draw a triangle abc in which ab=5 cm bc=6 cm and angle abc= 60 degrees
construct another triangle similar to triangle abc with scale factor 5/7 { 5 upon 7 }
Answer :
We follow these steps to construct angle ABC .
Step 1 : Draw a line BC = 6 cm .
Step 2 : Now take any radius ( Less than half of BC ) and take center " B " draw an arc that intersect our line BC at " P " , With same radius and center is " P " draw another arc that intersect our previous arc at " Q " .
Step 3 : Join BQ and extend it to D
Step 4 : Take radius = 5 cm and center " B " we draw an arcs that intersect our line BD at " A " Now we join AC , and get triangle ABC .
Now We construct another triangle with scale factor , As :
Step 5 : Draw a line BX As CBX is any acute angle and draw seven equal radius arc on line BX that intersect at X1 . X2 , X3 , X4 , X5 , X6 and X7 AS :
BX1 = X1 X2 = X2 X3 = X3X4 = X4X5 = X5X6 = X6 X7
Step 6 : Join X7 to C and than draw a line from X5 as parallel to X7C that intersect our line BC at C' .
Step 7 : Draw line from C' as parallel to AC that intersect line AB at A' .
We get triangle A'BC' whose sides are of corresponding sides of triangle ABC .
We follow these steps to construct angle ABC .
Step 1 : Draw a line BC = 6 cm .
Step 2 : Now take any radius ( Less than half of BC ) and take center " B " draw an arc that intersect our line BC at " P " , With same radius and center is " P " draw another arc that intersect our previous arc at " Q " .
Step 3 : Join BQ and extend it to D
Step 4 : Take radius = 5 cm and center " B " we draw an arcs that intersect our line BD at " A " Now we join AC , and get triangle ABC .
Now We construct another triangle with scale factor , As :
Step 5 : Draw a line BX As CBX is any acute angle and draw seven equal radius arc on line BX that intersect at X1 . X2 , X3 , X4 , X5 , X6 and X7 AS :
BX1 = X1 X2 = X2 X3 = X3X4 = X4X5 = X5X6 = X6 X7
Step 6 : Join X7 to C and than draw a line from X5 as parallel to X7C that intersect our line BC at C' .
Step 7 : Draw line from C' as parallel to AC that intersect line AB at A' .
We get triangle A'BC' whose sides are of corresponding sides of triangle ABC .