Math, asked by HUSNASHAIKH, 9 months ago

draw a ∆ABC with BC =6cm, AB=5 cm,&right angleABC=60° then construct a ∆ whose sides are 3/5 of the corresponding side of ∆ABC​

Answers

Answered by jchcdhjggdfjjgbjccc
0

Answer:

Draw a line segment BC with measurement of 6 cm.

(ii) Now construct angle 60o from point B and draw AB = 5 cm.

(iii) Join the point C with point A. Thus

ABC is the required triangle.

(iv) Draw a line BX which makes an acute angle with BC and is opposite of vertex A.

(v) Cut four equal parts of line BX namely BB1, BB2, BB3, BB4.

(vi) Now join B4 to C. Draw a line B3C' parallel to B4C.

(vii) And then draw a line B'C' parallel to BC.

Hence

AB'C' is the required triangle.

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Answered by sanjayaim85
0

Answer:

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