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# Land in the shape of an isosceles triangle has a base of 130 m

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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

Dec 28, 2017

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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

Dec 28, 2017