Construct a triangle ABC, whose perimeter is 10.5 cm and base angles are 60. and 45.. Construct another triangle whose sides are 4/3 of the corresponding sides of the triangle ABC.
Answer:
Given : Perimeter of triangle = 10.5 cm
And
Base angles , Let B = 60 , So C = 45
To construct required triangle there are steps of construction :
Step 1 : Draw PQ as length of the perimeter = 10.5 cm
Step 2 : Draw QPR an angle of 60 , As : Take any radius ( Less than half of PQ ) and take center " P " draw a arc that intersect our line PQ at " X " . Now with same radius and center is " X " draw another arc that intersect our previous arc at " Y " Now we join PY and extend it to R , SO QPR = 60
Step 3 : Draw PQS an angle of 45 , As : Take any radius ( Less than half of PQ ) and take center " Q " draw a semicircle that intersect our line PQ at " T " . Now with same radius and center is " T " draw another arc that intersect our semicircle at " U " . Again with same radius and center is " U " draw an arcs that intersect our semicircle at " D " .With same radius and center " U " and " D " draw two arcs that intersect each other at " E " .
Step 4 : Join QE that intersect our semicircle at " F " , With same radius and center " T " and " F " draw two arcs that intersect each other " V " Now we join QV and extend it to S , SO PQS = 45 .
Step 5 : Now we bisect QPR and PQS bisector , these meet at point A.
Step 6 : Draw perpendicular bisector : LM perpendicular bisector of PA and NO is perpendicular bisector of QA .
Step 7 : Line LM intersect line PQ at B and line NO intersect line PQ at C .
Step 8 : Join AB and AC
Then form triangle ABC is our required triangle , Here B = 60 , So C = 45 and AB + BC + CA = 10.5 cm
Now to construct another triangle whose sides are of the corresponding sides of triangle ABC , We follow these steps :
Step 1 : Draw a line BZ , As ZBC is any acute angle . Now draw four equal radius arc on line BZ that intersect at Z1 . Z2 and Z3 , Z4 AS :
B Z1 = Z1 Z2 = Z2 Z3 = Z3 Z4
Step 2 : Join Z3 to C and than draw a line from Z4 as parallel to Z3C that intersect our line PQ at C' .
Step 3 : Draw line From C' as parallel to AC that intersect line AB at A'
Step 4 : Join line C'A' and form a triangle A'BC' that is our required triangle. Whose sides are of the corresponding sides of triangle ABC .
Given : Perimeter of triangle = 10.5 cm
And
Base angles , Let B = 60 , So C = 45
To construct required triangle there are steps of construction :
Step 1 : Draw PQ as length of the perimeter = 10.5 cm
Step 2 : Draw QPR an angle of 60 , As : Take any radius ( Less than half of PQ ) and take center " P " draw a arc that intersect our line PQ at " X " . Now with same radius and center is " X " draw another arc that intersect our previous arc at " Y " Now we join PY and extend it to R , SO QPR = 60
Step 3 : Draw PQS an angle of 45 , As : Take any radius ( Less than half of PQ ) and take center " Q " draw a semicircle that intersect our line PQ at " T " . Now with same radius and center is " T " draw another arc that intersect our semicircle at " U " . Again with same radius and center is " U " draw an arcs that intersect our semicircle at " D " .With same radius and center " U " and " D " draw two arcs that intersect each other at " E " .
Step 4 : Join QE that intersect our semicircle at " F " , With same radius and center " T " and " F " draw two arcs that intersect each other " V " Now we join QV and extend it to S , SO PQS = 45 .
Step 5 : Now we bisect QPR and PQS bisector , these meet at point A.
Step 6 : Draw perpendicular bisector : LM perpendicular bisector of PA and NO is perpendicular bisector of QA .
Step 7 : Line LM intersect line PQ at B and line NO intersect line PQ at C .
Step 8 : Join AB and AC
Then form triangle ABC is our required triangle , Here B = 60 , So C = 45 and AB + BC + CA = 10.5 cm
Now to construct another triangle whose sides are of the corresponding sides of triangle ABC , We follow these steps :
Step 1 : Draw a line BZ , As ZBC is any acute angle . Now draw four equal radius arc on line BZ that intersect at Z1 . Z2 and Z3 , Z4 AS :
B Z1 = Z1 Z2 = Z2 Z3 = Z3 Z4
Step 2 : Join Z3 to C and than draw a line from Z4 as parallel to Z3C that intersect our line PQ at C' .
Step 3 : Draw line From C' as parallel to AC that intersect line AB at A'
Step 4 : Join line C'A' and form a triangle A'BC' that is our required triangle. Whose sides are of the corresponding sides of triangle ABC .