NCERT Solutions For Class 7 Maths Chapter 10 Practical Geometry Exercise 10.4

Class 7, Maths, Chapter 10, Exercise 10.4 Solutions

Q.1. Construct ∆ABC, given mA= 60°, mB = 30° and AB = 5.8 cm.

Ans:

 

Steps of construction:

  1. Draw a line segment AB = 5.8 cm.
  2. At point A, draw an angle of 60o . i.e. ∠PAB = 60o.
  3. At point B, draw an angle of 30oi.e. ∠QBA = 30o.
  4. Now, AP and BQ intersect at the point C.

Then, ΔABC is the required triangle.

Q.2. Construct ∆PQR if PQ = 5 cm, mPQR = 105° and mQRP = 40°. (Hint: Recall angle-sum property of a triangle).

Ans:

NCERT Solutions For Class 7 Maths Chapter 10 Practical Geometry Exercise 10.4 Q.2

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:

  1. Draw a line segment PQ = 5 cm.
  2. At point P, draw an angle of 35oi.e. ∠XPQ = 35o.
  3. At point Q, draw an angle of 105oi.e. ∠YQP = 105o.
  4. 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, mE = 110° and mF = 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)