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
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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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