9) Separate the polygon into 2 shapes (a rectangle and triangle). Then plug in the formula for those 2 shapes separately and add the 2 areas.
First, the triangle's area = bh x 1/2 or bh/2.
b=5 and h=5, so 5x5/2 or 25/2 = 12.5.
Next, solve the area for the rectangle(lw=A). l=5 and 2=8, so 5x8=40.
Add 40 +12.5 = 52.5cm^2 (always include units).
10) It is similar to the last problem. You will separate the polygon into 2 shapes (a circle and a square), but this time you will subtract the area of the sector formed from the circle (draw a dotted line where the circle would be formed in the square).
So, the area of the square is 2x2=4ft^2.
area of a circle is (pi)r2 or (pi)2^2= 4pi or about 12.5ft^2.
4+12.5 = 16.5 ft^2.
Now, the area of the sector must be subtracted from the combination of the 2 areas found above.
The area of a sector is 2(pi)r x (C/360).
The angle is given as 90°, the radius is 2 ft. Now plug these into the equation.
2(pi)(2)x(90/360) = 3.14ft^2.
16.5-3.14= 13.4 ft^2.
Answer A and E
Step-by-step explanation:
I did this question
To find the IQR you first need to write it in numerical order
77, 81, 83, 84, 86, 86, 88, 92
IQR is just Q3 - Q1
Q1 is the middle of the first have, since it has an even set of 4 numbers in the first half you need to take the average of the two middle ones.. which is 82
Q3 done the same process would be 87.
87 - 82 = 5
IQR = 5
This is the best i could find
