Question

The perimeter of the given triangle abc is 41.36 cm.given ab=10cm,ac=15cm,dc=12cm,angle abc=90degrre. Find the area of triangle abc

Answer 1

the remaining side bc of the given triangle is bc = 41.36-10-15 =16.36
since AC is hypotenuse of the given triangle the sides ab and bc is the height and base of it
so area of the triangle = 1/2*base*height
1/2*16.36*10 = 81.8sqcm

Answer 2

The area of traingle is 73.62 cm²

1) The last side of triangle ABC i.e. BC can be calculated as 41.36-10-15 = 16.36 cm.

2) Now, we can use Heron's Formula to calculate the area of given triangle.

3) Area = $\sqrt{s(s-a)(s-b)(s-c)}$ where s = perimeter/2

4) s= 41.36/2 = 20.68 cm.

5) Hence, Area would be $\sqrt{20.68(20.68 - 10)(20.68-15)(20.68-16.36)}$

6) Simplifying, $\sqrt{20.68*10.68*5.68*4.32}$

7) Hence area is $\sqrt{5419.43}$ = 73.61 cm^2