draw a triangle with side bc =6cm ab=5 cm and <abc =60° then construct a triangle whose sides are 3/4 of corresponding sides of the triangle abc
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yuvsingh1705:
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first draw a line with side BC as 6cm
Now from point B construct 60 degree and name that ray as Bl
Now measure the 5cm in your rounder and put the end at point B and cut the ray Bl with the pencil Name that intersection point as A thus ABC
Similarly make another ray BM from point B
Now make equal 4 arcs at ray BM starting from B and name that b1 b2 etc
Now join b3 and c and make parallel of the same line
when the line intersects Bq which was strengthen by BC name that point as c'
now make parallel of AC
thus triangle is divided in given ratio
Now from point B construct 60 degree and name that ray as Bl
Now measure the 5cm in your rounder and put the end at point B and cut the ray Bl with the pencil Name that intersection point as A thus ABC
Similarly make another ray BM from point B
Now make equal 4 arcs at ray BM starting from B and name that b1 b2 etc
Now join b3 and c and make parallel of the same line
when the line intersects Bq which was strengthen by BC name that point as c'
now make parallel of AC
thus triangle is divided in given ratio
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