Marcus wants to decorate his box that is i the shape of a cube with a color. he decides to cover the entire box with red color p
aper except the circular window, shown with his name on it. the box is show. rounded to nearest square centimeter, how much red paper is needed to cover Marcus box ? Marcus uses 3.14 to estimate pi
1 answer:
*see attached picture for the diagram of the box
Answer:
130cm²
Step-by-step explanation
The amount of red appear needed by Marcus to cover his box = Surface Area of the cube - Area of the circle.
Surface area of cube = 6s² (where s is the side length of the cube. s² therefore is S.A of 1 face of the cube)
Given that s = 5cm,
S.A of cube = 6s² = 6*5² = 150cm²
Area of circle = πr²;
Given that π = 3.14, and r = ½ of diameter of circle (i.e. ½ of side length of cube)
Area of circle = 3.14*(½*5)² = 3.14*(2.5)² = 3.14*6.25 =19.625
Amount of red paper needed = S.A of the cube - Area of the circle = 150cm² - 19.625cm² = 130.375cm² = 130cm²
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Explanation:
a. The line joining the midpoints of the parallel bases is perpendicular to both of them. It is the line of symmetry for the trapezoid. This means the angles and sides on one side of that line of symmetry are congruent to the corresponding angles and sides on the other side of the line. The diagonals are the same length.
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b. We observe that adjacent pairs of points have the same x-coordinate, so are on vertical lines, which have undefined slope. KN is a segment of the line x=1; LM is a segment of the line x=3. If the trapezoid is isosceles, the midpoints of these segments will be on a horizontal line. The midpoint of KN is at y=(3-2)/2 = 1/2. The midpoint of LM is at y=(1+0)/2 = 1/2. These points are on the same horizontal line, so the trapezoid <em>is isosceles</em>.
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c. We observed in part (b) that the parallel sides are KN and LM. The coordinate difference between K and L is (1, 3) -(3, 1) = (-2, 2). That is, segment KL is the hypotenuse of an isosceles right triangle with side lengths 2, so the lengths of KL and MN are both 2√2.
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For part (c), we used the shortcut that the hypotenuse of an isosceles right triangle is √2 times the leg length.
