
A triangle is inscribed in a circle of radius $1$. The distance between the orthocentre and the circumcentre of the triangle cannot be
A. $\dfrac{1}{2}$
B. $2$
C. $\dfrac{3}{2}$
D. $4$
Answer
495.9k+ views
Hint:The orthocentre is the point at which the triangle's three altitudes meet or converge. There are three altitudes since the triangle has three vertices and three sides. The point where the perpendicular bisectors of the triangle's sides converge is known as the circumcentre.
Complete step by step answer:
If each vertex of a form lies on the circle, it is said to be inscribed in the circle. The triangle is a right triangle if it is inscribed inside of a circle and the base of the triangle is also a diameter of the circle. An obtuse-angled triangle's circumcentre is outside the triangle. The circumcentre of a right-angled triangle is on the hypotenuse.
In the case of an acute triangle, the orthocentre is located inside the triangle. It is located on the outside of an obtuse triangle. Orthocentre is located at the right angle's vertex in a right-angled triangle.
The inequality for a triangle is given by,
$HO \leqslant 3R$
where, $H$ denotes the orthocentre of the triangle, $O$ denotes the circumcentre of the triangle and $R$ denotes the circumradius of the triangle.
Here, it is given that the radius of the circumcircle in which the triangle is inscribed is $R = 1$. Substituting $R = 1$ in the inequality we have,
\[ \Rightarrow HO \leqslant 3\left( 1 \right)\]
\[ \therefore HO \leqslant 3\]
So, HO line segment should be less than or equal to $3$. Hence, the distance between the orthocentre and the circumcentre of the triangle cannot be $4$.
So, option D is the correct answer.
Note: We must have knowledge about the terms orthocentre and circumcentre in order to solve the given question. We must also know the inequality used in the problem. We should take care of the calculations as it affects the final answer of the problem. One must have a clear understanding about the differences between the circumradius and inradius of a triangle.
Complete step by step answer:
If each vertex of a form lies on the circle, it is said to be inscribed in the circle. The triangle is a right triangle if it is inscribed inside of a circle and the base of the triangle is also a diameter of the circle. An obtuse-angled triangle's circumcentre is outside the triangle. The circumcentre of a right-angled triangle is on the hypotenuse.
In the case of an acute triangle, the orthocentre is located inside the triangle. It is located on the outside of an obtuse triangle. Orthocentre is located at the right angle's vertex in a right-angled triangle.
The inequality for a triangle is given by,
$HO \leqslant 3R$
where, $H$ denotes the orthocentre of the triangle, $O$ denotes the circumcentre of the triangle and $R$ denotes the circumradius of the triangle.
Here, it is given that the radius of the circumcircle in which the triangle is inscribed is $R = 1$. Substituting $R = 1$ in the inequality we have,
\[ \Rightarrow HO \leqslant 3\left( 1 \right)\]
\[ \therefore HO \leqslant 3\]
So, HO line segment should be less than or equal to $3$. Hence, the distance between the orthocentre and the circumcentre of the triangle cannot be $4$.
So, option D is the correct answer.
Note: We must have knowledge about the terms orthocentre and circumcentre in order to solve the given question. We must also know the inequality used in the problem. We should take care of the calculations as it affects the final answer of the problem. One must have a clear understanding about the differences between the circumradius and inradius of a triangle.
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