(1) In triangle JKL, we have JK = JL = 25 and KL = 30. Find the area.
(2) In triangle ABC, we have AB = 5, AC = 12, BC = 13. Find the circumradius.
(1) In triangle JKL, we have JK = JL = 25 and KL = 30. Find the area.
Let J be the apex
And the triangle is isosceles
Let the height = JM
Then we have right triangle JML such that
JM =sqrt ( JL^2 - (KL/2)^2 ) = sqrt ( 25^2 - 15^2) = sqrt (625 - 225) = sqrt 400 = 20
So....the area = (1/2)KL * JM = (1/2)(30) (20) = 300 units^2
(2) In triangle ABC, we have AB = 5, AC = 12, BC = 13. Find the circumradius.
This is a right triangle.......the circumradius = (1/2) (hypotenuse) = (1/2)(13) = 6.5
1) you could also use Heron's formula
semi perimeter = s = (25 + 25 + 30)/2 = 40
area = sqrt (s (s-JK)(s-JL)(s-KL) )
area = sqrt ( 40(15)(15)(10)) = 300 units2