On sports day, flags \(S\) and \(F\) are fixed 80 m apart along a sideline. Tent \(T\) sits on the perpendicular bisector of \(SF\), 30 m from midpoint \(M\...
Question 1 Report
On sports day, flags \(S\) and \(F\) are fixed 80 m apart along a sideline. Tent \(T\) sits on the perpendicular bisector of \(SF\), 30 m from midpoint \(M\), shown in the diagram.
- State the property a point on the bisector has, relative to \(S\) and \(F\). (1)
- Work out the distance from \(T\) to \(S\). (3)
- State the radius that would fix \(T\) by construction, centred at \(S\). (1)
- The tent must be at least 45 m from each flag. Determine whether \(T\) satisfies this. (2)
Answer Details
Every point on the perpendicular bisector of a line segment is the same distance from both endpoints, and this shared distance can be found using Pythagoras' theorem once the perpendicular offset and the half-length of the segment are known.
(a) By definition, a point on the perpendicular bisector of \(SF\) is equidistant from \(S\) and \(F\). [1 mark]
(b) Since \(SF=80\) m, the midpoint distance is \(MS=80\div2=40\) m. \(T\) sits 30 m from \(M\) along the perpendicular, so triangle \(TMS\) is right-angled at \(M\), giving:
\[TS=\sqrt{TM^{2}+MS^{2}}=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50 \text{ m}\] [3 marks](c) An arc of radius equal to \(TS\), drawn centred at \(S\), would pass through \(T\), so the radius that fixes \(T\) by construction, centred at \(S\), is 50 m. [1 mark]
(d) By the perpendicular bisector property from part (a), \(TF=TS=50\) m as well. Since the requirement is at least 45 m from each flag, and \(50 \geq 45\), the tent's position satisfies the rule. [2 marks]
Because \(T\) lies exactly on the perpendicular bisector, finding its distance to just one flag, \(S\), using Pythagoras' theorem automatically gives its distance to the other flag, \(F\), too, without a second calculation.
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