NCERT Solutions Class 7th Maths Chapter – 6 The Triangle and its Properties Example

Example 1. Find angle x in Fig 6.11.

Solution – Sum of interior opposite angles = Exterior angle
or 50° + x = 110°
or x = 60°

Example 2. In the given figure (Fig 6.18) find m≤P.

Solution – By angle sum property of a triangle,
m∠P+47° + 52° = 180°
Therefore
m∠P= 180°- 47°-52°
= 180°-99° = 81°

Example 3. Is there a triangle whose sides have lengths 10.2 cm, 5.8 cm and 4.5 cm?

Solution – Suppose such a triangle is possible. Then the sum of the lengths of any two
sides would be greater than the length of the third side. Let us check this.
Is 4.5+ 5.8 >10.2? Yes
Is 5.8 + 10.2 > 4.5? Yes
Is 10.2 + 4.5 > 5.8? Yes
Therefore, the triangle is possible.

Example 4. The lengths of two sides of a triangle are 6 cm and 8 cm. Between which
two numbers can length of the third side fall?

Solution – We know that the sum of two sides of a triangle is always greater the third.

Therefore, third side has to be less than the sum of the two sides. The third s less than 8 +6 = 14 cm.

The side cannot be less than the difference of the two sides. Thus, the third side has to be more than 8-6=2 cm.
The length of the third side could be any length greater than 2 and less than 14 cm.

Example 5. Determine whether the triangle whose lengths of sides are 3 cm, 4 cm, 5 cm is a right-angled triangle.

Solution – 3²=3×3=9; 4² = 4 x4 = 16; 5² = 5 x 5 = 25
We find 3²+ 4² = 5².
Therefore, the triangle is right-angled.

Note: In any right-angled triangle, the hypotenuse happens to be the longest side. In this example, the side with length 5 cm is the hypotenuse.

Example 6. A ABC is right-angled at C. If AC = 5 cm and BC = 12 cm find the length of AB.

Solution – A rough figure will help us (Fig 6.28).
By Pythagoras property,
to bee
AB² = AC² + BC²
=5²+12² = 25+ 144 = 169 = 13²
or AB² = 13². So, AB = 13
or the length of AB is 13 cm.

Note: To identify perfect squares, you may use prime factorisation technique.