The area of a segment is:
Area of Segment=Area of Sector−Area of Triangle
Given:
Step 1: Area of the sector
36060×π(8)2
=61×64π
=\frac{32\pi}{3}\ \text{cm}^2<br></p><pdata−pxy="true"style="color:rgb(0,0,0);margin−top:2px;margin−bottom:2px;">Areaofthetriangle</p><pdata−pxy="true"style="color:rgb(0,0,0);margin−top:2px;margin−bottom:2px;">Thetworadiiandthechordforman<b>equilateraltriangle</b>(sincethecentralangleis(60∘)),withside(8)cm.</p><pdata−pxy="true"style="color:rgb(0,0,0);margin−top:2px;margin−bottom:2px;">\text{Area}=\frac{\sqrt3}{4}(8)^2<br>=16\sqrt3\ \text{cm}^2<br></p><pdata−pxy="true"style="color:rgb(0,0,0);margin−top:2px;margin−bottom:2px;">Areaofthesegment</p><pdata−pxy="true"style="color:rgb(0,0,0);margin−top:2px;margin−bottom:2px;">\frac{32\pi}{3}-16\sqrt3\ \text{cm}^2<br></p><pdata−pxy="true"style="color:rgb(0,0,0);margin−top:2px;margin−bottom:2px;"><b>Answer:</b></p><pdata−pxy="true"style="color:rgb(0,0,0);margin−top:2px;margin−bottom:2px;">\boxed{\frac{32\pi}{3}-16\sqrt3\ \text{cm}^2}<br></p><pdata−pxy="true"style="color:rgb(0,0,0);margin−top:2px;margin−bottom:2px;">Approximatevalue:</p><pstyle="color:rgb(0,0,0);"></p><pdata−pxy="true"style="color:rgb(0,0,0);margin−top:2px;margin−bottom:2px;">\frac{32\pi}{3}-16\sqrt3$