Exercise 5.1 Practice
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Overview
This page provides comprehensive Ch 5: Circles - Exercise Set 5.1 Practice. Solve questions on triangle circumcircle construction, circumcentre location properties in acute and obtuse triangles, circumradius measurements, and the minimum radius of a circle passing through two points with step-by-step solutions.
Constructing Circumcircles & Circumcentre Locations
Q1: Circumcentre of Acute Triangle
Draw $\Delta ABC$ with $AB = 5\text{ cm}$, $\angle A = 70^\circ$ and $\angle B = 60^\circ$. Draw the circumcircle of $\Delta ABC$. Is the centre inside or outside the triangle?
Step 1: Determine the third angle:
$$\angle C = 180^\circ - (\angle A + \angle B) = 180^\circ - (70^\circ + 60^\circ) = 50^\circ$$
Since all three angles of $\Delta ABC$ ($70^\circ, 60^\circ, 50^\circ$) are acute (less than $90^\circ$), the triangle is an **acute-angled triangle**.
$$\angle C = 180^\circ - (\angle A + \angle B) = 180^\circ - (70^\circ + 60^\circ) = 50^\circ$$
Since all three angles of $\Delta ABC$ ($70^\circ, 60^\circ, 50^\circ$) are acute (less than $90^\circ$), the triangle is an **acute-angled triangle**.
Step 2: Circumcentre location rule:
• For any acute-angled triangle, the perpendicular bisectors of the sides intersect at a point (the circumcentre) that lies **inside** the triangle.
• Therefore, the circumcentre $O$ of this triangle lies **inside** $\Delta ABC$.
• For any acute-angled triangle, the perpendicular bisectors of the sides intersect at a point (the circumcentre) that lies **inside** the triangle.
• Therefore, the circumcentre $O$ of this triangle lies **inside** $\Delta ABC$.
Circumcentre lies INSIDE the triangle.
Q2: Circumcentre of Obtuse Triangle
Draw $\Delta ABC$ with $AB = 5\text{ cm}$, $\angle A = 100^\circ$, $AC = 4\text{ cm}$. Draw the circumcircle of $\Delta ABC$. Is the centre inside or outside the triangle?
Step 1: Classify the triangle:
Since $\angle A = 100^\circ$ is greater than $90^\circ$, $\Delta ABC$ is an **obtuse-angled triangle**.
Since $\angle A = 100^\circ$ is greater than $90^\circ$, $\Delta ABC$ is an **obtuse-angled triangle**.
Step 2: Circumcentre location rule:
• For any obtuse-angled triangle, the perpendicular bisectors of the sides intersect at a point (the circumcentre) that lies **outside** the triangle, positioned opposite to the obtuse vertex.
• Therefore, the circumcentre $O$ of this triangle lies **outside** $\Delta ABC$.
• For any obtuse-angled triangle, the perpendicular bisectors of the sides intersect at a point (the circumcentre) that lies **outside** the triangle, positioned opposite to the obtuse vertex.
• Therefore, the circumcentre $O$ of this triangle lies **outside** $\Delta ABC$.
Circumcentre lies OUTSIDE the triangle.
Q3: Circumradius of Isosceles Triangle
Draw $\Delta ABC$, with $AB = 6\text{ cm}$, $BC = 7\text{ cm}$ and $CA = 7\text{ cm}$. Draw the circumcircle of $\Delta ABC$. Let the circumcentre be $O$. Measure $OA, OB, OC$.
By definition, the circumcentre $O$ is equidistant from all three vertices of the triangle. Thus, the segments $OA$, $OB$, and $OC$ represent the circumradius $R$ of the circle:
$$OA = OB = OC = R$$
$$OA = OB = OC = R$$
We can compute the exact circumradius $R$ using the side lengths $a = 7$, $b = 7$, $c = 6$:
• Semiperimeter $s = \frac{7 + 7 + 6}{2} = 10\text{ cm}$.
• Area of triangle (Heron's Formula):
$$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{10 \times 3 \times 3 \times 4} = \sqrt{360} = 6\sqrt{10} \approx 18.97\text{ sq. cm}$$
• Semiperimeter $s = \frac{7 + 7 + 6}{2} = 10\text{ cm}$.
• Area of triangle (Heron's Formula):
$$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{10 \times 3 \times 3 \times 4} = \sqrt{360} = 6\sqrt{10} \approx 18.97\text{ sq. cm}$$
• Circumradius formula $R = \frac{abc}{4\Delta}$:
$$R = \frac{7 \times 7 \times 6}{4 \times 6\sqrt{10}} = \frac{49}{4\sqrt{10}} = \frac{49\sqrt{10}}{40} \approx \mathbf{3.87\text{ cm}}$$
Therefore, the measured values are:
$$OA = OB = OC \approx \mathbf{3.87\text{ cm}}$$
$$R = \frac{7 \times 7 \times 6}{4 \times 6\sqrt{10}} = \frac{49}{4\sqrt{10}} = \frac{49\sqrt{10}}{40} \approx \mathbf{3.87\text{ cm}}$$
Therefore, the measured values are:
$$OA = OB = OC \approx \mathbf{3.87\text{ cm}}$$
OA = OB = OC ≈ 3.87 cm
Q4: Least Radius through Two Points
What is the least possible radius of a circle through two points $A$ and $B$?
Let $A$ and $B$ be two points separated by a distance $d$.
• For any circle passing through both $A$ and $B$, the segment $AB$ forms a chord of the circle.
• The length of a chord in a circle can never exceed the diameter of the circle:
$$\text{Diameter } D \ge AB \implies 2r \ge AB \implies r \ge \frac{AB}{2}$$
• For any circle passing through both $A$ and $B$, the segment $AB$ forms a chord of the circle.
• The length of a chord in a circle can never exceed the diameter of the circle:
$$\text{Diameter } D \ge AB \implies 2r \ge AB \implies r \ge \frac{AB}{2}$$
The minimum radius is achieved when the diameter is exactly equal to the distance $AB$. In this case, the center of the circle is the midpoint of $AB$, and the radius is:
$$r_{min} = \mathbf{\frac{AB}{2}}$$
$$r_{min} = \mathbf{\frac{AB}{2}}$$
Least radius is half the distance AB (i.e. AB/2)