Todd Helmenstine A square is a quadrangle where all four sides (s) are of equal length. In the case of a right triangle, the base and height are the two sides that are perpendicular or at 90 degrees to each other. Area bh Square Perimeter and Surface Area Formulas Squares are four-sided figures where each side is of equal length. Once we have the base and height, we can use the following formula: Okay, this is a little tricky, but it makes more sense looking at the picture below. The height is the distance from the vertex opposite to the base at a 90 degree angle to the base. To figure out the surface area of a triangle, we need to know the base and the height. Try this and see if you get the same answer. We could have also divided it up into these two different rectangles. Now we can add the surface area of the two rectangles: Determine the lateral surface area of the outer cylindrical surface using the formula A 2rh. This looks confusing at first, but we can make this easier by dividing it up into two rectangles like this: To calculate the surface area of a hollow cylinder with an inner radius, r, outer radius, r, and a height, h, follow these steps: Calculate the surface area of the rings at the top and bottom using the formula A 2 (r - r). What is the surface area using yards for the units? Since this is a rectangle we can use the rectangle formula: The perimeter of this football field is the sum of all the sides 100 + 50 + 100 + 50 = 300 yards. We used this same example to demonstrate how to figure the perimeter (see perimeter for kids). Let's take the more complicated example of this football field. When we give the answer for surface area we used squared to indicate that it's a surface area and not just a straight line. Note: if the units were feet for this problem, the answer would be 16 feet squared. Let's try that and see if we get the same answer: Thus, the volume function is increasing for x < 10, and decreasing thereafter.With a rectangle and square we can also get the surface area by multiplying width (W) x length (L). You can prove that this critical value, x = 10, yields a MAX for volume by showing that the derivative goes from being + to - there. x = 10 The dims of the rectangular box of max volume are 10" x 10" x 5". So, we use the constraint, and solve for h to get h in terms of x: SA = x^2 + 4xh = 300 4xh = 300 - x^2 h =(300 - x^2) / (4x) Next, substitute that expression in for h in the volume equation: V = x^2 (300 - x^2) / (4x) Simplifying, V = 75x - (1/4)x^3 Take deriv, set = 0, solve: V' = 75 - (3/4)x^2 = 0. The problem with doing that for V = x^2h is that we have TWO variables. When you have a function in only one variable, it is relatively straightforward to take its derivative, set that deriv = 0, and solve to find the critical pt (either a max or a min). You are given a quantity to maximize or minimize (in this case the volume of the box), and you are given a constraint (in this case, the SA = 300 in^2). This is a very common question type for Calc AB, known as an optimization question. Thus, the volume function is increasing for x < 10, and decreasing thereafter. The dims of the rectangular box of max volume are 10" x 10" x 5". Next, substitute that expression in for h in the volume equation: So, we use the constraint, and solve for h to get h in terms of x: the volume of a square prism of high 2 dm wherein the base is: Rectangle with. This is a very common question type for Calc AB, known as an optimization question. Calculate the surface area of a four-sides 2-m high prism whose base is a.
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