right triangle
3^2 + 4^2 = 5^2
so
(1/2) base* altitude = (1/2)(4)(3) = 6
The sides of a triangle have lengths of 3, 4, and 5. Find the area of the triangle.
3 answers
This is a right triangle.
A = 1/2bh
A = (1/2)(3)(4)
A = 6 square units
A = 1/2bh
A = (1/2)(3)(4)
A = 6 square units
Given that you did not know that these three side lengths formed the smallest Pythagorean Triangle;
The area of a triangle can be derived from
1) A = bh/2 where b = the base and h = the altitude to the base.
2) A = ab(sinC)/2.
3) A = sqrt[s(s - a)(s - b)(s - c)] where a, b, and c are the three sides and s = the
semi-perimeter = (a + b + c)/2.
4) A = [4a^2b^2 - (c^2 - a^2 - b^2)^2]/4
5) A = rs where s is as defined above, and r is the radius of the incircle inscribed in
the triangle.
6) A = sqrt[r(ra)(rb)(rc)] in terms of the incircle and excircle radii.
7) A = s(s - c) for right triangles.
8) A = abc/4R
The area of a triangle can be derived from
1) A = bh/2 where b = the base and h = the altitude to the base.
2) A = ab(sinC)/2.
3) A = sqrt[s(s - a)(s - b)(s - c)] where a, b, and c are the three sides and s = the
semi-perimeter = (a + b + c)/2.
4) A = [4a^2b^2 - (c^2 - a^2 - b^2)^2]/4
5) A = rs where s is as defined above, and r is the radius of the incircle inscribed in
the triangle.
6) A = sqrt[r(ra)(rb)(rc)] in terms of the incircle and excircle radii.
7) A = s(s - c) for right triangles.
8) A = abc/4R