NCERT Solutions Class 7 Mathematics Practical Geometry Ex 10.5

Class 6 - Mathematics
Chapter - Practical Geometry : Exercise 10.5

NCERT Solutions Class 7 Mathematics textbook
Top Block 1

Question : 1.Construct the right angled ΔPQR, where 𝓂∠Q = 90o, QR = 8 cm and PR = 10 cm.

Answer :
To construct: A right angled triangle PQR where 𝓂∠Q = 90o, QR = 8 cm and PQ = 10 cm.

NCERT Solutions Class 7 Mathematics Practical Geometry
Steps of construction:
(a) Draw a line segment QR = 8 cm.

(b) At point Q, draw QX ⊥ 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.

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

NCERT Solutions Class 7 Mathematics Practical Geometry
Mddle block 1
Steps of construction:
(a) Draw a line segment EF = 4 cm.

(b) At point Q, draw EX ⊥ 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.

Question : 3.Construct an isosceles right angled triangle ABC, where 𝓂∠ACB = 90o and AC = 6 cm.

Answer :
To construct: An isosceles right angled triangle ABC where 𝓂∠C = 90o, AC = BC = 6 cm.

NCERT Solutions Class 7 Mathematics Practical Geometry
Steps of construction:
(a) Draw a line segment AC = 6 cm.

(b) At point C, draw XC ⊥ 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.

Bottom Block 3
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