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ToggleClass 7, Maths, Chapter 10, Exercise 10.4 Solutions
Q.1. Construct ∆ABC, given m∠A= 60°, m∠B = 30° and AB = 5.8 cm.
Ans:
Steps of construction:
- Draw a line segment AB = 5.8 cm.
- At point A, draw an angle of 60o . i.e. ∠PAB = 60o.
- At point B, draw an angle of 30oi.e. ∠QBA = 30o.
- Now, AP and BQ intersect at the point C.
Then, ΔABC is the required triangle.
Q.2. Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°. (Hint: Recall angle-sum property of a triangle).
Ans:
We know that the sum of the angles of a triangle is 180o.
∴ m∠PQR + m∠QRP + m∠RPQ = 180o
⟹ 105o+ 40o+ ∠RPQ = 180o
⟹ 145o + ∠RPQ = 180o
⟹ ∠RPQ = 180o– 1450
⟹ ∠RPQ = 35o
Steps of construction:
- Draw a line segment PQ = 5 cm.
- At point P, draw an angle of 35oi.e. ∠XPQ = 35o.
- At point Q, draw an angle of 105oi.e. ∠YQP = 105o.
- Now PX and QY intersect at the point R.
Then, ΔPQR is the required triangle.
Q.3. Examine whether you can construct ∆DEF such that EF = 7.2 cm, m∠E = 110° and m∠F = 80°. Justify your answer
Ans: Given: in ΔDEF, ∠E = 110o and ∠F = 80o
We know that the sum of the angles of a triangle is 180o.
Then,
⟹ ∠D + ∠E + ∠F = 180o
⟹ ∠D + 110o + 80o= 180o
⟹ ∠D + 190o = 180o
⟹ ∠D = 180o– 1900
⟹ ∠D = -10o
Here, the sum of two angles is 190o is greater than 180o. So, it is not possible to construct a triangle.
NCERT Solutions For Class 7 Maths, Chapter 10, Practical Geometry (All Exercises)
Class 7, Maths, Chapter 9, Practical Geometry
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.1
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.2
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.3
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.4 ← You are here
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.5