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SOURCE:COMPETITION Number of Problems: 26. FOR PRINT ::: (Book)
A wire is cut into two pieces, one of length and the other of length . The piece of length is bent to form an equilateral triangle, and the piece of length is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is ?
Using the formulas for area of a regular triangle and regular hexagon and plugging and into each equation, you find that . Simplifying this, you get
A unit square is rotated about its center. What is the area of the region swept out by the interior of the square?
First, we need to see what this looks like. Below is a diagram.
For this square with side length 1, the distance from center to vertex is , hence the area is composed of a semicircle of radius , plus times a parallelogram with height and base . That is to say, the total area is .
A regular octagon has sides of length two. Find the area of .
The area of the triangle can be computed as . We will now find and .
Clearly, is a right isosceles triangle with hypotenuse of length , hence . The same holds for triangle and its leg . The length of is equal to . Hence , and .
Then the area of equals .
Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
The outer circle has radius , and thus area . The little circles have area each; since there are 7, their total area is . Thus, our answer is .
Equilateral has side length , is the midpoint of , and is the midpoint of . What is the area of ?
The area of a circle can be given by . because it is the midpoint of a side, and because it is twice the length of . Each angle of an equilateral triangle is so . The area is .
In order to calculate the area of , we can use the formula , where is the base. We already know that , so the formula now becomes . We can drop verticals down from and to points and , respectively. We can see that . Now, we establish the relationship that . We are given that , and is the midpoint of , so . Because is a triangle and the ratio of the sides opposite the angles are is . Plugging those numbers in, we have . Cross-multiplying, we see that Since is the height , the area is .
In the overlapping triangles and sharing common side , and are right angles, , , , and and intersect at . What is the difference between the areas of and ?
Since and , . By alternate interior angles and AA~, we find that , with side length ratio . Their heights also have the same ratio, and since the two heights add up to , we have that and . Subtracting the areas, .
Let represent the area of figure . Note that and .
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A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?
Since the area of the square is , the length of the side is . The distance between the center of the semicircle and one of the bottom vertecies of the square is half the length of the side, which is .
Using the Pythagorean Theorem to find the square of radius, . So, the area of the semicircle is .
A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?
The cylinder can be "unwrapped" into a rectangle, and we see that the stripe is a parallelogram with base 3 and height 80.
Rhombus is similar to rhombus . The area of rhombus is and . What is the area of rhombus ?
Using properties of a rhombus, and . It is easy to see that rhombus is made up of equilateral triangles and . Let the lengths of the sides of rhombus be .
The longer diagonal of rhombus is . Since is a side of an equilateral triangle with a side length of , . The longer diagonal of rhombus is . Since is twice the length of an altitude of of an equilateral triangle with a side length of ,
The ratio of the longer diagonal of rhombus to rhombus is . Therefore, the ratio of the area of rhombus to rhombus is
Let be the area of rhombus . Then , so .