RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties In Text Exercise is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties In Text Exercise.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 7 |
| Subject | Maths |
| Chapter | Chapter 8 |
| Chapter Name | Triangle and its Properties |
| Exercise | In Text Exercise |
| Number of Questions | 8 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties In Text Exercise
(Page 101)
Question
Does the measurement of other two angles change by changing (RBSESolutions.com) the measurement of one of the three angles? Try by drawing and testing various triangles and fill in the blanks :
| Name of Triangle | Angle |
| ∆ABC | ∠A= 50°, ∠B = 60°, ∠C = 70° |
| ∆ABC | ∠A= 30°, ∠B = …°, ∠C = …° |
| ∆ABC | ∠A= 100°, ∠B = …°, ∠C = ..° |
Solution:
Yes, on changing any angle of a triangle, the (RBSESolutions.com) measures of other angles change. It is cleared in table:
| Name of Triangle | Angles |
| ∆ABC | ∠A= 50°, ∠B = 60°, ∠C = 70° |
| ∆ABC | ∠A= 30°, ∠B = 120°, ∠C = 30° |
| ∆ABC | ∠A= 100°, ∠B = 50°, ∠C = 30° |
(Page 105)
Question
Construct the triangles according to (RBSESolutions.com) given measurement :
1. ∆XYZ having sides 5 cm, 4 cm and 6 cm.
2. ∆MNO having sides 6.5 cm, 5.5 cm and 3cm.
3. ∆PQR having sides 5 cm, 6 cm and 12 cm.
4. ∆UVW having sides 2.0 cm, 3.0 cm and 5.0
Were you able to construct triangles from all the given measurements ? If not, why ? Discuss. Write the measurement of the sides of triangles you constructed in the following table :
Solution:
1. In ∆XYZ , XY = 5 cm, YZ = 4 cm, ZX = 6 cm
2. In ∆MNO, MN = 6.5 cm, NO = 4.5 cm OM = 3 cm
3. In ∆PQR, PQ = 5 cm QR = 6 cm, RP = 12 cm. Triangle is (RBSESolutions.com) not possible.
4. In ∆UVW, UV = 2 cm VW = 3 cm, WV = 5 cm. Triangle is not possible
Therefore from table we conclude that sum of the (RBSESolutions.com) measures of all sides is always greater than the third side.
We could not construct some triangle because the sum of two sides should not be smaller or equal to the third side. In ∆PQR and ∆UVW, p + q < r and u + v = w. Therefore triangles can not be constructed.
Do and Learn
(Page 102)
Question
Find the value of x in each (RBSESolutions.com) of the following triangles :
Solution:
Sum of all angles (RBSESolutions.com) of a triangle is 180°
(i) 70° + 60° + x = 180°
⇒ 130° + x = 180°
∴ x = 180° – 130° = 50°
(ii) 30° + x + 90° = 180°
⇒ 120° + x = 180°
⇒ x = 180° – 120° = 60°
(iii) 40° + 40° + x = 180°
⇒ 80° + x = 180°
⇒ x = 180° – 80° = 100°
(Page 103)
Question 1
Find the value of exterior (RBSESolutions.com) angle x from the following triangles :
Solution:
The sum of two (RBSESolutions.com) opposite interior angles, is equal to the exterior angle,
(i) x = 60° + 70° = 130°
(ii) x = 60° + 40° = 100°
(iii) x = 60° + 90° = 150°
Question 2
Is it possible to construct a triangle with two right angles?
Solution:
There can not be such triangle where two angles are right angles because sum of all three angles is two right angles Therefore third angle will be 0°. Than such triangle is impossible.
Question 3
It is possible to construct (RBSESolutions.com) a triangle whose all the three angles are greater than 60°?
Solution:
There is no triangle is possible because if each angle is greater than 60° then the sum of all angles will be greater then 180°.
(Page 106)
Question 1
Construct a triangle (RBSESolutions.com) with side length 3.5 cm 4.5 cm and 6 cm.
Solution:
Draw a line AB = 6 cm. Draw an arc of length 3.5 cm from A and 4 cm from B.
Those arc intersect each other at C. Join AC and BC. This is the required ∆ABC.
Question 2
Is it possible to construct (RBSESolutions.com) a triangle with sides having length 4 cm, 5 cm and 9 cm?
Solution:
The sum of two sides of a triangle should be more than third side.
Here 4 + 5 ⊁ 9
∴ Such triangle is not possible.
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