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Construct the circumcentre of the triangle XYZ shown below, using only a compass and straight edge - Leaving Cert Mathematics - Question 6 - 2022
Question 6
Construct the circumcentre of the triangle XYZ shown below, using only a compass and straight edge. Label the circumcentre C. Show your construction lines clearly. ... show full transcript
Worked Solution & Example Answer:Construct the circumcentre of the triangle XYZ shown below, using only a compass and straight edge - Leaving Cert Mathematics - Question 6 - 2022
Step 1
Construct the circumcentre of the triangle XYZ
Answer
- Begin by drawing the triangle XYZ on your paper.
- Use the compass to construct the perpendicular bisector of segment XY.
- To do this, place the compass point on point X and draw an arc above and below the line.
- Without changing the compass width, place the compass point on point Y and draw two arcs, intersecting the first two arcs.
- Label the intersection points of the arcs as points P and Q.
- Draw a straight line through points P and Q. This will be the perpendicular bisector of XY.
- Repeat steps 2 to 4 for segment XZ to find its perpendicular bisector, and label its intersection points as R and S.
- The point where the two perpendicular bisectors intersect is the circumcentre C of triangle XYZ.
Step 2
Find |ADC|.
Answer
To find |ADC|:
- Note that since [AB] is the diameter, |ADB| = 90° (angles in a semi-circle).
- Since triangle ADB is isosceles, we have: |BAD| = |ABD|.
- Given that |DAC| = 40°, we can set up the equation: |ADC| = 90° - |BAD|.
- Since |BAD| and |ABD| are equal, we can denote |BAD| = x.
- The equation becomes: .
- Simplifying gives: .
- Thus, → .
- Therefore, |ADC| = 90° - 25° = 65°.
Step 3
Prove that O cannot be inside the triangle PQR.
Answer
Assume that O is inside the triangle PQR.
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The angle at the centre I, ∠[P O Q], must be twice the angle at point R: |PQR|, hence:
∠[POQ] must be > 180°.
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However, since we know that the angle at R is greater than 90°, it follows that: |PQR| must also be greater than 90°.
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This leads to a contradiction, as the circumcentre O cannot exist within the triangle when the angle exceeds 90°.
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Therefore, it must be concluded that O cannot be inside triangle PQR.