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3 years ago

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A trapezoid has base lengths of 12 centimeters and 13 centimeters. The other sides have lengths of 5 centimeters and 10 centimet

ers. A rectangle with side lengths of 2 centimeters and 5 centimeters is connected to the side with length 10 centimeters.
What is the area of the composite figure?

1 answer:

r-ruslan [8.4K]3 years ago

8 0

Answer:

The area of the composite figure is 392.12 $cm^{2}$ .

Step-by-step explanation:

The area of the composite figure = area of trapezoid + area of rectangle

Area of trapezium = $\frac{1}{2}$ ( a +b)h

Where: a is the length of the first base, b the length of the second base and h is the height of the trapzium.

Applying Pythagoras theorem, the height, h, is;

h = $\sqrt{5^{2} - 1^{2} }$

= $\sqrt{24}$

h = 2 $\sqrt{6}$

Area of trapezium = $\frac{1}{2}$ ( a +b)h

= $\frac{1}{2}$ (13 + 12) × 2 $\sqrt{6}$

= 156 $\sqrt{6}$

= 382.12 $cm^{2}$

Area of trapezium is 382.12 $cm^{2}$

Area of rectangle = length × width

= 5 × 2

= 10 $cm^{2}$

Area of rectangle = 10 $cm^{2}$

Therefore,

area of the composite figure = 382.12 + 10

= 392.12 $cm^{2}$

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Step-by-step explanation:

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<u>i) Using characteristic equation:</u>

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As $e^{rt}$ could never be zero, the term (r²+1) must be zero:

(r²+1)=0

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$y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]$

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$y'(t)_{p}=2ACos(2t)-2BSin(2t)$

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$-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)$

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(s²+1)·Y(s)-2s-1= $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)= $\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}$

Y(s)= $\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}$

Y(s)= $\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}$

Using partial franction method:

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2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

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We solve the equation system:

A+C=2

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A+4C=8

B+4D=6

The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

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