KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4
KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4
Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.4
Question 1.
Construct ∆ ABC, given m∠A= 30° and m∠B = 30° and AB = 5.8 cm.
Solution:
Steps of Construction.
- Draw a line segment AB of length 5.8 cm, at A draw a ray AM making an angle of 60° with AB.
- At ‘B’ draw a ray ABN making an angle of 30° with BA.
- Mark the point of intersection of two rays as ‘C’. Now we get the A ABC which is required.
Question 2.
Construct ∆ PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°. (Hint Recall angle-sum property of a triangle).
Solution:
By angle-sum property of a triangle ∠RPQ + ∠PQR + ∠QRP = 180°
∠RPQ + 105° + 40° = 180°
∠RPQ + 145° = 180°
∴ ∠RPQ = 180° – 145° = 35°,
Steps of Construction
- Draw a line segment PQ of length 5 cm. At P draw a ray PM making an angle of 35° with PQ.
- At Q draw a ray QN making an angle of 105° with QP.
- Mark the point of intersection of two rays as R. We get the required P Q R triangle.
Question 3.
Examine whether you can construct ∆ DEF such that E F = 7.2 cm. m∠E = 110° and m∠F = 80°. Justify your answer.
Solution:
∠E + ∠F = 110°+ 80°= 190°
This triangle is not possible to construct because the measures of two angles exceed 180°.
The sum of three angles of a triangle is always equal to 180°.
So it is not possible.
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