CBSE Class 7 Mathematics Chapter 6: The Triangles and its Properties NCERT Solutions

In the given triangle PQR, D is the midpoint of the side QR. This means that the line segment PD connects the vertex P to the midpoint D of the opposite side QR. By definition, a line segment from a vertex to the midpoint of the opposite side is called a median.

Therefore, PD is a median.

The question also mentions PM. Without further information about M, we cannot definitively state what PM is. However, if M were a point such that PM is perpendicular to QR, then PM would be an altitude. Based on the provided context and typical triangle properties, it's likely that the question intends to ask about the nature of PD.

Regarding the question 'Is <math>QM = MR</math>?', since D is the midpoint of QR, it divides QR into two equal segments. Therefore, QD = DR. The question asks about QM and MR, which implies M is a different point. If M is not the midpoint D, then QM is generally not equal to MR. However, if M is intended to be D, then QM = MR would be true because D is the midpoint. Assuming M is a distinct point, and D is the midpoint, then QM <math>\neq</math> MR unless M coincides with D.

Answer: PD is a median. The nature of PM depends on the definition of M. Generally, <math>QM \neq MR</math> unless M is the midpoint D.

1. In <math>\triangle</math> ABC, BE is a median:

   A median connects a vertex to the midpoint of the opposite side. For BE to be a median in <math>\triangle</math> ABC, E must be the midpoint of side AC. This means AE = EC. We draw a triangle ABC and mark point E on AC such that E is exactly in the middle of A and C. Then, we draw a line segment from vertex B to point E.

   

   E

2. In <math>\triangle</math> PQR, PQ and PR are altitudes of the triangle:

   An altitude is a line segment from a vertex that is perpendicular to the opposite side. For PQ and PR to be altitudes, PQ must be perpendicular to QR, and PR must be perpendicular to PQ. This condition is only possible if the angle at P is a right angle (<math>\angle P = 90^{\circ}</math>). In this case, PQ is the altitude from P to PR (or its extension), and PR is the altitude from P to PQ (or its extension). More accurately, if <math>\angle P = 90^{\circ}</math>, then PQ is the altitude from Q to PR, and PR is the altitude from R to PQ. If the question implies PQ and PR are altitudes *from* P, this is only possible if the triangle is degenerate or if P is a right angle and Q and R are on the lines forming the right angle. A more standard interpretation is that PQ is an altitude to side PR, and PR is an altitude to side PQ, which implies <math>\angle P = 90^{\circ}</math>. In a right-angled triangle at P, PQ is the altitude from Q to PR, and PR is the altitude from R to PQ.

   

   P

3. In <math>\Delta</math> XYZ, YL is an altitude in the exterior of the triangle:

   An altitude from a vertex is perpendicular to the opposite side. For YL to be an exterior altitude, it must be drawn from vertex Y and be perpendicular to the line containing the opposite side XZ. This typically happens in an obtuse-angled triangle where the angle at X or Z is obtuse. If <math>\angle X</math> is obtuse, the altitude from Y to XZ will fall outside the triangle. We draw a line segment YL such that L is on the extension of XZ, and YL is perpendicular to the line XZ.

   

An isosceles triangle is a triangle that has at least two sides of equal length. Let's consider an isosceles triangle ABC where AB = AC. We want to check if the median and the altitude drawn from the same vertex can be the same line segment.

Let's draw a median from vertex A to the base BC. Let this median be AL, where L is the midpoint of BC. By definition of a median, L is the midpoint of BC, so BL = LC.

Now, let's consider the altitude from vertex A to the base BC. Let this altitude be AM, where M is a point on BC such that AM is perpendicular to BC (<math>\angle AMB = 90^{\circ}</math>).

In an isosceles triangle ABC with AB = AC, the median AL and the altitude AM drawn from vertex A to the base BC are indeed the same line segment. This can be proven by considering the two triangles formed: <math>\triangle ABL</math> and <math>\triangle ACL</math> (if L is the midpoint) or <math>\triangle ABM</math> and <math>\triangle ACM</math> (if AM is the altitude).

Consider <math>\triangle ABM</math> and <math>\triangle ACM</math>. We have:

• AB = AC (Given, as <math>\triangle ABC</math> is isosceles)
• <math>\angle AMB = \angle AMC = 90^{\circ}</math> (AM is the altitude)
• AM = AM (Common side)

By the RHS (Right angle-Hypotenuse-Side) congruence criterion, <math>\triangle ABM \cong \triangle ACM</math>. This implies that BM = CM, so M is the midpoint of BC. Therefore, the altitude AM is also the median AL (since L and M are the same point).

Diagram: Α

B L/M C

Conclusion: Yes, the median and the altitude drawn from the same vertex to the base of an isosceles triangle can be the same line segment.