A rectangular picture frame is 30 cm long and 20 cm wide. A picture of area 459 cm2 is centrally placed inside the frame leaving a uniform margin of width of x cm. Find the dimensions of the picture

multipleQuestions newQuestion Find the dimensions of the picture newAnswer The dimensions of the picture are 27 cm by 17 cm. Steps: 1. Define variables: Let the length of the frame be L_f = 30 cm. Let the width of the frame be W_f = 20 cm. Let the area of the picture be A_p = 459 cm^2. Let the uniform margin width be x cm. Let the length of the picture be L_p. Let the width of the picture be W_p. 2. Express picture dimensions in terms of frame dimensions and margin: Since the picture is centrally placed with a uniform margin of x cm, the dimensions of the picture are: L_p = L_f - 2x = 30 - 2x W_p = W_f - 2x = 20 - 2x 3. Formulate an equation using the picture's area: The area of the picture is A_p = L_p × W_p. 459 = (30 - 2x)(20 - 2x) 4. Solve the quadratic equation for x: Expand the equation: 459 = 600 - 60x - 40x + 4x^2 459 = 4x^2 - 100x + 600 Rearrange into standard quadratic form (ax^2 + bx + c = 0): 4x^2 - 100x + 600 - 459 = 0 4x^2 - 100x + 141 = 0 Use the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a: Here, a=4, b=-100, c=141. x = -(-100) ± sqrt((-100)^2 - 4(4)(141))2(4) x = 100 ± sqrt(10000 - 2256)8 x = 100 ± sqrt(7744)8 x = (100 ± 88)/(8) Two possible values for x: x_1 = (100 + 88)/(8) = (188)/(8) = 23.5 x_2 = (100 - 88)/(8) = (12)/(8) = 1.5 5. Determine the valid value for x: The dimensions of the picture must be positive. If x = 23.5 cm: L_p = 30 - 2(23.5) = 30 - 47 = -17 cm (Not physically possible) If x = 1.5 cm: L_p = 30 - 2(1.5) = 30 - 3 = 27 cm W_p = 20 - 2(1.5) = 20 - 3 = 17 cm These dimensions are positive, so x = 1.5 cm is the correct margin width. 6. State the dimensions of the picture: The length of the picture is 27 cm. The width of the picture is 17 cm. Verification: Area = 27 cm × 17 cm = 459 cm^2, which matches the given area.