Draw , with , and . Draw the circumcircle of . Let the circumcentre be . Measure , , .

Understand the Question The circumcentre () of a triangle is the point of intersection of the perpendicular bisectors of its sides. The circumcentre is equidistant from all three vertices of the triangle, meaning (the circumradius). The circle drawn with centre and radius passes through all three vertices , , and , forming the circumcircle. Step 1 · Construct the Triangle 1. Draw a line segment . 2. With as centre and radius , draw an arc. 3. With as centre and radius , draw another arc intersecting the first arc at point . 4. Join and to complete . Step 2 · Locate the Circumcentre 1. Draw the perpendicular bisectors of sides , , and . 2. Mark the point where these perpendicular bisectors intersect as . Point is the circumcentre of . Step 3 · Draw the Circumcircle 1. With as centre and radius equal to , draw a circle. 2. The circle passes through all three vertices , , and . Step 4 · Measure the Radii Measuring the distances from the circumcentre to each vertex: All three lengths are equal: . Answer > Common Mistakes Incentre vs. Circumcentre: Using angle bisectors instead of perpendicular bisectors. Angle bisectors give the incentre (for an incircle), while perpendicular bisectors give the circumcentre (for a circumcircle). Compass Radius Shift: Letting the compass width slip while drawing arcs for the perpendicular bisectors or sides, which shifts the circumcentre away from being equidistant to all vertices. Inaccurate Arc Intersections: Not opening the compass to more than half the length of a side when constructing perpendicular bisectors, resulting in arcs that do not intersect.

Circles and Geometric Shapes | Exercise 5.1

Question 3

Draw $\Delta \text{ABC}$ , with $\text{AB} = 6\text{ cm}$ , $\text{BC} = 7\text{ cm}$ and $\text{CA} = 7\text{ cm}$ . Draw the circumcircle of $\Delta \text{ABC}$ . Let the circumcentre be $O$ . Measure $\text{OA}$ , $\text{OB}$ , $\text{OC}$ .

Solution

Understand the Question

The circumcentre ( $O$ ) of a triangle is the point of intersection of the perpendicular bisectors of its sides.
The circumcentre is equidistant from all three vertices of the triangle, meaning $\text{OA} = \text{OB} = \text{OC} = R$ (the circumradius).
The circle drawn with centre $O$ and radius $\text{OA}$ passes through all three vertices $A$ , $B$ , and $C$ , forming the circumcircle.

Step 1 · Construct the Triangle

Diagram 1

Draw a line segment $\text{AB} = 6\text{ cm}$ .
With $A$ as centre and radius $7\text{ cm}$ , draw an arc.
With $B$ as centre and radius $7\text{ cm}$ , draw another arc intersecting the first arc at point $C$ .
Join $\text{AC}$ and $\text{BC}$ to complete $\Delta \text{ABC}$ .

Step 2 · Locate the Circumcentre

Draw the perpendicular bisectors of sides $\text{AB}$ , $\text{BC}$ , and $\text{AC}$ .
Mark the point where these perpendicular bisectors intersect as $O$ .

Point $O$ is the circumcentre of $\Delta \text{ABC}$ .

Step 3 · Draw the Circumcircle

Diagram 3

With $O$ as centre and radius equal to $\text{OA}$ , draw a circle.
The circle passes through all three vertices $A$ , $B$ , and $C$ .

Step 4 · Measure the Radii

Measuring the distances from the circumcentre $O$ to each vertex:

\begin{aligned} \text{OA} &\approx 4\text{ cm} \\ \text{OB} &\approx 4\text{ cm} \\ \text{OC} &\approx 4\text{ cm} \end{aligned}

All three lengths are equal: $\text{OA} = \text{OB} = \text{OC} \approx 4\text{ cm}$ .

Answer

$\text{OA} \approx 4\text{ cm}, \quad \text{OB} \approx 4\text{ cm}, \quad \text{OC} \approx 4\text{ cm}$

Common Mistakes

Incentre vs. Circumcentre: Using angle bisectors instead of perpendicular bisectors. Angle bisectors give the incentre (for an incircle), while perpendicular bisectors give the circumcentre (for a circumcircle).
Compass Radius Shift: Letting the compass width slip while drawing arcs for the perpendicular bisectors or sides, which shifts the circumcentre away from being equidistant to all vertices.
Inaccurate Arc Intersections: Not opening the compass to more than half the length of a side when constructing perpendicular bisectors, resulting in arcs that do not intersect.

Question 3

More questions in Exercise 5.1