Question

For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perimeter. Give your answer as a completely simplified exact value in terms of π (no approximations).
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Answer 1

Answer:

Part 1)

a) The area is $A=36(\pi-2)\ cm^2$  

b) The perimeter is $P=6(\pi+2\sqrt{2})\ cm$

Part 2)

a) The area is $A=576\ cm^2$

b) The perimeter is $P=24(\pi+2)\ cm$

Step-by-step explanation:

see the attached figure to better understand the problem

Part 1)

a) Find the area of the figure N 1

we know that

The area of the figure N 1 is equal to the area of a quarter circle minus the area of the triangle

Find the area of the quarter circle

The area of quarter circle is equal to

$A=\frac{1}{4}\pi r^{2}$

we have

$r=12\ cm$

substitute

$A=\frac{1}{4}\pi (12)^{2}$

$A=36\pi\ cm^2$

Find the area of the triangle

The area of triangle is equal to

$A=\frac{1}{2}(b)(h)$

we have

$b=12\ cm\\h=12\ cm$

substitute

$A=\frac{1}{2}(12)(12)$

$A=72\ cm^2$

therefore

The area of the figure is

$A=(36\pi-72)\ cm^2$

Simplify

$A=36(\pi-2)\ cm^2$              

b) Find the perimeter of figure N 1

we know that

The perimeter of the figure N 1 is equal to the circumference of a quarter circle plus the side AC of triangle

Find the circumference of a quarter circle

The perimeter of a quarter of circle is equal to

$C=\frac{1}{4}2\pi r$

simplify

$C=\frac{1}{2}\pi r$

we have

$r=12\ cm$

substitute

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

$C=6\pi\ cm$

Find the length side AC

Applying the Pythagorean Theorem

$AC^2=12^2+12^2$

$AC^2=288$

$AC=\sqrt{288}\ cm$

simplify

$AC=12\sqrt{2}\ cm$

The perimeter of the figure is

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

simplify

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

Part 2)

a) Find the area of the figure N 2

we know that

The area of the figure N 2 is equal to the area of a semicircle plus the area of a square minus the area of semicircle

so

The area of the figure is equal to the area of the square

$A=24^2\\A=576\ cm^2$

b) Find the perimeter of the figure N 2

we know that

The perimeter of the figure N 2 is equal to the length side AB plus the length side DC plus the circumference of two semicircles

so

The perimeter of the figure N 2 is equal to two times the length side AB plus the circumference of one circle

$P=2(AB)+\pi D$

$P=2(24)+\pi (24)$

$P=(48+24\pi)\ cm$

simplify

$P=24(\pi+2)\ cm$