A circular piece of sheet metal has a diameter of 20 in. The edges are to be cut off to form a r

A circular piece of sheet metal has a diameter of 20 in. The...

Key Concepts

System of Equations

In this type of problem, multiple conditions such as the area of the rectangle and the relationship between the rectangle’s diagonal and the circle's diameter lead to a system of equations. Solving these simultaneous equations is key to determining the unknown dimensions of the rectangle. This often involves substitution or elimination methods to arrive at the solution.

Area of a Rectangle

The area of a rectangle is computed by multiplying its length by its width. In problems where the area is given along with other geometric constraints, this concept is used to set up one of the equations in the system, which then helps in determining the dimensions of the rectangle.

Pythagorean Theorem

The Pythagorean Theorem is crucial in relating the sides of the rectangle to its diagonal, especially when the rectangle is inscribed within a circle. Since the diagonal of a rectangle with length and width is given by the square root of the sum of the squares of these two dimensions, this theorem helps in formulating the equation connecting the rectangle’s sides and the circle’s diameter.

Inscribed Figures

This concept involves understanding how a rectangle can be positioned within a circle so that all its vertices lie on the circumference. In such cases, the rectangle's diagonal is equal to the circle’s diameter, which creates a direct relationship between the dimensions of the rectangle and the dimensions of the circle.

Transcript

00:01 If we have a situation where we have a circular piece of sheet metal that has a diameter of 20, and that is inches, and we're going to cut it so that it's a rectangle.

00:19 Well, the diameter is now from the endpoints of the rectangle.

00:31 And we can create a system of equations for things we know about the situation.

00:37 For instance, we know that in a rectangle, we have a length and a width, that the length times the width in this case should form a rectangle of area 160 square inches.

00:48 So length times width equals 160.

00:50 You also know that this is creating a right triangle.

00:53 We can use the pythagorean theorem.

00:56 But the length squared plus the width squared would equal the hypotenuse, which is 20 squared.

01:03 And 20 squared is 400.

01:07 If both of these things are true, we can use a substitution method.

01:11 For instance, we can say that the length is equal to 160 divided by the width if we just divide both sides by w.

01:21 And then we can substitute this in for l into the second equation.

01:26 So that becomes 160 over w squared plus w squared equals 400.

01:33 Now 160 over w squared is 160 squared, which is 25 ,600 over w squared plus w squared, plus w squared, equals 400.

01:46 One thing we can do here is scale this all by w squared.

01:54 That way we don't have it in the denominator.

01:57 And so what happens is we have 25 ,600 plus w to the fourth equals 400 w squared.

02:06 We really have a quadratic in disguise.

02:09 If i get this all on one side, we have w to the fourth minus 400 w squared plus 25 ,600 equals zero.

02:19 I'm getting this all on one side of the equation.

02:23 And if we let u be w squared, then this would be u squared minus 400 u plus 25 ,600.

02:35 If i've lost you, make sure you reference solving quadratics by a substitution...

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