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allochka39001 [22]
3 years ago
11

Find the area of the figure to the nearest tenth

Mathematics
1 answer:
damaskus [11]3 years ago
8 0
That's a part of a circle that has a radius of 9 inches.

Area of any circle = (pi) x (radius)²

Area of the full circle that this is a piece of = (pi) x (9)² = (81 pi)  square inches.

That's the whole circle.  How much of the whole circle is in the picture ?

A whole circle has 360 degrees in the middle. This piece has 105 degrees in the middle.
This piece is ( 105/360 ) of the whole circle.

So the area of the piece is (105/360) x (81 pi) = <u>74.22 square inches</u> (rounded)
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Ralph is painting the barn below, including the sides and roof.  He wants to know how much paint to purchase. 
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PART A

The barn is constructed by the following 2D shapes

Two triangles of height = 4' and base = 20'
Area = 2 × (0.5×4×20) = 80

Two rectangles of length = 20' and width = 15'
Area = 2 × 20 × 15 = 600

Two rectangles of length = 45' and width = 15'
Area = 2 × 45' × 15' = 1350

One base of length = 45' and width = 15'
Area = 45 × 15 = 675

Two rectangles on each on one side of the roof. We have the length = 45'  but not the width. We can work out the width by using Pythagoras theorem

w² = 4² + 10²
w² = 16 + 100
w² = 116
w = √116 = 10.77

Area of the two rectangles on the roof part is = 2 ×10.77 × 45 = 969.33

Total area to paint = 969.33+675+1350+600+80 = 3674.33 ≈ 3700 (to the nearest hundreth)


PART B

Numbers of paint cans needed = 3700 ÷ 57 = 64.9 ≈ 65 cans

PART C

Total cost of paint = 65 × 23.50 = $1527.50

PART D

The barn is constructed by a cuboid and a rectangular prism

V of cuboid = length × width × height
V of cuboid = 20 × 45 × 15
V of cuboid = 13500

V of triangular prism = Area of cross section × depth
V = [0.5×4×20] × 45
V = 1800

Total volume = 1800 + 13500 = 15300


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