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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 $\angle $ B = 60$\xb0$ , So $\angle $ C = 45$\xb0$

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$\xb0$ , 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$\xb0$

Step 3 : Draw $\angle $ PQS an angle of 45$\xb0$ , 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$\xb0$ .

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$\xb0$ , So $\angle $ C = 45$\xb0$ 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 Z

_{1}. Z

_{2}and Z

_{3}, Z

_{4}AS :

B Z

_{1}= Z

_{1}Z

_{2}= Z

_{2}Z

_{3}= Z

_{3}Z

_{4}

Step 2 : Join Z

_{3}

_{ }to C and than draw a line from Z

_{4}as parallel to Z

_{3}C 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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