Mathematics: I need help with my rsm hwrk pls help me

answer: h-2step-by-step explanation:...

I need help with my rsm hwrk pls help me - id10601691, blackwell0315

ashleymstep15

17.03.2022

$A= 32(\pi-2)\ cm^2, \ \ \ P=8\pi\ cm$

Step-by-step explanation:

When we consider Area under arc AC is is representing a quarter as ADBC is a square, $\angle D=90^{\circ}$ .

Area of quadrant = $\dfrac{\pi r^2}{4}$

here r= 8 cm

Area under Arc AC= $\dfrac{\pi (8)^2}{4}=16\pi\ cm^2$

Area of white region ABC = Area of square ADBC - Area under Arc AC

$=8^2-16\pi\ \ [\text{Area of square} = sides^2]\\\\= 64-16\pi\ \ =16(4-\pi)\ cm^2$

Similarly , Area of white region ADC = $16(4-\pi)\ cm^2$

Area of shaded region = Area of square - Area of white region ABC - Area of white region ADC

$=64-(64-16\pi)-(64-16\pi)\\\\= 32\pi-64\ cm^2 \\\\= 32(\pi-2)\ cm^2$

Area of shaded region = $= 32(\pi-2)\ cm^2$

Length of arc AC = $\dfrac{Circumference}{4}=\dfrac{2\pi r}{4}$

$\dfrac{\pi (8)}{2}=4\pi cm$

Perimeter of shaded region = 2(AC) = $8\pi\ cm$

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