6cm : 3cm in lowest form
1 answer:
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Answer:
1 - (π/6) - (√3/4)
Step-by-step explanation:
First, calculate the area of ABCD. That's just
Then, we are told that arcs BD and AC are circular, meaning they are portions of a circle. Therefore, let's redraw this image a little. Look below for the image.
Essentially, what I've done is created an equilateral triangle and two sectors. You may be wondering why the triangle is equilateral. Well, we firstly know it's isosceles because the two original arcs are circular and congruent. Their overlap creates two congruent lines that can be extend from point M to the vertices A and D. These two lines also happen to be radii of the quarter circles because the two sectors have AM AND AD as radii. AM and AD = 1, so AMD is an equilateral triangle.
Draw altitude OD. This is a right angle. Since equilateral triangles are also isosceles triangles, and isosceles triangles have the property of sharing the median to base and altitude to base, we can say that OD = AD =
. MD = 1, so this triangle is a 30-60-90 triangle. Why? This is because of the 30º angle converse theorem. The shorter leg is half of the hypotenuse MD, so <OMD is a 30º angle, making <ODM a 60º angle and inevitably <MDC a 30º angle (because <D is a right angle) and <ABM as well because the two sectors are congruent.
Because m<ABM = m<MDC = 30º, these two sectors are 30/360 of the individual circles or 1/12 of the original circles, which have areas of π (because 1^2*π = π). Each is 1/12π, combined they are 1/6π.
The area of the equilateral triangle is √3/8. This is because MO can be calculated, using the 30-60-90 ratios, to be √3/2. (1 * √3/2)/2 = (√3/2)/2 = √3/4; this uses the area of a triangle. Now, time to calculate the shaded area.
The area of the square is 1, the area of the sectors π/6, and the equilateral triangle √3/4. Subtract the equilateral triangle and sectors' areas from the area of the square to be left with the remaining shaded part. 1-(π/6)-(√3/4) is your final answer. To put that a little more cleanly below in the image.
Hope this helps, have a great day.
<span>Difference of squares method is a method that is used to evaluate the difference between two perfect squares.
For example, given an algebraic expression in the form:
can be factored as follows:
From the given expressions, the only expression containing two perfect squares with the minus sign in the middle is the expression in option A.
i.e.
which can be factored as follows:
.</span>
For example, given an algebraic expression in the form:
can be factored as follows:
From the given expressions, the only expression containing two perfect squares with the minus sign in the middle is the expression in option A.
i.e.
which can be factored as follows:
The box and whisker plot is attached.
We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21
The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median). This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median). This is between 15 and 19: (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.
We draw the middle line of the box at 14, the median. We draw the left side of the box at the lower quartile, 9. We draw the right side of the box at the upper quartile, 17. From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.
We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21
The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median). This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median). This is between 15 and 19: (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.
We draw the middle line of the box at 14, the median. We draw the left side of the box at the lower quartile, 9. We draw the right side of the box at the upper quartile, 17. From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.