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Allisa [31]
2 years ago
9

What are the dimensions of a storage tank

Mathematics
1 answer:
motikmotik2 years ago
4 0

Answer:

Step-by-step explanation:

550 48 6-0 3/16 3/16 800

1000 48 10-10 3/16 3/16 1300

1100 48 11-11 3/16 3/16 1400

1500 48 15-8 3/16 3/16 1650

65 9-0 3/16 3/16 1500

2000 65 11-10 3/16 3/16 2050

2500 65 14-10 3/16 3/16 2275

3000 65 17-8 3/16 3/16 2940

4000 65 23-8 3/16 3/16 3600

5000 72 23-8 1/4 1/4 5800

84 17-8 1/4 1/4 5400

7500 84 26-6 1/4 1/4 7150

96 19-8 1/4 1/4 6400

10000 96 26-6 1/4 5/16 8540

120 17-0 1/4 5/16 8100

12000 96 31-6 1/4 5/16 10500

120 20-8 1/4 5/16 9500

15000 108 31-6 5/16 5/16 13300

120 25-6 5/16 5/16 12150

20000 120 34-6 5/16 5/16 15500

25000 120 42-6 3/8 3/8 22300

30000 120 51-3 3/8 3/8 28000

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<u>Solution-</u>

From the figure,

AE = 2.4

EB = 2.8

BC = 11.7


Area of rectangle 1 = 8.68 sq.in

\Rightarrow FH \times HI=8.68

\Rightarrow EB \times HI=8.68  (∵ sides of the rectangle 2)

\Rightarrow 2.8 \times HI=8.68

\Rightarrow HI=3.1


Area of Triangle 1 = 6.48 sq.in

\Rightarrow \frac{1}{2}\times AE \times EG= 6.48

\Rightarrow \frac{1}{2}\times AE \times (EF+FG)= 6.48

\Rightarrow \frac{1}{2}\times AE \times (EF+HI)= 6.48  (∵ sides of the rectangle 1)

\Rightarrow EF+3.1= 5.4

\Rightarrow EF=2.3

\Rightarrow BH=2.3  (∵ sides of the rectangle 2)


BC = BH+HI+IC

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The area of Rectangle 2,

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The area of Triangle 2,

\frac{1}{2}\times GI \times IC=\frac{1}{2}\times EB \times IC=\frac{1}{2}\times 2.8 \times 6.3=8.82\ sq.in


The area of the whole figure = Area of Triangle 1 + Area of rectangle 1 + Area of Triangle 2 + Area of rectangle 2

= 6.48+8.68+8.82+6.44=30.42 sq.in


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