Ncert Solutions For Class 7 Maths Ex 10.3

Ncert Solutions For Class 7 Maths Chapter 10 Ex 10.3

 

Q1: Construct   a Δ  ABC so that AC = 5 cm, AB = 3 cm and CAB = 90O .

Solution:

Constructing a Δ DEF where DE = 5 cm, DF = 3 cm and m EDF = 90O .

Steps of construction:

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

(b) At point A, draw an angle of 90O with the help of compass i.e., XAB = 900.

(c) Taking A as the center, draw an arc of radius 5 cm, which cuts AX at the point E.

(d)Connect BC.

This  is the required right angled triangle ACB:

3.1

 

Q2: Construct an isosceles triangle which has equal sides of length  6.5 cm and with a angle of 110 ° between them.

Solution:

Constructing an isosceles triangle where AC = AB = 6.5 cm and A = 110 ° .

Steps of construction:

(a) Make a line segment AB= 6.5 cm.

(b) At  A, make a  110° with the help of protractor, i.e., XAB = 110 ° .

(c) Take A as the  center, and draw an arc of radius 6.5 cm, which cuts AX at point  C.

(d) Connect BC

This is the required isosceles Δ ABC:

3.2

 

Q3:

Construct a Δ ZXC with XC = 7.5 cm, ZC = 5 cm and C = 60O .

Solution:

Constructing a  Δ ZXC where XC = 7.5 cm, ZC = 5 cm and C = 60O.

Steps of construction:

(a) Make a line segment XC = 7.5 cm.

(b) At C, make an angle of 600 with the help of protractor, i.e., XCB = 60O .

(c) Take C as the center  and draw an arc of 5cm, which cuts the line YC at the point Z.

(d)Connect  AB

Thus, this is the required Δ ZXC:

3.3

 

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