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Construct a \Delta ABC with sides BC = 6 cm, AB = 5 cm and \angle ABC= 60^{\circ}. Then construct a triangle whose sides are  \frac{4}{3}  of the corresponding sides of \Delta ABC

 

 

Answers (1)

(i) Construct the required triangle ABC

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

(iii) Cut four equal parts of line BX namely BB , BB , BB , BB 4.

(iv) Now join B to C. Draw a line B C' parallel to B C.

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

Hence \Delta AB'C' is the required triangle.

 

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