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$pts.) Find the area of the shaded region, given that m/_ACB=90^(@),AO=10cm, and AC=12cm {:[AB=20cm],[AC=12cm],[CB=],[a^(2)+b^(2)=c^(2)]:}$

$2$ . $p t s$ .) Find the area of the shaded region, given that $m \angle A C B=90^{\circ}, A O=10 \mathrm{~cm}$ , and $A C=12 \mathrm{~cm}$ $\newline$ $\begin{array}{c} A B=20 \mathrm{~cm} \\ A C=12 \mathrm{~cm} \\ C B= \\ a^{2}+b^{2}=c^{2} \end{array}$

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2

p t s

.) Find the area of the shaded region, given that

m \angle A C B=90^{\circ}, A O=10 \mathrm{~cm}

, and

A C=12 \mathrm{~cm}

\newline

\begin{array}{c} A B=20 \mathrm{~cm} \\ A C=12 \mathrm{~cm} \\ C B= \\ a^{2}+b^{2}=c^{2} \end{array}

Find CB Length: First, we need to find the length of CB using Pythagoras' theorem since we have a right-angled triangle at A. $\newline$ $AC^2 + CB^2 = AB^2$ $\newline$ $12^2 + CB^2 = 20^2$ $\newline$ $144 + CB^2 = 400$ $\newline$ $CB^2 = 400 - 144$ $\newline$ $CB^2 = 256$ $\newline$ $CB = \sqrt{256}$ $\newline$ $CB = 16$ cm
Calculate Triangle Area: Now, we calculate the area of the triangle ABC. $\newline$ Area = $(1/2) \times \text{base} \times \text{height}$ $\newline$ Area = $(1/2) \times AC \times CB$ $\newline$ Area = $(1/2) \times 12 \, \text{cm} \times 16 \, \text{cm}$ $\newline$ Area = $6 \, \text{cm} \times 16 \, \text{cm}$ $\newline$ Area = $96 \, \text{cm}^2$
Find Sector Area: Next, we find the area of the sector AOB. The radius of the sector is AO which is $10\,\text{cm}$ . The angle of the sector is $90$ degrees because $m/\_ACB=90^{\circ}$ . Area of sector AOB = $(\text{angle}/360) \times \pi \times \text{radius}^2$ Area of sector AOB = $(90/360) \times \pi \times (10\,\text{cm})^2$ Area of sector AOB = $(1/4) \times \pi \times 100\,\text{cm}^2$ Area of sector AOB = $25\pi\,\text{cm}^2$
Find Shaded Region Area: Finally, we subtract the area of the sector $AOB$ from the area of the triangle $ABC$ to find the area of the shaded region. $\newline$ Area of shaded region = Area of triangle $ABC$ - Area of sector $AOB$ $\newline$ Area of shaded region = $96\,\text{cm}^2 - 25\pi\,\text{cm}^2$