Plz help me on all of the LCM

7. The figure seems most easily seen as a trapezoid with a triangle removed. The trapezoid has bases of 26 and 14, and a height of 15. So, its area is ...

A = (1/2)(b1 +b2)(h) = (1/2)(26+14)(15) = 300 . . . . . . square inches

The triangle has a base of 26 and a height of 9, so an area of ...

A = (1/2)bh = (1/2)(26)(9) = 117 . . . . . . square inches

Then the shaded area is ...

300 in^2 -117 in^2 = 183 in^2

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8. The area of the donut can be found different ways. One way is to multiply the width of the shaded area (3.5 m) by the circumference of its centerline.

The centerline diameter is 21 m - 3.5 m = 17.5 m, so the circumference of that circle is ...

C = πd = π(17.5 m) = 17.5π m

Then the area of the shaded part is ...

(3.5 m)(17.5π m) = 61.25π m^2 ≈ 192.42 m^2

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9. As in problem 7, this can be treated as a parallelogram less a triangle. Here, we need to do a little work to find the area of the triangle.

The triangle of interest is a right triangle with hypotenuse 23.8 ft and one leg 21 ft. The Pythagorean theorem can be used to find the other leg:

h = √(23.8^2 -21^2) = √125.44 = 11.2

Then the area of the triangle is ...

A = (1/2)bh = (1/2)(21)(11.2) = 117.6 . . . ft^2

The area of the parallelogram (including the triangle) is ...

A = bh = (23.8 ft)(15 ft) = 357 ft^2

So, the shaded area is the difference ...

357 ft^2 -117.6 ft^2 = 239.4 ft^2

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<em>Comment on significant digits</em>

The area in problem 8 may be considered to be 192.325 m^2 if you are required to use 3.14 as the value for pi. Since that value is only good to 3 significant digits, it is quite pointless to report the answer to 6 significant digits when using that value. It is quite reasonable to use 192 m^2 for the answer, especially if you're required to round to whole numbers.