NCERT Solutions Class 7 Mathematics Practical Geometry Ex 10.3

Class 6 - Mathematics
Chapter - Practical Geometry : Exercise 10.3

NCERT Solutions Class 7 Mathematics textbook
Top Block 1

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

Answer :
To construct: ΔDEF where DE = 5 cm, DF = 3 cm and m∠EDF = 90o.

NCERT Solutions Class 7 Mathematics Practical Geometry
Steps of construction:
(a) Draw a line segment DF = 3 cm.

(b) At point D, draw an angle of 90o with the help of compass i.e., ∠XDF = 90o.

(c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E.

(d) Join EF.

It is the required right angled triangle DEF.

Question : 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.

Answer :
To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and ∠Q = 110o.

Steps of construction:
(a) Draw a line segment QR = 6.5 cm.

(b) At point Q, draw an angle of 110o with the

help of protractor, i.e., ∠YQR = 110o.

(d) Taking Q as centre, draw an arc with radius

cm, which cuts QY at point P. 110o

(e) Join PR

It is the required isosceles triangle PQR.

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

Answer :
To construct: ΔABC where BC = 7.5 cm, AC = 5 cm and m∠C = 60o.

NCERT Solutions Class 7 Mathematics Practical Geometry
Mddle block 1
Steps of construction:
(a) Draw a line segment BC = 7.5 cm.

(b) At point C, draw an angle of 60o with the help of protractor, i.e., ∠XCB = 60o.

(c) Taking C as centre and radius 5 cm, draw an arc, which cuts XC at the point A.

(d) Join AB

It is the required triangle ABC.

Bottom Block 3
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