{"id":795709,"date":"2024-03-02T14:00:42","date_gmt":"2024-03-02T08:30:42","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=795709"},"modified":"2024-03-02T14:00:42","modified_gmt":"2024-03-02T08:30:42","slug":"ncert-class-7-maths-chapter-6-the-triangle-and-its-properties","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-class-7-maths-chapter-6-the-triangle-and-its-properties\/","title":{"rendered":"NCERT Solutions and Notes for Class 7 Maths Chapter 6: The Triangle and Its Properties (Free PDF)"},"content":{"rendered":"\n

In NCERT Class 7 Maths Chapter 6, The Triangle and Its Properties, we will study the concept of different triangles, their properties, and how to determine certain aspects of a triangle such as its median, altitude (or height of a triangle) and the exterior & interior angles of a triangle. Students will also learn the angle sum property of a triangle. In the later part of this chapter, students will learn about the elation and differences between an equilateral and an isosceles triangle in detail. Let us now have a look at the NCERT Class 7 Maths Chapter 6 The Triangle and Its Properties Notes and Solutions (PDF).\u00a0<\/p>\n\n\n\n

Download NCERT Class 7 Maths Chapter 6 – The Triangle and Its Properties Notes and Solutions PDF<\/strong><\/a><\/p>\n\n\n\n

Explore all the Chapters of Class Mathematics:-<\/strong><\/p>\n\n\n\n

Chapter 1<\/strong><\/a><\/td>Chapter 2<\/strong><\/a><\/td>Chapter 3<\/strong><\/a><\/td>Chapter 4<\/strong><\/a><\/td>Chapter 5<\/strong><\/a><\/td><\/tr>
Chapter 6<\/td>Chapter 7<\/strong><\/a><\/td>Chapter 8<\/strong><\/a><\/td>Chapter 9<\/strong><\/a><\/td>Chapter 10<\/strong><\/a><\/td><\/tr>
Chapter 11<\/strong><\/a><\/td>Chapter 12<\/strong><\/td>Chapter 13<\/strong><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n\n\n\n

NCERT Class 7 Maths Chapter 6 Notes – PDF Available<\/h2>\n\n\n\n

Check the topic-wise notes for NCERT Maths Class 7 – Chapter 6 below. You can also download the PDF of the notes and take a printout to study later when you need quick revision before going to the e\ud835\udc65am hall. <\/p>\n\n\n\n

Top 10 Short Maths Tricks for Comple\ud835\udc65 Problems<\/strong><\/p>\n\n\n\n

Topic 1: Medians of a Triangle<\/h3>\n\n\n\n

A median connects a vertex of a triangle to the mid-point of the opposite side.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

In the above diagram, AD is the median of the triangle ABC. <\/p>\n\n\n\n

Maths Books for Competitive E\ud835\udc65ams<\/strong><\/a><\/p>\n\n\n\n

Topic 2: Altitudes of a Triangle  <\/h3>\n\n\n\n

The height is given by the line segment that starts from the vertex of the triangle, comes straight down to the opposite side and is perpendicular to it.<\/p>\n\n\n\n

Topic 3: Exterior Angle of a Triangle and its Property<\/h3>\n\n\n\n

In the figure given below, \u2220ACD is the exterior angle. <\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Property of the exterior angle of a triangle: <\/strong>An exterior angle of a triangle is equal to the sum of its interior opposite angles.<\/p>\n\n\n\n

Topic 4: Angle Sum Property of a Triangle<\/h3>\n\n\n\n

This property establishes the relation between all the three angles of a triangle. The angle sum property of a triangle states that – the sum of the measures of the three angles of a triangle is 180\u00b0.<\/p>\n\n\n\n

Maths Formulas for Class 10<\/strong><\/p>\n\n\n\n

Topic 5: Case of Special Triangles <\/h3>\n\n\n\n
    \n
  1. Equilateral Triangle: <\/strong>A triangle in which all three sides are of equal lengths is called an equilateral triangle. In this triangle, each angle measures 60\u00b0. <\/li>\n\n\n\n
  2. Isosceles Triangle:<\/strong> A triangle in which two sides are of equal lengths is called an isosceles triangle. In this triangle, the base angles opposite to the equal sides are equal.<\/li>\n<\/ol>\n\n\n\n

    Topic 6: Sum of the Lengths of Two Sides of a Triangle<\/h3>\n\n\n\n

    The sum of the lengths of any two sides of a triangle is greater than the third side.<\/p>\n\n\n\n

    Topic 7: Right-Angled Triangles and Pythagoras Property<\/h3>\n\n\n\n
      \n
    1. Hypotenuse:<\/strong> In a right-angled triangle, the side opposite to the right angle is called the hypotenuse; the other two sides are known as the legs of the right-angled triangle.<\/li>\n\n\n\n
    2. Pythagoras Property:<\/strong> In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the legs. For any right-angled triangle, the Pythagoras property holds.<\/li>\n<\/ol>\n\n\n\n

      This property can also give a very useful result – if the Pythagoras property holds, the triangle must be right-angled.<\/p>\n\n\n\n

      Maths for Competitive E\ud835\udc65ams<\/strong><\/a><\/p>\n\n\n\n

      Download NCERT Class 7 Maths Chapter 6 – The Triangle and Its Properties Notes and Solutions PDF<\/strong><\/a><\/p>\n\n\n\n

      Explore all the Chapters of Class Mathematics:-<\/strong><\/p>\n\n\n\n

      Chapter 1<\/strong><\/a><\/td>Chapter 2<\/strong><\/a><\/td>Chapter 3<\/strong><\/a><\/td>Chapter 4<\/strong><\/a><\/td>Chapter 5<\/strong><\/a><\/td><\/tr>
      Chapter 6<\/td>Chapter 7<\/strong><\/a><\/td>Chapter 8<\/strong><\/a><\/td>Chapter 9<\/strong><\/a><\/td>Chapter 10<\/strong><\/a><\/td><\/tr>
      Chapter 11<\/strong><\/a><\/td>Chapter 12<\/strong><\/td>Chapter 13<\/strong><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

      NCERT Solutions of Class 7 Mathematics Chapter 6: The Triangle and Its Properties – Free PDF Download!\u00a0<\/h2>\n\n\n\n

      Below we have provided solutions for NCERT Class 7 Maths Chapter 6, The Triangle and Its Properties. Go through for answers to some important questions. <\/p>\n\n\n\n

      Courses and Career Options in Commerce Without Maths in 2023<\/strong><\/a><\/p>\n\n\n\n

      E\ud835\udc65ercise 6.1 Solutions<\/h3>\n\n\n\n

      Q 1. In \u0394 PQR, D is the mid-point of.<\/strong><\/p>\n\n\n\n

      \"\"<\/figure>\n\n\n\n

      <\/strong> is _________________.<\/strong><\/p>\n\n\n\n

      PD is _________________.<\/strong><\/p>\n\n\n\n

      Is QM = MR?<\/strong><\/p>\n\n\n\n

      Solutions.<\/strong> The answers are given below.<\/p>\n\n\n\n

        \n
      1. PM is the altitude of the triangle as it starts from the vertex of the triangle, comes straight down to the opposite side, and is perpendicular to it.<\/li>\n\n\n\n
      2. PD is the median of the triangle as one end is connected to the vertex P and another to the mid-point of the opposite side. <\/li>\n\n\n\n
      3. Since the median of the triangle, \u0394PQR cuts QR at point D, so, D is the mid-point of the line segment. <\/li>\n<\/ol>\n\n\n\n

        \u2234 QM \u2260 MR. <\/p>\n\n\n\n

        SAT Maths Subject Test<\/strong><\/p>\n\n\n\n

        Exercise 6.2 Solutions<\/h3>\n\n\n\n

        Q 1. Find the value of the unknown exterior angle \ud835\udc65 in the following diagrams:<\/strong><\/p>\n\n\n\n

          \n
        1. <\/strong><\/li>\n\n\n\n
        2. <\/strong><\/li>\n<\/ol>\n\n\n\n

          Solutions.<\/strong> The answers are given below.<\/p>\n\n\n\n

            \n
          1. We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.<\/li>\n<\/ol>\n\n\n\n

            \u2234 \ud835\udc65 = 65\u00b0 + 45\u00b0 \u21d2 \ud835\udc65 = 130\u00b0<\/p>\n\n\n\n

              \n
            1. Similar to (a), \ud835\udc65 = 50\u00b0 + 50\u00b0 \u21d2 \ud835\udc65 = 100\u00b0<\/li>\n<\/ol>\n\n\n\n

              NCERT Class 8 Maths Syllabus<\/strong><\/p>\n\n\n\n

              Q 3. Find the value of the unknown interior angle \ud835\udc65 in the following figures:<\/strong><\/p>\n\n\n\n

                \n
              1. <\/strong><\/li>\n\n\n\n
              2. <\/strong><\/li>\n\n\n\n
              3. <\/strong><\/li>\n\n\n\n
              4. <\/strong><\/li>\n<\/ol>\n\n\n\n

                Solutions. <\/strong>We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles. Now, the solutions to the given questions are given below.<\/p>\n\n\n\n

                  \n
                1. From the property stated above, we get:<\/li>\n<\/ol>\n\n\n\n

                  \ud835\udc65 + 50\u00b0 = 115\u00b0 \u21d2 \ud835\udc65 = 115\u00b0 – 50\u00b0 = 65\u00b0<\/p>\n\n\n\n

                  \u2234 \ud835\udc65 = 65\u00b0<\/p>\n\n\n\n

                    \n
                  1. From the property stated above, we get:<\/li>\n<\/ol>\n\n\n\n

                    \ud835\udc65 + 30\u00b0 = 80\u00b0 \u21d2 \ud835\udc65 = 80\u00b0 – 30\u00b0 = 50\u00b0<\/p>\n\n\n\n

                    \u2234 \ud835\udc65 = 50\u00b0 <\/p>\n\n\n\n

                      \n
                    1. From the property stated above, we get:<\/li>\n<\/ol>\n\n\n\n

                      \ud835\udc65 + 60\u00b0 = 120\u00b0 \u21d2 \ud835\udc65 = 120\u00b0 – 60\u00b0 = 60\u00b0<\/p>\n\n\n\n

                      \u2234 \ud835\udc65 = 60\u00b0<\/p>\n\n\n\n

                        \n
                      1. One of the interior angles (opposite to the exterior angle of the given triangle) measures 90\u00b0 (right-angled). So, from the property stated above, we get:<\/li>\n<\/ol>\n\n\n\n

                        \ud835\udc65 + 90\u00b0 = 125\u00b0 \u21d2 \ud835\udc65 = 125\u00b0 – 90\u00b0 = 35\u00b0<\/p>\n\n\n\n

                        \u2234 \ud835\udc65 = 35\u00b0<\/p>\n\n\n\n

                        Career after BSc Maths<\/strong><\/a><\/p>\n\n\n\n

                        E\ud835\udc65ercise 6.3 Solutions<\/h3>\n\n\n\n

                        Q 1. Find the value of the unknown \ud835\udc65 in the following diagrams:<\/strong><\/p>\n\n\n\n

                          \n
                        1. <\/strong><\/li>\n\n\n\n
                        2. <\/strong><\/li>\n<\/ol>\n\n\n\n

                          Solutions. <\/strong>The answers are given below.<\/p>\n\n\n\n

                            \n
                          1. We know that the sum of all three interior angles of a triangle is 180\u00b0. So, from the given figure, we get:<\/li>\n<\/ol>\n\n\n\n

                            \ud835\udc65 + 50\u00b0 + 60\u00b0 = 180\u00b0<\/p>\n\n\n\n

                            \u2234 \ud835\udc65 =  180\u00b0 – 50\u00b0 – 60\u00b0<\/p>\n\n\n\n

                            \u21d2 \ud835\udc65 =  180\u00b0 – 110\u00b0 = 70\u00b0<\/p>\n\n\n\n

                            Hence, \ud835\udc65 =  70\u00b0<\/p>\n\n\n\n

                              \n
                            1. The given triangle is right-angled. So, from the given figure, we get:<\/li>\n<\/ol>\n\n\n\n

                              \ud835\udc65 + 2\ud835\udc65 + 90\u00b0 = 180\u00b0<\/p>\n\n\n\n

                              \u2234 3\ud835\udc65 =  180\u00b0 – 90\u00b0<\/p>\n\n\n\n

                              \u21d2 3\ud835\udc65 =  90\u00b0 <\/p>\n\n\n\n

                              \u21d2 \ud835\udc65 = 90\u00b0\/3 = 30\u00b0<\/p>\n\n\n\n

                              So, \ud835\udc65 =  30\u00b0 and 2\ud835\udc65 =  30\u00b0 \u00d7 2 = 60\u00b0<\/p>\n\n\n\n

                              Hence, the three angles in the given triangle measure 30\u00b0, 60\u00b0 and 90\u00b0. <\/p>\n\n\n\n

                              \n