In figure 7.43, A is the centre of the circle. ∠ABC = 45° and AC = 7√2 cm. Find the area of segment BXC.
Figure: B and C are points on the circle with centre A; X is on the minor arc BC; segment BXC is the minor segment cut off by chord BC.
Answer: ≈ 27.93 cm² (π = 3.14), or 28 cm² with π = 22/7
AB = AC (radii), so \(\angle ACB=\angle ABC=45^\circ\) and \(\angle BAC=90^\circ\). Radius \(r=7\sqrt2\), \(r^2=98\).
Area of segment = area of sector − area of triangle formed by the two radii and the chord.
\[A(A\text{-}BXC)=\frac{90}{360}\times3.14\times98=76.93\]
\[A(\Delta ABC)=\tfrac12\times7\sqrt2\times7\sqrt2=49\]
\[A(\text{segment})=76.93-49=27.93\ \text{cm}^2\]
Note: the textbook answer key prints 3.92 sq.cm, which does not match these data; the working above gives about 27.93 cm².
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