Land in the shape of an isosceles triangle has a base of 130 m. An altitude from one of the legs of the triangle is 120 m. What is the area of the property?
Can someone please solve this without using trignometry.
a) 3000
b) 9840
c) 9512
d) 10 140
e) 10 200
+128460 We can equate the area as follows
(1/2)base * height = A ⇒ 65 * height = A (1)
120 * (1/2)side = A ⇒ 60 * side = A (2)
Set (1) = (2) and solve for the side
60 * side = 65 * height
side = (65/60) * height
side = (13/12)* height = (13/12)*h
The side is the hypotenuse of a right triangle with the height and 1/2 base being the legs
So.......Using the Pythagorean Theorem, we have that
sqrt [ side^2 - height^2 ] = 1/2 base
Substitute
sqrt [ ( (13/12)h)^2 - h^2] = 1/2 base
sqrt [ (169/ 144)h^2 - (144/144)h^2 ] = 1/2 base
sqrt [ (169 - 144) h^2 / 144 ] = 1/2 base
sqrt [ 25h^2 / 144 ] = 1/2 base
(h / 12) sqrt (25) = 1/2 base
(5/12)h = 1/2 base
(5/12)h = 65
h = (65)(12) / 5
h = 13 * 12 = 156 m
So....Area =
(1/2)base * height
(1/2)130m * 156m
65m * 156m =
10140 m^2
