mi six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h 2). then its area A = h x w (unit * For our discussion, the “width” is the horizontal length. 3 mi 2x3 = 6 mi2 w
  • 13. Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). * For our discussion, the “width” is the horizontal length.
  • 14. Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s s A square * For our discussion, the “width” is the horizontal length.
  • 15. Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s The perimeter of a square is s + s + s + s = 4s. The area of a square is s*s = s2. s A square * For our discussion, the “width” is the horizontal length.
  • 16. Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s The perimeter of a square is s + s + s + s = 4s. The area of a square is s*s = s2. s So the perimeter of a 10 m x 10 m square is 40 m. A square 2 = 100 m2. and its area is 10 * For our discussion, the “width” is the horizontal length.
  • 17. Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s The perimeter of a square is s + s + s + s = 4s. The area of a square is s*s = s2. s So the perimeter of a 10 m x 10 m square is 40 m. A square 2 = 100 m2. and its area is 10 s An area with four equal sides in general is called a rhombus (diamond shapes). s A rhombus * For our discussion, the “width” is the horizontal length.
  • 18. Area 3 mi A 2 mi x 3 mi rectangle may be cut into 2x3 2 mi six 1 x 1 squares so it covers an area of = 6 mi2 2 x 3 = 6 mi2 (square miles). In general, given the rectangle with w height = h (units) width* = w (units), h A = h x w (unit2) then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown. s The perimeter of a square is s + s + s + s = 4s. The area of a square is s*s = s2. s So the perimeter of a 10 m x 10 m square is 40 m. A square 2 = 100 m2. and its area is 10 s An area with four equal sides in general is called a rhombus (diamond shapes). The perimeter of s a rhombus is 4s, but its area depends on its shape. A rhombus * For our discussion, the “width” is the horizontal length.
  • 19. Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 4 R 12 12
  • 20. Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 4 R 12 12
  • 21. Area Example A. Find the area of the following shape R. Assume the unit is meter. There are two basic approaches. 4 4 R 12 12
  • 22. Area Example A. Find the area of the following shape R. Assume the unit is meter. There are two basic approaches. I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2 corner removed. 8 4 R 12 12 4 4 4 R 12 12
  • 23. Area Example A. Find the area of the following shape R. Assume the unit is meter. There are two basic approaches. I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2 corner removed. Hence the area of R is 144 – 32 = 112 m2 8 4 R 12 12 4 4 4 R 12 12
  • 24. Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 R There are two basic approaches. 12 12 I. We may view R as a 12 x 12 = 144 m2 square Il. We may dissect R into two 2 with a 4 x 8 = 32 m rectangles I and II as shown. corner removed. Hence the area of R is 144 – 32 = 112 m2 8 8 4 R 12 4 4 I 12 12 4 II 12 4
  • 25. Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 R There are two basic approaches. 12 12 I. We may view R as a 12 x 12 = 144 m2 square Il. We may dissect R into two 2 with a 4 x 8 = 32 m rectangles I and II as shown. corner removed. Area of I is 12 x 8 = 96, Hence the area of R is area of II is 4 x 4 = 16. 144 – 32 = 112 m2 8 8 4 R 12 4 4 I 12 12 4 II 12 4
  • 26. Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 R 4 There are two basic approaches. 12 12 I. We may view R as a 12 x 12 = 144 m2 square Il. We may dissect R into two 2 with a 4 x 8 = 32 m rectangles I and II as shown. corner removed. Area of I is 12 x 8 = 96, Hence the area of R is area of II is 4 x 4 = 16. 144 – 32 The area of R is the sum of the = 112 m2 two or 96 + 16 = 112 m2. 8 8 4 R 12 4 4 I 12 12 4 II 12
  • 27. Area Example A. Find the area of the following shape R. Assume the unit is meter. 4 4 R There are two basic approaches. 12 12 I. We may view R as a 12 x 12 = 144 m2 square Il. We may dissect R into two 2 with a 4 x 8 = 32 m rectangles I and II as shown. corner removed. Area of I is 12 x 8 = 96, Hence the area of R is area of II is 4 x 4 = 16. 144 – 32 The area of R is the sum of the = 112 m2 two or 96 + 16 = 112 m2. 8 4 R 12 4 4 I 12 12 8 8 iii 4 II 12 12 4 4 iv (We may also cut R into iii and iv as shown here.) 12
  • 28. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. 2 ft
  • 29. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. I II III 2 ft
  • 30. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, I II III 2 ft
  • 31. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, I II III 2 ft
  • 32. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. I II III 2 ft
  • 33. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. I II III 2 ft
  • 34. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? I 2 ft II III 20 ft 4 ft 25 ft 6 ft
  • 35. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? The area of the larger strip is 25 x 6 = 150 ft2 I 2 ft II III 20 ft 4 ft 25 ft 6 ft
  • 36. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. I 2 ft II III 20 ft c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? 25 ft 2 The area of the larger strip is 25 x 6 = 150 ft and the area of the smaller strip is 20 x 4 = 80 ft2. 4 ft 6 ft
  • 37. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. I 2 ft II III 20 ft c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? 25 ft 2 The area of the larger strip is 25 x 6 = 150 ft and the area of the smaller strip is 20 x 4 = 80 ft2. Their sum 150 + 80 = 230 includes the 4 ft x 6 ft (= 24 ft2) rectangular overlap twice. 4 ft 6 ft
  • 38. Area b. Find the area of the following shape R. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft2. I 2 ft II III 20 ft c. Two rectangular walk-ways crosses each other perpendicularly as shown. The larger one is 25 ft x 6 ft and the smaller one is 20 ft x 4 ft. What is the total area they cover? 25 ft 2 The area of the larger strip is 25 x 6 = 150 ft and the area of the smaller strip is 20 x 4 = 80 ft2. Their sum 150 + 80 = 230 includes the 4 ft x 6 ft (= 24 ft2) rectangular overlap twice. Hence the total area covered is 230 – 24 = 206 ft2. 4 ft 6 ft
  • 39. Area A parallelogram is a shape enclosed by two sets of parallel lines. h b
  • 40. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h b
  • 41. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h b
  • 42. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b
  • 43. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height.
  • 44. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. 8 ft 12 ft
  • 45. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. 8 ft 12 ft 8 ft 12 ft
  • 46. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. 8 ft 12 ft 8 ft 8 ft 12 ft 12 ft
  • 47. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. 8 ft 12 ft 8 ft 8 ft 12 ft 12 ft 8 ft 12 ft
  • 48. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h h b b Hence the area of the parallelogram is A = b x h where b = base and h = height. For example, the area of all the parallelograms shown 8 ft 8 ft 12 ft here is 8 x 12 = 96 ft2, 12 ft so they are the same size. 8 ft 12 ft 8 ft 12 ft
  • 49. Area A triangle is half of a parallelogram.
  • 50. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, h b
  • 51. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, h b h b
  • 52. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. h b h b
  • 53. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. h b h b
  • 54. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. 8 ft 12 ft h b h b
  • 55. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. 8 ft 8 ft 12 ft 12 ft h b h b
  • 56. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. 8 ft 8 ft 12 ft 12 ft 8 ft 12 ft h b h b
  • 57. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. h b h b 8 ft 8 ft 12 ft 12 ft 8 ft 8 ft 12 ft 12 ft
  • 58. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. Therefore the area of a triangle is A = (b x h) ÷ 2 or A = b x h 2 where b = base and h = height. 8 ft 8 ft 12 ft 12 ft h b h b For example, the area of all the triangles shown here is (8 x 12) ÷ 2 = 48 ft2, i.e. they are the same size. 8 ft 8 ft 12 ft 12 ft
  • 59. Area A trapezoid is a 4-sided figure with one set of opposite sides parallel.
  • 60. Area A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 Example B. Find the area of the following trapezoid R. Assume the unit is meter. 8 12
  • 61. Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle.
  • 62. Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5,
  • 63. Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2.
  • 64. Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 4 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2. The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m2.
  • 65. Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 4 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2. The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m2. Therefore the area of the trapezoid is 40 + 10 = 50 m2.
  • 66. Area 8 A trapezoid is a 4-sided figure with one set of opposite sides parallel. 5 12 Example B. Find the area of the following trapezoid R. 4 8 Assume the unit is meter. By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2. The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m2. Therefore the area of the trapezoid is 40 + 10 = 50 m2. We may find the area of any trapezoid by slicing it one parallelogram and one triangle. A direct formula for the area of a trapezoid may be obtained by pasting two copies together as shown on the next slide.
  • 67. Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. a T h b
  • 68. Area a T Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. h b a h b
  • 69. Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) a T h b a b h b a a+b
  • 70. Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) so its area is (a + b)h. a T h b a b h b a a+b
  • 71. Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) so its area is (a + b)h. Therefore the area of a trapezoid T is half of (a + b)h or that T = (a + b)h ÷ 2 a T h b a b h b a a+b
  • 72. Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) so its area is (a + b)h. Therefore the area of a trapezoid T is half of (a + b)h or that T = (a + b)h ÷ 2 a T h b a b h b a a+b For example, applying this formula in the last example, we have the same answer: 8 5 12
  • 73. Area Suppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known. By pasting another copy of the T to itself, we obtain a parallelogram. The base of the parallelogram is (a + b) so its area is (a + b)h. Therefore the area of a trapezoid T is half of (a + b)h or that T = (a + b)h ÷ 2 a T h b a b h b a a+b For example, applying this formula in the last example, we have the same answer: 8 (12 + 8) 5÷2 = 100÷2 = 50. 5 12