Lesson Problems on the area and the dimensions of a rectangle surrounded by a strip
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<H2>Problems on the area and the dimensions of a rectangle surrounded by a strip</H2> The problems collected in this lesson are a) for rectangular garden surrounded by a path of a uniform width, b) for rectangular pool surrounded by a walkway of uniform way, and c) for rectangular picture concluded in a frame of uniform wide. <H3>Problem 1</H3>A rectangular garden has dimensions of 18 feet by 13 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 516 square feet? <B>Solution</B> <pre> Let "x" be the path uniform width in feet. Then the garden and the surrounding path represent the rectangle with dimensions (18+2x) and (13+2x). Total area of the garden and the path is (18+2x)*(13+2x) square feet, while the area of the garden is 18*13 square feet. Thus the area of the path itself is the difference (18+2x)*(13+2x) - 18*13, and you have an equation (18+2x)(13+2x) - 18*13 = 516. Simplify and solve it for x: 18*13 + 36x + 26x + 4x^2 - 18*13 = 516, 4x^2 + 62x - 516 = 0, 2x^2 + 31x - 258 = 0, {{{x[1,2]}}} = {{{(-31 +- sqrt(31^2-4*2*(-258)))]/4}}} = {{{(-31 +- 55)/4}}}. Only positive solution is acceptable: x = {{{(-31+55)/4}}} = 6 ft. <B>Answer</B>. The uniform path width should be 6 ft. </pre> <H3>Problem 2</H3>A path of uniform width surrounds a rectangular garden that is 5 m wide and 12 m long. The area of the path is 168 {{{m^2}}}. Find the width of the path. <B>Solution</B> <pre> Your equation is (5+2x)*(12+2x) - 5*12 = 168, where x is an unknown width of the path. The equation's left side is the difference between the areas of the larger rectangle and the area of smaller rectangle which represents the garden itself. The right side is the given area of the surrounding path. Simplify the equation: {{{4x^2 + 10x + 24x + 60 - 60}}} = {{{168}}}, or {{{4x^2 + 34x - 168}}} = {{{0}}}. Use the quadratic formula to find the roots. The only root which fits is positive x = 3.5. <B>Check</B>: (5+2*3.5)*(12+2*3.5) - 60 = 12*19 - 60 = 168. (OK!) <B>Answer</B>. The width of the path is 3.5 m. </pre> <H3>Problem 3</H3>A 17 ft x 33 ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway is 216 {{{ft^2}}}, how wide is the walkway? <B>Solution</B> <pre> Let x = the width of the walkway. Then you have this equation to determine x: (17 + 2x)*(33 + 2x) - 17*33 = 216. Do you understand why it is your equation to determine x? Because it is the difference of the areas of two rectangles. OK. Now let us solve it. Open the parentheses, and you will get {{{17*33 + 2*33x + 2*17x + 4x^2}}} - {{{17*33}}} = {{{216}}}, or {{{4x^2 + 2*33x + 2*17x}}} = {{{216}}}, or {{{4x^2 + 100x - 216}}} = {{{0}}}, or dividing by 4 both sides {{{x^2 + 25x - 54}}} = {{{0}}}, Factor the left side {{{x^2 + 25x - 54}}} = (x-2)*(x+27). So the roots of the last quadratic equation are 2 and -27. Only positive x = 2 is acceptable. <B>Answer</B>. The width of the walkway is 2 ft. </pre> <H3>Problem 4</H3>A metal sleeve of length 20 cm has rectangular cross-section 10 cm by 8 cm (outer dimensions). The metal has uniform thickness, x cm, along the sleeve, and the total volume of metal in the sleeve is 495 cm^3. Derive the equation 16x^2 - 144x + 99 = 0 and find the solution for the unknown value of thickness x in this problem. <B>Solution</B> <pre> In cross-section, we have empty rectangle of the size (10-2x) cm by (8-2x) cm surrounded by metallic strip. So, the area of the metal in cross-section is area = 10*8 - (10-2x)*(8-2x) cm^2. +---------------------------------------------------------+ | Since the length is 20 cm, the volume of the metal | | is 20 times the area of the cross-section. | +---------------------------------------------------------+ So, your equation for x is -20*(4x^2 - 36x) = 495 cm^3. Divide both sides by 5 and simplify step by step. You will get precisely the equation which you request 16x^2 - 144x + 99 = 0. Solve it using the quadratic formula. You will get the roots {{{x[1]}}} = 8.25 and {{{x[2]}}} = 0.75. 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