Find the area of the Composite Shape.

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Step-by-step explanation:

Important Information:

Area of a circle=

Area of a triangle=

Diameter of a cirlce=

Method:

The first step is to divide the shape into smaller shapes. In this case, we have a right -angled triangle and a semi-circle.

The next step is to work out the area of the triangle. Since we know the radius of the semi-circle is 3 inches, to work out the diameter we would simply multiply the radius of 3 by 2, this gives us 6 inches. This is because the radius is half of the diameter.

Now that we know the diameter of the semi-cirlce, we can begin to work out the area of the triangle. We can do this by first multiplying the base of 6 inches by the height of 9 inches, this gives us 54 inches squared. This however gives us the area of 2 of those triangles. To work out the area of that triangle, we would need to divide the value of 54 by 2, this gives us 27 inches squared.

The next step is to work out the area of the semi-circle. We can do this by multiplying pi by the radius of 3, squared, this gives us 28.27 inches squared (2 d.p). However, this is for a full circle and we need to work out the area of the semi-circle. We can do this by dividing the area of the circle, which is 28.27, by 2, this gives us 14.14 inches squared (2 d.p). This is because a semi-cirlce is half of a circle.

The final step is to add the Areas of the two shapes together. You can do this by adding the area of the triangle, which is 27 inches squared, to the area of the semi circle, which is 14.14 inches squared (2 d.p), this gives us 41.14 inches squared (2 d.p).

Step by Step:

1) Multiply 3 by 2.

2) Multiply 6 by 9.

3) Divide 54 by 2.

4) Multiply pi by 3 squared.

5) Divide 28.27 by 2.

6) Add 27 to 14.14.

step-by-step explanation:

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: