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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

Guest 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

 

 

cool cool cool

CPhill  Dec 28, 2017

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