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9. Find the area of each figure to the nearest tenth 140,110,180,50 8,8,10

9. Find the area of each figure to the nearest tenth 140,110,180,50 8,8,10-example-1
9. Find the area of each figure to the nearest tenth 140,110,180,50 8,8,10-example-1
9. Find the area of each figure to the nearest tenth 140,110,180,50 8,8,10-example-2
User ScottyB
by
5.3k points

1 Answer

2 votes

Answer:

  1. 18550 cm²
  2. 88 ft²

Explanation:

1. There are several ways the area can be divided up so that formulas for common figures can be used to find the areas of the pieces. In the attached figure, we have identified an overall rectangle ABXE and a trapezoid BXDC that is subtracted from it.

The area of the rectangle is the product of length and width:

area ABXE = (180 cm)(140 cm) = 25,200 cm²

The area of a trapezoid is the product of its height (DX = 70 cm) and the average of its base lengths ((BX +DC)/2 = 95 cm).

area BXDC = (70 cm)(95 cm) = 6650 cm²

Then the area of figure ABCDE is the difference of these areas:

area ABCDE = area ABXE - area BXDC = (25,200 - 6,650) cm²

area ABCDE = 18,550 cm²

__

2. In order to find the area of the figure, we need to know the length DE. That length is one leg of right triangle DEA, so we can use the Pythagorean theorem. That theorem tells us ...

DE² + EA² = AD²

DE² + (8 ft)² = (10 ft)² . . . . . substitute the given values

DE² = 36 ft² . . . . . . . . . . . . .subtract 64 ft²

DE = 6 ft . . . . . . . . . . . . . . . take the square root

Now, we can choose to add the area of triangle DEA to that of square ABCE, or we can treat the whole figure as a trapezoid with bases AB=8 ft and DC=14 ft. In the latter case, the average base length is ...

(8 ft + 14 ft)/2 = 11 ft

and the area is the product of this and the 8 ft height:

area ABCD = (11 ft)(8 ft) = 88 ft²

9. Find the area of each figure to the nearest tenth 140,110,180,50 8,8,10-example-1
User Rizzy
by
4.6k points
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