Mathematics – Class 7 – Chapter 10 – Practical Geometry – Exercise 10.3 – NCERT Exercise Solution

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

Solution:

Mathematics - Class 7 - Chapter 10 - Practical Geometry - Exercise 10.3 - NCERT Exercise Solution

Step 1. Draw a line segment DF = 3 cm.

Step 2. Consider point D as centre and make an angle of 90o then draw a ray DX i.e. ∠XDF = 90o.

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

Step 4. Join EF.

Hence, 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:

Mathematics - Class 7 - Chapter 10 - Practical Geometry - Exercise 10.3 - NCERT Exercise Solution

Step 1. Draw a line segment AB = 6.5 cm.

Step 2. Consider point A as centre to make an angle of 110o. Now, draw a ray AX, ∠XAB = 110o.

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

Step 4. Join CB.

Hence, ΔABC is the required isosceles triangle.

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

Solution:

Mathematics - Class 7 - Chapter 10 - Practical Geometry - Exercise 10.3 - NCERT Exercise Solution

Step 1. Draw a line segment BC = 7.5 cm.

Step 2. Consider point C as centre to make an angle of 60o. Now, draw a ray CX, i.e. ∠XCB = 60o.

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

Step 4. Join AB.

Hence, ΔABC is the required triangle.

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