# 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.

Given :   Perimeter of triangle = 10.5 cm
And
Base angles , Let $\angle$ B = 60$°$ , So $\angle$ 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 $\angle$ 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  $\angle$ QPR  = 60$°$

Step 3 : Draw $\angle$ 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  $\angle$ PQS  = 45$°$ .

Step 5 : Now  we bisect $\angle$ QPR  and $\angle$ 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  $\angle$ B = 60$°$ , So $\angle$ C  =  45$°$ and AB +  BC  +  CA  =  10.5 cm

Now to construct another triangle whose sides are $\frac{4}{3}$ of the corresponding sides of triangle ABC , We follow these steps :

Step 1 : Draw a line BZ , As $\angle$ 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 $\frac{4}{3}$ of the corresponding sides of triangle ABC .

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