Find the perimeter of triangle ABC to the nearest tenth.

Question

20 points

Answer 1

Triangle ACD is a 45°-45°-90° triangle. This right triangle has sides n,n, and n√2.

Triangle BCD is a 30°-60°-90° triangle. This right triangle has sides m.m√3, and 2m.

The perimeter of triangle ABC is AD+DB+BC+CA = 8 + 8√3 + 16 + 8√2 = 49.170

To the nearest tenth, the perimeter of triangle ABC is 49.2.

Answer 2

CB = 16; AC = 8√2; AB = √(16² + 8²·2) = 8√6

So, perimeter P = 16 + 8√2 + 8√6 = 46.9

Answer 3

To find Perimeter , first find length of three sides .

BC = 8/Sin 30 = 16 cm .

AC = 8/Sin45 = 11·314 cm

AB = AD + DB

AB = 8/Tan45 + 8/Tan30 = 8 + 13·856 = 21· 856 cm

Perimeter= AB + AC + BC

Perimeter triangle ABC = 21·856 +11·314 + 16 =49·17 cm