Question

17. Construct a ∆ ABC in which AB=5cm , BC=5.6cm and AC=4.2cm .​

Answer 1

Step-by-step explanation:

(i) Draw BC = 4.6 cm.(ii) With B as centre and radius equal to 5 cm draw an arc and with C ascentre and radius equal to 5.8 cm draw another arc to cut the previous arc at A.(iii) Join AB and AC.(iv) Make an acute angle ∠CBE.

(i) Draw BC = 4.6 cm.(ii) With B as centre and radius equal to 5 cm draw an arc and with C ascentre and radius equal to 5.8 cm draw another arc to cut the previous arc at A.(iii) Join AB and AC.(iv) Make an acute angle ∠CBE.(v) Set off three equal distances along BE at B1, B2 and B3.

(i) Draw BC = 4.6 cm.(ii) With B as centre and radius equal to 5 cm draw an arc and with C ascentre and radius equal to 5.8 cm draw another arc to cut the previous arc at A.(iii) Join AB and AC.(iv) Make an acute angle ∠CBE.(v) Set off three equal distances along BE at B1, B2 and B3.(vi) Join B3C.

(i) Draw BC = 4.6 cm.(ii) With B as centre and radius equal to 5 cm draw an arc and with C ascentre and radius equal to 5.8 cm draw another arc to cut the previous arc at A.(iii) Join AB and AC.(iv) Make an acute angle ∠CBE.(v) Set off three equal distances along BE at B1, B2 and B3.(vi) Join B3C.(vii) From B2 draw B2C' || B3C,meeting BC at C'.

(i) Draw BC = 4.6 cm.(ii) With B as centre and radius equal to 5 cm draw an arc and with C ascentre and radius equal to 5.8 cm draw another arc to cut the previous arc at A.(iii) Join AB and AC.(iv) Make an acute angle ∠CBE.(v) Set off three equal distances along BE at B1, B2 and B3.(vi) Join B3C.(vii) From B2 draw B2C' || B3C,meeting BC at C'.(viii) Join AC'.

(i) Draw BC = 4.6 cm.(ii) With B as centre and radius equal to 5 cm draw an arc and with C ascentre and radius equal to 5.8 cm draw another arc to cut the previous arc at A.(iii) Join AB and AC.(iv) Make an acute angle ∠CBE.(v) Set off three equal distances along BE at B1, B2 and B3.(vi) Join B3C.(vii) From B2 draw B2C' || B3C,meeting BC at C'.(viii) Join AC'.Then,ABC' is the required triangle.