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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
Answer
- Draw the triangle XYZ on your paper. 2. Use a compass to construct the perpendicular bisector of side XY. To do this, place the compass point on point X, adjust the compass width to more than half the length of XY, and draw arcs above and below the line segment. Repeat this step from point Y. 3. Label the intersection points of the arcs as points P and Q. 4. Draw a straight line through points P and Q; this is the perpendicular bisector of XY. 5. Repeat the step for side YZ to find the perpendicular bisector, labeling its intersection points R and S, and drawing the line through R and S. 6. The point where the two perpendicular bisectors intersect is the circumcentre C of triangle XYZ.
Step 2
Find ∠ADC
Answer
Given that AB is a diameter and ∠LADBI = 90° (Inscribed angle theorem), we conclude:
- Since ∠LADBI is 90°, we know that triangle ADB is isosceles:
- Hence, ∠ABD = ∠ADB.
- The interior angles of triangle ABD sum to 180°:
egin{align*} ∠A + ∠B + ∠D & = 180°
ightarrow 40° + ∠ABD + ∠ADB = 180°
ightarrow 40° + 2∠ABD = 180°
ightarrow 2∠ABD = 140°
ightarrow ∠ABD = 70°
ightarrow ∠ADC = 90° - ∠LAC = 90° - 40° = 50°
ext{Thus, } ∠ADC = 50°
\end{align*}
Step 3
Prove that O cannot be inside the triangle PQR
Answer
Assume that O is inside the triangle PQR.
- The angle at O must be such that it creates subtended angles, thus: ∠PQR must be less than 180°.
- Given that ∠R is greater than 90°, we state that ∠PQR + ∠R > ∠OQR, which leads to a contradiction as the angles in triangle PQR cannot sum to more than 180° considering that ∠R itself exceeds 90°.
- Therefore, our initial assumption that O is inside the triangle PQR must be incorrect.