PLEASE help!!!
Find the area and the perimeter of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximations). The figures below are based on semicircles or quarter circles and problems b), c), and d) are involving portions of a square.

PLEASE help Find the area and the perimeter of the shaded regions below Give your answer as a completely simplified exact value in terms of π no approximations class=

Respuesta :

Answer:

 

Step-by-step explanation:

The shaded regions consist of a triangle and a semicircle

Area of shaded regions = Area of Triangle + Area of semicircle

Area of Triangle = (h x b) ÷ 2

Height, h = 10 cm

Base, b = 8 cm

Are of the triangle component = (10 x 8) ÷ 2

= 80 ÷ 2

= 40 cm^2

 Area of semicircle = πr^2 ÷ 2

Diameter, D = 8 cm

Radius, r = 8 cm ÷ 2 = 4 cm

Are of the semicircle component = [π(4^2) ÷ 2)

= 16π ÷ 2

= 16π cm^2

 Total area of shaded regions

= (40+16π) cm^2

= 8 (5 +2π) cm^2

Answer:

[tex]\boxed{\sf Area = 8\pi + 40\ cm^2}[/tex]

[tex]\boxed{\sf Perimeter = 4\pi + 22\ cm}[/tex]

Step-by-step explanation:

Area of the figure:

Firstly: Area of semicircle:

[tex]\sf \frac{\pi r^2}{2} \\Where\ r = 4 \ cm\\\frac{\pi (4)^2}{2} \\\frac{16 \pi}{2}\\8 \pi \ cm^2[/tex]

Then Area of Triangle

[tex]\sf 1/2 (Base)(Height)\\1/2(10)(4)\\10*2\\20\ cm^2[/tex]

Area of Figure = Area of Semicircle + 2(Area of triangle)

=> 8π + 2(20)

=> 8π + 40 cm²

Perimeter of Semicircle:

Firstly, we'll have to find the hypotenuse

[tex]\sf c^2 = a^2+b^2\\c^2 = 4^2+10^2\\c^2 = 16+100\\c^2 = 116\\c = 11\ cm[/tex]

Then, Perimeter of the semi-circle:

=> πr

Where r = 4 cm

=> 4π

Now, the perimeter of the whole figure:

=> 4π + 2(11)

=> 4π + 22 cm