Answer
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Hint: In the above problem, we are asked to find the number of altitudes that a triangle has. So, the number of possible altitudes can be seen as a triangle that has three vertices and three opposite sides with respect to those three vertices. And we know that altitude is a line segment from any vertex to the opposite side in such a way so that the line segment will make a right angle with the opposite vertex. In this way, you can draw the altitudes on a triangle and then count those altitudes.
Complete step-by-step solution:
Let us draw a triangle first. So, below, we have drawn a triangle ABC as follows:
Now, we are asked to find the altitudes on this triangle. For that, we should know how altitudes are drawn inside the triangle. An altitude is drawn from any of the three vertices of a triangle onto the opposite side of the triangle and the angle that altitude is making with the opposite side is ${{90}^{\circ }}$. Keeping this concept of altitude in the mind, we are going to draw altitudes in the above triangle. First of all, we are going to draw the altitude from vertex A onto the side BC as follows:
From the above figure, you can see that we have drawn AD on the side BC and AD is the altitude. Similarly, we are drawing altitudes from the vertices B and C we get,
In the above figure, we have drawn two other altitudes BH and CI. Hence, a total of 3 altitudes are possible in the above triangle.
Hence, in a triangle a total of 3 altitudes are possible.
Note: An important point that we are sharing here is that the intersection point of the three altitudes is the orthocenter of the triangle is known as the orthocenter of the triangle. In geometry, you can find various problems related to the orthocenter so this is information which you will find in numerous applications.
Complete step-by-step solution:
Let us draw a triangle first. So, below, we have drawn a triangle ABC as follows:
Now, we are asked to find the altitudes on this triangle. For that, we should know how altitudes are drawn inside the triangle. An altitude is drawn from any of the three vertices of a triangle onto the opposite side of the triangle and the angle that altitude is making with the opposite side is ${{90}^{\circ }}$. Keeping this concept of altitude in the mind, we are going to draw altitudes in the above triangle. First of all, we are going to draw the altitude from vertex A onto the side BC as follows:
From the above figure, you can see that we have drawn AD on the side BC and AD is the altitude. Similarly, we are drawing altitudes from the vertices B and C we get,
In the above figure, we have drawn two other altitudes BH and CI. Hence, a total of 3 altitudes are possible in the above triangle.
Hence, in a triangle a total of 3 altitudes are possible.
Note: An important point that we are sharing here is that the intersection point of the three altitudes is the orthocenter of the triangle is known as the orthocenter of the triangle. In geometry, you can find various problems related to the orthocenter so this is information which you will find in numerous applications.
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