# The Triangle and Its Properties Chapter 6 Class 7 Maths NCERT PDF

NCERT Solutions for Class 7 Maths Chapter 6′ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 7 Maths Chapter 6 Exercise Solution’ using the download button.

### Chapter 3: The Triangle and its Properties Textbook With Solution

#### 6.8 Right angled Triangles And Pythagoras Property

Pythagoras, a Greek philosopher of sixth century B.C. is said to have found a very important and useful property of right-angled triangles given in this section.

The property is, hence, named after him. In fact, this property was known to people of many other countries too.

The Indian mathematician Baudhayan has also given an equivalent form of this property. We now try to explain the Pythagoras property.

In a right-angled triangle, the sides have some special names. 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.

In ΔABC (Fig 6.23), the right-angle is at B. So, AC is the hypotenuse. AB and BC are the legs of ΔABC.
Make eight identical copies of a right-angled triangle of any size you prefer.

For example, you make a right-angled triangle whose hypotenuse is a unit long and the legs are of lengths b units and c units (Fig 6.24).

Draw two identical squares on a sheet with sides of lengths b + c. You are to place four triangles in one square and the remaining four triangles in the other square, as shown in the following diagram (Fig 6.25)

### NCERT Solutions Class 7 Maths Chapter 3 The Triangle and its Properties

Q1. Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm (ii) 3 cm, 6 cm, 7 cm (iii) 6 cm, 3 cm, 2 cm

This problem is based on a very simple property that if the sum of lengths of any two sides of a triangle is greater than the third side, then it is possible to draw a triangle.

By using this property, you just have to check by taking the sum of any two sides of a triangle whether it is greater than the third side or not.

For better understanding take any two sides one by one and check if some of these two sides is greater than the third side or not, if it is greater than the third side then it is possible to draw a triangle.
(i) 2 + 3 > 5 No
3 + 5 > 2 Yes
5 + 2 > 3 Yes
No, the triangle is not possible
(ii) 3 + 6 > 7 yes
6 + 7 > 3 yes
7 + 3 > 3 yes
Yes, the triangle is possible
(iii) 6 + 3 > 2 yes
3 + 2 > 6 No
2 + 6 > 3 Yes
No, the triangle is not possible

Q 2. Draw rough sketches for the following:
(i) In ∆ABC, BE is a median.
(ii) In ∆PQR, PQ and PR are altitudes of the triangle.
(iii) In ∆XYZ, YL is an altitude in the exterior of the triangle.

Difficulty level: Easy
What is known:
Statement.
What is unknown:
Rough sketch for given statement.

Reasoning:
Read the statement carefully and draw a rough figure according to the statement.
Solution:
(i) Here BE is a median in triangle ABC therefore AE = EC.