Practical Geometry Ex-10.5

NCERT solutions for Maths Practical Geometry 

NCERT Solutions for Class 7 Maths Practical Geometry

Class –VII Mathematics (Ex. 10.5)

Question 1.Construct the right angled $\Delta$PQR, where $m\angle$Q = ${90^ \circ },$ QR = 8 cm and PR = 10 cm.

Answer:

To construct: A right angled triangle PQR where $m\angle$Q = ${90^ \circ },$ QR = 8 cm and PQ = 10 cm.

Steps of construction:

(a) Draw a line segment QR = 8 cm.

(b) At point Q, draw QX $\bot$ QR.

(c) Taking R as centre, draw an arc of radius 10 cm.

(d) This arc cuts QX at point P.

(e) Join PQ.

It is the required right angled triangle PQR.

NCERT Solutions for Class 7 Maths Exercise 10.5

Question 2.Construct a right angled triangle whose hypotenuse is 6 cm long and one the legs is 4 cm long.

Answer:

To construct: A right angled triangle DEF where DF = 6 cm and EF = 4 cm

Steps of construction:

(a) Draw a line segment EF = 4 cm.

(b) At point Q, draw EX $\bot$ EF.

(c) Taking F as centre and radius 6 cm, draw an arc. (Hypotenuse)

(d) This arc cuts the EX at point D.

(e) Join DF.

It is the required right angled triangle DEF.

NCERT Solutions for Class 7 Maths Exercise 10.5

Question 3.Construct an isosceles right angled triangle ABC, where $m\angle$ACB = ${90^ \circ }$ and AC = 6 cm.

Answer:

To construct: An isosceles right angled triangle ABC where $m\angle$C = ${90^ \circ },$ AC = BC = 6 cm.

Steps of construction:

(a) Draw a line segment AC = 6 cm.

(b) At point C, draw XC $\bot$ CA.

(c) Taking C as centre and radius 6 cm, draw an arc.

(d) This arc cuts CX at point B.

(e) Join BA.

It is the required isosceles right angled triangle ABC.