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AREA PROBLEMS

AREA PROBLEMS. LET’S USE POLYNOMIALS – Area Problem Solving. An oil painting is 10 cm longer than it is wide and is bordered on all sides by a 3 cm wide frame. If the area of the frame alone is 402 cm 2 , what are the dimensions of the painting? What are our problem solving strategies????.

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AREA PROBLEMS

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  1. AREA PROBLEMS

  2. LET’S USE POLYNOMIALS – Area Problem Solving An oil painting is 10 cm longer than it is wide and is bordered on all sides by a 3 cm wide frame. If the area of the frame alone is 402 cm2, what are the dimensions of the painting? What are our problem solving strategies????

  3. Area Problem Solving An oil painting is 10 cm longer than it is wide and is bordered on all sides by a 3 cm wide frame. If the area of the frame alone is 402 cm2, what are the dimensions of the painting? STEP 1 : What is your unknown? What is the question asking for? STEP 2: State your knowns STEP 3: Solve

  4. Area Problem Solving – Example 1 An oil painting is 10 cm longer than it is wide and is bordered on all sides by a 3 cm wide frame. If the area of the frame alone is 402 cm2, what are the dimensions of the painting? STEP 1: What is your unknown? What is the question asking for? Dimensions of the painting. x = width x+10 = length x+10 x x

  5. Area Problem Solving – Example1 An oil painting is 10 cm longer than it is wide and is bordered on all sides by a 3 cm wide frame. If the area of the frame alone is 402 cm2, what are the dimensions of the painting? STEP 1: Unknown is width and length of painting, width = x, length = x + 10 STEP 2 : State your knowns Area = (length)(width) x+10 x x

  6. Area Problem Solving – Example 1 An oil painting is 10 cm longer than it is wide and is bordered on all sides by a 3 cm wide frame. If the area of the frame alone is 402 cm2, what are the dimensions of the painting? STEP 1: Unknown is width and length of painting, width = x, length = x + 10 STEP 2 : State your knowns Area = (length)(width) Painting: width = x, length = x+10 Painting + Frame: width = x + 6, length = x + 16 3cm x+10 x 3cm 3cm x x+6 3cm x+10+6 or x +16

  7. Area Problem Solving – Example 1 An oil painting is 10 cm longer than it is wide and is bordered on all sides by a 3 cm wide frame. If the area of the frame alone is 402 cm2, what are the dimensions of the painting? STEP 1: Unknown is width and length of painting, width = x, length = x + 10 STEP2: State your knowns Area = (length)(width) Painting: width = x, length = x+10 Frame: width = x + 6, length = x + 16 Area of frame = 402 cm2 3cm x+10 x 3cm 3cm x x+6 3cm x+10+6 or x +16

  8. Area Problem Solving – Example 1 An oil painting is 10 cm longer than it is wide and is bordered on all sides by a 3 cm wide frame. If the area of the frame alone is 402 cm2, what are the dimensions of the painting? STEP 1: Unknown is width and length of painting width = x, length = x + 10 STEP 2 : State your knowns Area = (length)(width) Painting: width = x, length = x+10 Frame: width = x + 6, length = x + 16 Area of frame = 402 cm2 STEP 3: SOLVE Total Area – Area of Painting = Area of Border Total Area – Area of Painting = 402cm2 3cm x+10 x 3cm 3cm x x+6 3cm x+10+6 or x +16

  9. x+10 x SOLVE x x+6 x +16 MULTIPLY SIMPLIFY SIMPLIFY SUBTRACT 96 SIMPLIFY DIVIDE by 12 SOLVE Width = 25.5cm Length = 35.5cm

  10. Area Problem Solving – Example 2 • A rectangle is twice as long as it is wide. If its length is decreased by 4 and its width is decreased by 2, its area is decreased by 24. Find its original dimensions. STEP 1 : What is your unknown? What is the question asking for? STEP 2: State your knowns STEP 3: Solve

  11. Area Problem Solving – Example 2 • A rectangle is twice as long as it is wide. If its length is decreased by 4 and its width is decreased by 2, its area is decreased by 24. Find its original dimensions. STEP 1 : What is your unknown? What is the question asking for? Dimensions of original rectangle width = x, length = 2x 2x x

  12. Area Problem Solving – Example 2 • A rectangle is twice as long as it is wide. If its length is decreased by 4 and its width is decreased by 2, its area is decreased by 24. Find its original dimensions. STEP 1 : What is your unknown? What is the question asking for? Dimensions of original rectangle width = x, length = 2x STEP 2: What are your knowns? Dimensions of new rectangle width = x -2, length = 2x – 4 Area = (length)(width) Area of original rectangle: or Area of new rectangle: Area of new rectangle = area of original rectangle - 24 = 2x 2x - 4 x - 2 x

  13. Area Problem Solving – Example 2 • A rectangle is twice as long as it is wide. If its length is decreased by 4 and its width is decreased by 2, its area is decreased by 24. Find its original dimensions. STEP 1 : What is your unknown? What is the question asking for? Dimensions of original rectangle width = x, length = 2x STEP 2: What are your knowns? Dimensions of new rectangle width = x -2, length = 2x – 4 Area = length x width Area of original rectangle: or Area of new rectangle: Area of new rectangle = area of original rectangle - 24 = STEP 3: SOLVE Area of new rectangle = area of original rectangle - 24 = 2x 2x - 4 x - 2 x

  14. SOLVE = MULTIPLY SIMPLIFY SUBTRACT SUBTRACT 8 DIVIDE SOLVE Dimensions of original rectangle width = x, length = 2x Width = 4, Length = 8

  15. AREA PROBLEMS – Part II

  16. Area Problem Solving – #1 A rectangle is three times as long as it is wide. If its length and width are both decreased by 2 cm, its area is decreased by 36 cm2. Find its original dimensions. STEP 1 : What is your unknown? What is the question asking for? Dimensions of original rectangle width = x, length = 3x 3x x

  17. Area Problem Solving – #1 A rectangle is three times as long as it is wide. If its length and width are both decreased by 2 cm, its area is decreased by 36 cm2. Find its original dimensions. STEP 1 : What is your unknown? What is the question asking for? Dimensions of original rectangle width = x, length = 3x STEP 2: What are your knowns? Dimensions of new rectangle width = x -2, length = 3x – 2 Area = (length)(width) Area of original rectangle: or Area of new rectangle: Area of new rectangle = area of original rectangle - 36 3x 3x - 2 x - 2 x

  18. Area Problem Solving – #1 A rectangle is three times as long as it is wide. If its length and width are both decreased by 2 cm, its area is decreased by 36 cm2. Find its original dimensions. STEP 1 : What is your unknown? What is the question asking for? Dimensions of original rectangle width = x, length = 2x STEP 2: What are your knowns? Dimensions of new rectangle width = x -2, length = 2x – 4 Area = length x width Area of original rectangle: or Area of new rectangle: Area of new rectangle = area of original rectangle - 36 STEP 3: SOLVE Area of new rectangle = area of original rectangle - 36 = 3x 3x - 2 x - 2 x

  19. SOLVE = MULTIPLY SIMPLIFY SUBTRACT SUBTRACT 8 DIVIDE SOLVE Dimensions of original rectangle width = x, length = 2x Width = 5, Length = 15

  20. LET’S USE POLYNOMIALS – Area Problem Solving A house has two rooms of equal area. One room is square and the other room is a rectangle 4 ft narrower and 5 ft longer than the square one. Find the area of each room STEP 1 – Area of square room Area of rectangle room So, I will need dimensions of each, length and width

  21. Area Problem Solving – #1 A rectangle is three times as long as it is wide. If its length and width are both decreased by 2 cm, its area is decreased by 36 cm2. Find its original dimensions. STEP 1 – Area of square room Area of rectangle room So, I will need dimensions of each, length and width x x+5 x x-4

  22. Area Problem Solving – #1 A rectangle is three times as long as it is wide. If its length and width are both decreased by 2 cm, its area is decreased by 36 cm2. Find its original dimensions. STEP 1 – Area of square room Area of rectangle room So, I will need dimensions of each, length and width STEP 2 – Area = (length)(width) Area = x2 Area = (x+5)(x-4) x x+5 x-4 x

  23. Area Problem Solving – #1 A rectangle is three times as long as it is wide. If its length and width are both decreased by 2 cm, its area is decreased by 36 cm2. Find its original dimensions. STEP 1 – Area of square room Area of rectangle room So, I will need dimensions of each, length and width STEP 2 - Area = (length)(width) Area = x2 Area = (x+5)(x-4) AREAS ARE EQUAL, SO x2=(x+5)(x-4) x x+5 x x-4

  24. SOLVE MULTIPLY SIMPLIFY SUBTRACT ADD SOLVE AREA of each room = 400 ft2 Dimensions of square room width = 20 ft, length = 20 ft Dimensions of rectangular room Width = 16 ft, Length = 25 ft

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