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shusha [124]
4 years ago
14

The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f

igure and give your answers as a completely simplified exact value in terms of π (no approximations).

Mathematics
1 answer:
WITCHER [35]4 years ago
4 0

Answer:

Part 1) A=36(\pi-2)\ cm^2

Part 2) P=6(\pi+2\sqrt{2})\ cm

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

so

A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)

we have that the base and the height of triangle is equal to the radius of the circle

r=12\ cm\\b=12\ cm\\h=12\ cm

substitute

A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2

simplify

Factor 36

A=36(\pi-2)\ cm^2

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r

substitute the given values

C=\frac{1}{2}\pi (12)\\C=6\pi\ cm

Applying the Pythagorean Theorem

The hypotenuse of right triangle is equal to

AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm

simplify

AC=12\sqrt{2}\ cm

Find the perimeter

P=(6\pi+12\sqrt{2})\ cm

simplify

Factor 6

P=6(\pi+2\sqrt{2})\ cm

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4 years ago
Which coordinate pair identifies the center of the circle represented by 4x2 + 4y2 − 16x − 24y + 36 = 0.
anastassius [24]

Answer:

Option A is correct

Coordinate pair identifies the center of the circle is, (2 , 3)

Step-by-step explanation:

the general equation of the circle is given by :

(x-a)^2+(y-b)^2=r^2 ; where (a, b) represents the coordinates of the circle  and r is the radius of the circle.

Given : 4x^2+4y^2-16x-24y+36=0

4x^2-16x+4y^2-24y+36=0

Take common 4 from above equation we have:

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Divide both sides by 4 we get;

x^2-4x+y^2-6y+9=0

Add and subtract 4 in above equation:

x^2-4x+y^2-6y+9+4-4=0

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Add and subtract 9 in above equation:

x^2-4x+4+y^2-6y+5+9-9=0

x^2-4x+4+y^2-6y+9-4=0

or

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Using identity: (a-b)^2 = a^2 - 2ab + b^2

(x-2)^2+(y-3)^2 = 2^2

On comparing with the general equation of the circle we have;

a = 2 , b= 3 and r = 2

Therefore, the coordinate pair identifies the center of the circle is, (2 , 3)





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4 years ago
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