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AoPS

AoPS. Introduction to Geometry. Chapter 4. Perimeter and Area. Problem 4.2. D. 8. Find the perimeter of ΔDEF. 30°. 30°. E. F. 8√3. Problem 4.2. D. 8. Find the perimeter of ΔDEF. Since <E = <F, we have DE = DF = 8. Therefore the perimeter is DE + DF + EF = 16 + 8√3. 30°. 30°.

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AoPS

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  1. AoPS Introduction to Geometry

  2. Chapter 4 Perimeter and Area

  3. Problem 4.2 D 8 Find the perimeter of ΔDEF. 30° 30° E F 8√3

  4. Problem 4.2 D 8 Find the perimeter of ΔDEF. Since <E = <F, we have DE = DF = 8. Therefore the perimeter is DE + DF + EF = 16 + 8√3. 30° 30° E F 8√3

  5. Problem 4.3 E D Given that each little square in the grid is a 1 x 1 square, find the perimeter of ABCDEFGHIJ. G F B I H C J A

  6. Problem 4.3 E D Given that each little square in the grid is a 1 x 1 square, find the perimeter of ABCDEFGHIJ. G F B I H C J A Let’s take the slick approach: The horizontal side on the bottom has length 8; therefore, the sum of the horizontal sides on top, which together cover the same horizontal distance without bactracking, must be 8. Similarly, AB + CD = 5, so the vertical sides on the left have total length 5. The ones on the right have the same total length, so they contribute 5 to the perimeter. P = 2(8 + 5) = 26.

  7. Problem 4.4 The length of each leg of an isosceles triangle is three times the length of the base of the triangle. The perimeter of the triangle is 91. What is the length of the base of the triangle?

  8. Problem 4.4 The length of each leg of an isosceles triangle is three times the length of the base of the triangle. The perimeter of the triangle is 91. What is the length of the base of the triangle? Let the base of the triangle have length x. Then the length of each length is 3x. Since the perimeter of the triangle is 91, we must have x + 3x + 3x = 91. Solving this equation for x, we find that x = 13. The length of the base of the triangle is 13.

  9. Exercise 4.1.5 A triangle with perimeter 45 has one side that is twice as long as the shortest side and another side that is 50% longer than the shortest side. Find the length of the shortest side of the triangle. Let the shortest side have length x. Then the other two sides have lengths 1.5x and 2x. The perimeter of the triangle is 45, so 45 = x + 1.5x + 2x, or 45 = 4.5x. Solving the equation, x = 10, which means the shortest side has length of 10.

  10. Problem 4.6 The length of one side of a rectangle is 4 less than 3 times an adjacent side. The perimeter of the rectangle is 64. Find the area of the rectangle.

  11. Problem 4.6 The length of one side of a rectangle is 4 less than 3 times an adjacent side. The perimeter of the rectangle is 64. Find the area of the rectangle. Let the lengths of the two sides be x and y. We are given that x = 3y – 4. Since the opposite sides of a rectangle are equal, the perimeter of the rectangle is 2x + 2y = 2(3y – 4) + 2y = 8y – 8. We are given that the perimeter is 64, so we have 8y – 8 = 64. Therefore, y = 9 and x = 3(9) – 4 = 23, so the area of the rectangle is 9(23) = 207.

  12. Exercise 4.2.5 A square poster is replaced by a rectangular poster that is 2 inches wider and 2 inches shorter. What is the difference in the number of square inches between the area of the larger poster and the smaller poster? (MATHCOUNTS)

  13. Exercise 4.2.5 A square poster is replaced by a rectangular poster that is 2 inches wider and 2 inches shorter. What is the difference in the number of square inches between the area of the larger poster and the smaller poster? (MATHCOUNTS) Let the side of the square poster be x, in inches. Then the area of the square poster is x2. The area of the rectangular poster is (x + 2) (x – 2) = x2 – 4, so their positive difference is 4 in2.

  14. Exercise 4.2.6 The perimeter of a square garden is 64 meters. The path surrounding the garden has a uniform width and has an area of 228 m2. How many meters of fencing are needed to surround the outer edge of the path? (MATHCOUNTS) Path P Garden

  15. Exercise 4.2.6 The perimeter of a square garden is 64 meters. The path surrounding the garden has a uniform width and has an area of 228 m2. How many meters of fencing are needed to surround the outer edge of the path? (MATHCOUNTS) Let x and y be the side length of the garden and the width of the path, respectively, in meters. The perimeter of the garden is 4x = 64, so x=16. Path P Garden

  16. Exercise 4.2.6 The perimeter of a square garden is 64 meters. The path surrounding the garden has a uniform width and has an area of 228 m2. How many meters of fencing are needed to surround the outer edge of the path? (MATHCOUNTS) Let x and y be the side length of the garden and the width of the path, respectively, in meters. The perimeter of the garden is 4x = 64, so x=16. Path P Garden

  17. Problem 4.10 Given that BC = 40 and CD = 20, what are the ratios of the area of triangle ABC to area of triangle ACD and area of triangle ABC to area of triangle ABD?

  18. Problem 4.10 Given that BC = 40 and CD = 20, what are the ratios of the area of triangle ABC to area of triangle ACD and area of triangle ABC to area of triangle ABD? The altitude from A is the same for both ACD and ABC. Let h be the length of this altitude. Hence, the area of ABC = ½ (BC)(h) = 20h and the area of ACD = ½ (CD)(h) = 10h.

  19. Problem 4.10 Given that BC = 40 and CD = 20, what are the ratios of the area of triangle ABC to area of triangle ACD and area of triangle ABC to area of triangle ABD? The altitude from A is the same for both ACD and ABC. Let h be the length of this altitude. Hence, the area of ABC = ½ (BC)(h) = 20h and the area of ACD = ½ (CD)(h) = 10h. Therefore, area ABC/area ACD = 20h/10h = 2. Notice that area ABC/area ACD = (BC)/(CD).

  20. Problem 4.10 Given that BC = 40 and CD = 20, what are the ratios of the area of triangle ABC to area of triangle ACD and area of triangle ABC to area of triangle ABD? Similarly, altitude from A is the same for both ABC and ABD, h. So, the area of ABC = ½ (BC)(h) and the area of ABD = ½ (BD)(h). Therefore, area ABC/area ABD = 20h/30h = 2/3. Notice that area ABC/area ABD = (BC)/(BD).

  21. Problem 4.10 Given that BC = 40 and CD = 20, what are the ratios of the area of triangle ABC to area of triangle ACD and area of triangle ABC to area of triangle ABD? A key stepin our solution is noting that the three triangles have an altitude in common. Notice that in each case the ratio of the areas of the triangles equals the ratios of the lengths of the sides to which this common altitude is drawn.

  22. Problem 4.11 Suba and Sam have been hired to paint a triangle with base AB on the weirdly shaped wall in the diagram. They are told to choose either C or D as the third vertex of the triangle. Suba thinks using D will make the triangle look cooler, but Sam thinks using C will result in a smaller triangle to paint. How can Suba convince Sam to choose D?

  23. Problem 4.11 Suba and Sam have been hired to paint a triangle with base AB on the weirdly shaped wall in the diagram. They are told to choose either C or D as the third vertex of the triangle. Suba thinks using D will make the triangle look cooler, but Sam thinks using C will result in a smaller triangle to paint. How can Suba convince Sam to choose D? To figure out which of ABC or ABD will use more paint, we need to consider the areas of the two triangles. Since area of ABD = ½ (AB)(DE) and area of ABC = ½ (AB)(BC) we see that DE/BC = ¾.

  24. Problem 4.11 So Suba should point that triangle ABD will require less paint because its area is less than the area of triangle ABC. To figure out which of ABC or ABD will use more paint, we need to consider the areas of the two triangles. Since area of ABD = ½ (AB)(DE) and area of ABC = ½ (AB)(BC) we see that DE/BC = ¾.

  25. !! IMPORTANT !! 1. If 2 triangles share an altitude, then the ratio of their areas = the ratio of the bases to which that altitude is drawn. This is particularly useful in problems in which 2 triangles have bases along the same line. 2. If 2 triangles share a base, then the ratio of their areas = the ratio of the altitudes to that base.

  26. Problem 4.12 AC and BD meet at X. Given [ABX] = 24, [BCX] = 15, and [CDX] = 10, find [ADX]. Note: [ADX] means the area of triangle ADX.

  27. Problem 4.12 AC and BD meet at X. Given [ABX] = 24, [BCX] = 15, and [CDX] = 10, find [ADX]. Note: [ADX] means the area of triangle ADX. Since triangles ABX and CBX share an altitude from B, we have AX[ABX]8 CX [CBX] 5 = =

  28. Problem 4.12 AC and BD meet at X. Given [ABX] = 24, [BCX] = 15, and [CDX] = 10, find [ADX]. Note: [ADX] means the area of triangle ADX. Turning to triangles ADX and CDX, we have AX[ADX]8 CX [CDX] 5 = =

  29. Problem 4.12 AC and BD meet at X. Given [ABX] = 24, [BCX] = 15, and [CDX] = 10, find [ADX]. Note: [ADX] means the area of triangle ADX. Turning to triangles ADX and CDX, we have AX[ADX]8 CX [CDX] 5 Therefore, [ADX] = (8/5)[CDX] = 16. = =

  30. Exercise 4.3.3 Euclid Apartment Building has a wall that is 30 ft long and 10 ft high. Jean is supposed to paint a triangle with one vertex at the top of the wall, and a base that runs 8 ft along the bottom of the wall. Jean wants to put the vertex at one corner and the base at the other side of the wall as shown on the left.

  31. Exercise 4.3.3 Euclid Apartment Building has a wall that is 30 ft long and 10 ft high. Jean is supposed to paint a triangle with one vertex at the top of the wall, and a base that runs 8 ft along the bottom of the wall. Jean wants to put the vertex at one corner and the base at the other side of the wall as shown on the left. George the building owner wants to save money on paint. He insists that the top vertex be above the middle of the base as shown on the bottom left so the triangle won’t be so big. However, to show he’s a nice guy, he says that she can make the base 10 ft wide instead. Jean argues that George’s design will use more paint than hers. Is she right?

  32. Exercise 4.3.3 Jean’s triangle has a base of 8 ft and height of 10 ft, so the area is ½ (8)(10) = 40 ft2. George’s triangle has base 10 ft and height 10 ft, so the area is ½ (10)2 = 50 ft2. We could have seen this without computing the areas by noting that the heights of the 2 triangles are the same, but George’s triangle has a longer base than Jean. s. George the building owner wants to save money on paint. He insists that the top vertex be above the middle of the base as shown on the bottom left so the triangle won’t be so big. However, to show he’s a nice guy, he says that she can make the base 10 ft wide instead. Jean argues that George’s design will use more paint than hers. Is she right?

  33. Exercise 4.3.4 By what factor is the area of triangle multiplied if the length of its base is doubled and the height is tripled? (Source: MATHCOUNTS)

  34. Exercise 4.3.4 By what factor is the area of triangle multiplied if the length of its base is doubled and the height is tripled? (Source: MATHCOUNTS) Solution 1: Let the triangle have base b and height h, so it has area bh/2. If the base is doubled to 2b and the height is tripled to 3h, then the area of the new triangle is (2b)(3h)/2 = 3bh. Therefore, the area increases by a factor of 3bh ½ bh = 6.

  35. Exercise 4.3.4 By what factor is the area of triangle multiplied if the length of its base is doubled and the height is tripled? (Source: MATHCOUNTS) Solution 2: Doubling the base multiplies the area by 2 and tripling the height multiplies the area by 2. Doing both of these will multiply the area by 2 x 3 = 6.

  36. Review Problem 4.15 Find the perimeter of a square that has as area of 75. Express your answer in radical form.

  37. Review Problem 4.15 Find the perimeter of a square that has as area of 75. Express your answer in radical form. Let the side of the square be s. Then the area is s2. Since the area of the square is 75, we have s2. Solving s2 = 75, gives us √75 = √25 * 3 = 5√3 as the side length of the square. There are four sides, so the perimeter is 4(5√3) = 20√3.

  38. Problem 4.17 The length of a given rectangle is 2 less than 4 times the width of the rectangle. Its perimeter is 51. What is its area?

  39. Problem 4.17 The length of a given rectangle is 2 less than 4 times the width of the rectangle. Its perimeter is 51. What is its area? Let the width of the rectangle be w, so the length is 4w – 2. In terms of w, the perimeter then is 2(w) + 2(4w – 2) = 10w – 4. Since the perimeter = 51, we have 10w – 4 = 51, so w = 5.5. Therefore, the length is 4w – 2 = 20, so the area is (5.5) (20) = 110.

  40. Problem 4.20 A rectangle has a perimeter of 28 cm and an area of 48 cm2. Find the dimensions of the rectangle?

  41. Problem 4.20 A rectangle has a perimeter of 28 cm and an area of 48 cm2. Find the dimensions of the rectangle? Let x and y be the dimension of the rectangle. Then the perimeter of the rectangle is 2(x + y) = 28, so x + y = 14. The area of the rectangle is xy = 48.

  42. Problem 4.20 A rectangle has a perimeter of 28 cm and an area of 48 cm2. Find the dimensions of the rectangle? Let x and y be the dimension of the rectangle. Then the perimeter of the rectangle is 2(x + y) = 28, so x + y = 14. The area of the rectangle is xy = 48. Substituting y = 14 – x into xy = 48, we get x(14 – x) = 48. Multiplying we get x2 – 14x + 48 = 0 which factors to (x – 6)(x – 8) = 0. The solutions are x = 6 and x = 8, which lead to y = 8 and y = 6, respectively.

  43. Problem 4.20 A rectangle has a perimeter of 28 cm and an area of 48 cm2. Find the dimensions of the rectangle? Let x and y be the dimension of the rectangle. Then the perimeter of the rectangle is 2(x + y) = 28, so x + y = 14. The area of the rectangle is xy = 48. Substituting y = 14 – x into xy = 48, we get x(14 – x) = 48. Multiplying we get x2 – 14x + 48 = 0 which factors to (x – 6)(x – 8) = 0. The solutions are x = 6 and x = 8, which lead to y = 8 and y = 6, respectively. Therefore, the rectangle is 6 cm by 8 cm. [We could also have used trial and error to solve x + y = 14 and xy = 48. ]

  44. Problem 4.25 In the figure below, AB = 12 and BC = AD = 8. BC | AB and DA | AB. How many square units are in the area of the shaded region? (Source: MATHCOUNTS) D C B A

  45. Problem 4.25 In the figure below, AB = 12 and BC = AD = 8. BC | AB and DA | AB. How many square units are in the area of the shaded region? (Source: MATHCOUNTS) D C Let E be the point on AB where the 2 triangles meet. Then the area of the shaded region is area ΔADE + ΔBCE = ½ AE * AD + ½ BE * BC = ½AE * 8 + ½BE * 8 = 4AE + 4BE = 4AB = 48 sq units. B A E

  46. Problem 4.26 In the figure below, [WOZ] = 12, [ZOY] = 18, and [WXYZ] = 50. Find [WOX].

  47. Problem 4.26 In the figure below, [WOZ] = 12, [ZOY] = 18, and [WXYZ] = 50. Find [WOX]. W Z O X Y

  48. Problem 4.26 In the figure below, [WOZ] = 12, [ZOY] = 18, and [WXYZ] = 50. Find [WOX]. W Z ΔWOZ and ΔWYZ share altitude from Z, so WO/WY = [WOZ]/[WZY] = 12/30 = 2/5. O X Y

  49. Problem 4.26 In the figure below, [WOZ] = 12, [ZOY] = 18, and [WXYZ] = 50. Find [WOX]. W Z ΔWOZ and ΔWYZ share altitude from Z, so WO/WY = [WOZ]/[WZY] = 12/30 = 2/5. O X Y We also have [WXY] = [WXYZ] – [WZY] = 20. Since [WOZ]/[WZY] = WO/WY = 2/5, we have [WOX] = (2/5)[WXY] = 8.

  50. Problem 4.29 We begin with an equilateral triangle. We divide each side into three segments of equal length, and add an equilateral triangle to each side using the middle third as a base. We then repeat this, to get Given that the perimeter of the 1st figure is 12, what is the perimeter of the 2nd figure? the 3rd figure?

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