Find the area of a triangle with sides a = 2 cm. b = 3 cm. and c = 3 cm.

Question

find the area of a triangle with sides a = 2 cm. b = 3 cm. and c = 3 cm.

Answer 1

Because this is an isoceles triangle, we can apply some special methods. If we drop a perpendicular from the vertex where the congruent sides join, the result is two congruent right triangles.

Half of the base side is one cm long. The hypotenuse of either right triangle is 3 cm long. That means the height of these triangles is

√(3² - 1²) = √(9 - 1) = √8 = 2√2 cm

There are two congruent right triangles which comprise our original triangle. Each has a 1 cm short leg and a 2√2 cm long leg. So,

2×(1/2)×a×b = a×b = 2√2 square centimeters.

You could also use Heron's formula, which works on any triangle:

Area of triangle = √[s(s-a)(s-b)(s-c)]

where a,b and c are the lengths of sides of triangle

and s is the semi-perimeter of triangle

s = (a+b+c)/2

s = (2+3+3)/2 = 4

s-a = 4-2 = 2

s-b = 4-3 = 1

s-c = 4-3 = 1

So, the area of this triangle is

√(2×4×1×1) = √8 = 2√2 cm, just like before.

Answer 2

Could use Herron's Formula, yet that is rather messy.

Note the triangle is isosceles, Therefore the height, h, sotisfies the Pythagorean Theorem applied

3² = h² + (1)²

Can you solve for h and put into area formular and answer?