A rectangular area can be completely tiled with 200 square tiles. If the side length of each tile was increased by 1 cm, it would take only 128 tiles to tile the area. Find the side length of each tile.
let the length of square tiles be x. thus, 200*x^2=128(x+1)^2 =>25x^2=16x^2+16+32x =>9x^2-32x-16=0 =>(9x+4)(x-4)=0 =>x=4 (answer)
while they ask you to degree the element of a rectangle, they desire to comprehend how a lot of a unit will extra healthful interior that rectangle. In different words, for a 9 in via 2 in rectangle, they desire to comprehend how many a million via a million (insert unit) squares will extra healthful interior this rectangle. BTW, A=LW section=LengthxWidth answer to above question is: A=LW A=(9)(2) A=18
here goes... Conceptually, if you knew the area of the rectangle was, say 5000cm squared, and you knew that the length of the tile was, say 5cm, then you would know that the number of tiles necessary would be (5000cm sqrd)/5^2 = 5000/25 = 200 tiles. Using this conceptual knowledge, we can set up equations to answer this question. Let A represent the area of the rectangle to be tiled (this is a constant in this question, we just need to give it a label here) Let x represent the length of the original square tile. We know from the question the following: A / (x^2) = 200 tiles rearrange for A = 200(x^2) This is Equation 1. The question says that if the length increases by 1cm (i.e. x+1) then only 128 tiles are needed, so A / [(x+1)^2] = 128 tiles rearrange for A = 128(x+1)^2 This is Equation 2. Substitute Equation 1 into Equation 2 and solve for x! So the first step is: 200(x^2) = 128(x+1)^2 ---> divide both sides by 8 25(x^2) = 16(x+1)^2 You might be tempted at this point to expand and collect like terms, but you'd end up with an equation with which it appears difficult to solve for x. Here however, you should notice that all parts of this equation are very easily square rooted...so, square root both sides to simplify and solve for x: sqrt [25(x^2)]= sqrt[16(x+1)^2] 5x = 4(x+1) 5x = 4x + 4 x = 4 Therefore the side length of each tile is 4 cm. Of course, you can check your answer by plugging it into any of the equations above, and making sure that the left side of the equation acutally does equal the right side!
Let A be the area of the rectangle and l be the length of tile in cm. 200*l*l= A then, 128(l+1)*(l+1)= A So, 200*l*l= 128(l+1)*(l+1) 9l*l-32l-16=0 If you solve this equation the result will come l= 4cm so side of each tile is 4cm.