We can divide the above figure into two rectangles MNSR and OPQS.
The dimension of rectangle MNSR is 3 x 4 (Since, MR = 3 and MN = 4).
So, area of rectangle MNSR = 3* 4 = 12 square units.
Similarly dimension of rectangle OPQS is 3 x 1 because
PQ = MR - NO = 3 - 2 = 1
SQ = RQ - MN = 7 - 4 = 3.
So, area of OPQS = 3* 1 = 3 square units.
Hence, the area of polygon MNOPQR = Area of rectangle MNSR that is 12 square units + Area of rectangle OPQS that is 3 square units.
= 12 + 3
= 15 square units.
Hope this helps you!
They give the formula as:
Surface Area =<span> (2 • <span>π <span>• r²) + (2 • <span>π • r • height)</span></span></span></span>
However the 2*PI*r^2 part of the formula is used to calculate the 2 "ends" of a cylinder. Since the problem states that you are NOT to count any of surface area of the "ends" then you only need the <span>(2 • <span>π • r • height) part of the formula.
So, r = 3 inches and height = 8 * 3 inches, the side area equals
2 * PI * r * height
2 * 3.14159 * 3 * 24 =
</span></span>
<span>
<span>
<span>
452.39 cubic inches which is the lateral area.</span></span></span>
You would want to add all of the angles together to find the sum. Set this equal to 360.
2x-9 + 2x-9 + x + x + x = 360
Combine like terms,
7x-18 = 360
Add 18 to both sides,
7x = 378
Divvied both sides by 7,
x = 54
D. I’m not sure but try that
