NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4
These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.4
Question 1.
Construct ∆ABC given m∠A = 60° , m ∠B = 30 and AB = 5.8 cm.
Answer:
Steps of Construction
(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line segment AB = 5.8 cm.
(iii) Construct ∠BAX = 60° at A.
(iv) At B, construct ∠ABY = 30° .
(v) Let the rays AX and BY intersect at C. Thus, ∆ABC is the required triangle.
Question 2.
Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m ∠QRP = 40°.
(Hint : Recall angle-sum property of a triangle)
Answer:
Here the line segment PQ is given. We can construct the triangle, if the measure of ∠QPR is known.
∠QPR = 180° – [∠PQR + ∠QRP]
= 180°-(105°+ 40°)
= 180° – 145°
∠QPR = 35°
Steps of Construction
Question 3.
Examine whether you can construct ∆DEF such that EF = 7.2 cm, m∠E = 110 and m∠F = 80°. Justify your answer.
Answer:
Here m∠E =110 and m∠F = 80°
i. e. m∠E + m∠F= 110°+ 80°= 190°>180°
Since, sum of the three angles of a triangle is 180°.
∆DEF is not possible, having sum of two angles > 180°
Thus, ∆DEF cannot be constructed.
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