# 1.12 area w

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Published on February 6, 2014

Author: Tzenma

Source: slideshare.net

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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