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  • Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and Ð ABC = 60°. Then construct a triangle whose sides are 3 by 4 of the corresponding sides of the triangle ABC.

5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \angle ABC = 60°. Then construct a triangle whose sides are 3/4  of the corresponding sides of the triangle ABC.

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Steps of construction:-  

(i) 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 \DeltaABC 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 \Delta AB'C' is the required triangle.

 

Posted by

Devendra Khairwa

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