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

In <math>\triangle</math> PQR, D is given as the mid-point 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 asks about PM. Without further information about point M, we cannot definitively state what PM is. However, if PM were an altitude, it would be a line segment from P perpendicular to QR. If M were the midpoint of QR, then PM would be a median, but D is already defined as the midpoint.

Regarding the question 'Is <math>QM = MR</math>?', since D is the midpoint of QR, it divides QR into two equal segments. Thus, QD = DR. The question asks about QM and MR, which relate to a point M. If M is the same point as D, then QM would not necessarily equal MR unless M is the midpoint. However, if the question implicitly assumes M is the midpoint (like D), then QM = MR would be true. Given D is the midpoint, the statement QM = MR is only true if M is also the midpoint, which is D.

Answer: PD is a median. The nature of PM depends on the definition of M. If M is the midpoint of QR, then QM = MR. However, D is explicitly stated as the midpoint, so QD = DR.

1. For BE to be a median in <math>\triangle</math> ABC: A median connects a vertex to the midpoint of the opposite side. So, E must be the midpoint of side AC. This means AE = EC. Draw a triangle ABC and mark point E on AC such that AE = EC. Then draw the line segment BE.
2. For PQ and PR to be altitudes of <math>\triangle</math> PQR: An altitude is perpendicular to the opposite side. If PQ is an altitude, it must be perpendicular to QR. If PR is an altitude, it must be perpendicular to PQ. This implies that <math>\angle&space;PQR&space;=&space;90^{\circ}</math> and <math>\angle&space;RPQ&space;=&space;90^{\circ}</math>. This is only possible if the triangle is degenerate or if P is the vertex with the right angle and Q and R are on the other two vertices. However, the standard interpretation is that PQ is perpendicular to QR and PR is perpendicular to PQ. This means <math>\angle&space;Q&space;=&space;90^{\circ}</math> and <math>\angle&space;R&space;=&space;90^{\circ}</math>, which is impossible in a Euclidean triangle as the sum of angles would exceed 180 degrees. A more likely interpretation is that PQ is an altitude from P to QR (so PQ <math>\perp</math> QR) and PR is an altitude from P to QR (so PR <math>\perp</math> QR). This is also impossible unless Q, R, and the foot of the altitude coincide. If PQ and PR are altitudes, it usually means they are altitudes from different vertices to opposite sides. Let's assume PQ is the altitude from P to QR, and PR is the altitude from R to PQ. This implies <math>\angle&space;PQR&space;=&space;90^{\circ}</math> and <math>\angle&space;RPQ&space;=&space;90^{\circ}</math>, which is impossible. The sketch usually implies a right-angled triangle at P, where PQ is one side and PR is another side, and the altitudes from Q and R would fall outside. If PQ and PR are altitudes, it means PQ <math>\perp</math> QR and PR <math>\perp</math> PQ. This implies <math>\angle&space;Q&space;=&space;90^{\circ}</math> and <math>\angle&space;P&space;=&space;90^{\circ}</math>, which is impossible. The sketch provided in the source shows a right-angled triangle at P, where PQ is perpendicular to PR. In this case, PQ is the altitude from Q to PR, and PR is the altitude from R to PQ.
3. For YL to be an altitude in the exterior of <math>\Delta</math> XYZ: An exterior altitude occurs in obtuse triangles. Draw an obtuse triangle XYZ, where the angle at X is obtuse. To draw the altitude from Y to the side XZ, you need to extend the side XZ beyond X. Then, draw a line segment YL from Y perpendicular to the extended line XZ. L will be outside the triangle.

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 need to check if the median and the altitude drawn from the same vertex can be the same line segment.

Let's draw the median from vertex A to the opposite side BC. Let D be the midpoint of BC. Then AD is the median. By definition of a midpoint, BD = DC.

Now, let's consider the altitude from vertex A to the opposite side BC. Let AM be the altitude. By definition of an altitude, AM must be perpendicular to BC, meaning <math>\angle&space;AMB&space;=&space;90^{\circ}</math>.

In an isosceles triangle ABC with AB = AC, the median AD drawn to the base BC is also the altitude to BC. This can be proven using congruent triangles. Consider <math>\triangle</math> ABD and <math>\triangle</math> ACD:

• AB = AC (Given, isosceles triangle)
• BD = DC (D is the midpoint, so AD is a median)
• AD = AD (Common side)

By SSS congruence criterion, <math>\triangle</math> ABD <math>\cong</math> <math>\triangle</math> ACD.

Since the triangles are congruent, their corresponding angles are equal. Therefore, <math>\angle&space;ADB&space;=&space;\angle&space;ADC</math>.

Also, <math>\angle&space;ADB&space;+&space;\angle&space;ADC&space;=&space;180^{\circ}</math> (linear pair).

Substituting <math>\angle&space;ADC</math> with <math>\angle&space;ADB</math>, we get:

<math>\angle&space;ADB&space;+&space;\angle&space;ADB&space;=&space;180^{\circ}</math>

<math>2&space;\angle&space;ADB&space;=&space;180^{\circ}</math>

<math>\angle&space;ADB&space;=&space;90^{\circ}</math>

Since <math>\angle&space;ADB&space;=&space;90^{\circ}</math>, AD is perpendicular to BC. Thus, AD is the altitude from A to BC.

We have shown that AD is both the median and the altitude from vertex A to the base BC in the isosceles triangle ABC.

Verification: Yes, by drawing a diagram of an isosceles triangle ABC with AB = AC, we can verify that the median AD drawn from vertex A to the base BC is also the altitude from A to BC, as it is perpendicular to BC. A

B D C