# Class 7 Math NCERT Solutions for Chapter – 10 Practical Geometry Ex – 10.3

## Practical Geometry

1. Construct ΔDEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90o.

Solution:-

Steps of construction:

1. Draw a line segment DF = 3 cm.

2. At point D, draw a ray DX to making an angle of 90o i.e. ∠XDF = 90o.

3. Along DX, set off DE = 5cm.

4. Join EF.

Then, ΔEDF is the required right angled triangle.

2. Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110o.

Solution:-

Steps of construction:

1. Draw a line segment AB = 6.5 cm.

2. At point A, draw a ray AX to making an angle of 110o i.e. ∠XAB = 110o.

3. Along AX, set off AC = 6.5cm.

4. Join CB.

Then, ΔABC is the required isosceles triangle.

3. Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.

Solution:-

Steps of construction:

1. Draw a line segment BC = 7.5 cm.

2. At point C, draw a ray CX to making an angle of 60o i.e. ∠XCB = 60o.

3. Along CX, set off AC = 5cm.

4. Join AB.

Then, ΔABC is the required triangle.