Question

construct triangle ABC in which BC =3.2 Cm angle ACB = 45 degree and perimeter of triangle ABC is 10 cm????????​

Answer 1

Given : triangle ABC,in which BC=3.2 cm angle ACB=45° and perimeter of triangle ABC is 10 cm  

To find : Construct triangle

Solution:

Step 1: Draw a line segment BC 3.2 cm.

Step 2 : Draw ∠C = 45° using protractor and draw a ray CX.

Step 3 : With Center C and radius 10-3.2= 6.8 cm, make an arc which intersects CX at D.

Step 4 : Join BD

Step 5 : Draw a perpendicular bisector of segment DB it intersect the ray CX at point A.

Step 6 : Join C to B

Triangle is constructed

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Answer 2

Step-by-step explanation:

Perimeter of ∆ABC = AB + BC + AC

∴ 10 = AB + 3.2 + AC

∴ AB + AC = 10 – 3.2

∴ AB + AC = 6.8 cm

Now, In ∆ABC

BC = 3.2 cm, ∠ACB = 45° and AB + AC

= 6.8 cm ….(i)

As shown in the rough figure draw j seg BC = 3.2 cm

Draw a ray CT making an angle of 45° with CB

Take a point D on ray CT, such that CD = 6.8 cm

Now, CA + AD = CD [C - A - D]

∴ CA + AD = 6.8 cm …(ii)

Also, AB + AC = 6.8 cm ….(iii) [From (i)]

∴ CA + AD = AB + AC [From (ii) and (iii)]

∴ AD = AB

∴ Point A is on the perpendicular bisector of seg DB

∴ The point of intersection of ray CT and perpendicular bisector of seg DB is point A.