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3 years ago

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What is the area of the polygon below

2 answers:

algol133 years ago

5 0

Answer: The correct option is (D) 147 sq. units.

Step-by-step explanation: We Re given to find the area of the polygon shown in the figure.

Let us divide the given polygon in two rectangles A and B as shown in the attached figure below.

The dimensions of rectangle A are

length, l = 10 units and breadth, b = 3 units.

So, the area of rectangle A is given by

$AREA_A=l\times b=10\times3=30~\textup{sq. units}.$

And, the dimensions of rectangle B are

length, l' = 13 units and breadth, b' = 12 - 3 = 9 units.

So, the area of rectangle B is given by

$AREA_B=l'\times b'=13\times9=117~\textup{sq. units}.$

Therefore, the area of the polygon is

$AREA=AREA_A+AREA_B=30+117=147~\textup{sq. units}.$

Thus, the required area of th given polygon is 147 sq. units.

Option (D) is CORRECT.

Fudgin [204]3 years ago

3 0

Separate it into two different shapes

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3 years ago

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

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Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

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