Practical Geometry Ex-10.3

NCERT solutions for Maths Practical Geometry 

NCERT Solutions for Class 7 Maths Practical Geometry

Class –VII Mathematics (Ex. 10.3)

Question 1.Construct $\Delta$DEF such that DE = 5 cm, DF = 3 cm and $m\angle$EDF = ${90^ \circ }.$

Answer:

To construct: $\Delta$DEF where DE = 5 cm, DF = 3 cm and $m\angle$EDF = ${90^ \circ }.$

Steps of construction:

(a) Draw a line segment DF = 3 cm.

(b) At point D, draw an angle of ${90^ \circ }$ with the help of compass i.e., $\angle$XDF = ${90^ \circ }$.

(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.

NCERT Solutions for Class 7 Maths Exercise 10.3

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 $110^\circ .$

Answer:

To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and $\angle$Q = $110^\circ .$

Steps of construction:

(a) Draw a line segment QR = 6.5 cm.

(b) At point Q, draw an angle of $110^\circ$ with the

help of protractor, i.e., $\angle$YQR = $110^\circ .$

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

1. cm, which cuts QY at point P. 110$^\circ$

(e) Join PR

It is the required isosceles triangle PQR.

NCERT Solutions for Class 7 Maths Exercise 10.3

Question 3.Construct $\Delta$ABC with BC = 7.5 cm, AC = 5 cm and $m\angle$C = ${60^ \circ }.$

Answer:

To construct: $\Delta$ABC where BC = 7.5 cm, AC = 5 cm and $m\angle$C = ${60^ \circ }.$

Steps of construction:

(a) Draw a line segment BC = 7.5 cm.

(b) At point C, draw an angle of ${60^ \circ }$ with the help of protractor, i.e., $\angle$XCB = ${60^ \circ }.$

(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.