Question 1132035

 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?



given:

 base {{{base=130}}}
altitude  {{{a=120}}}


We can equate the area as follows:


{{{A=(1/2)base * height }}}

{{{A=(130/2) * height }}} 

{{{A =65*height}}}.............. (1)  



area of the triangle using given altitude:


{{{A[1] =120 * (1/2)*side}}} 

{{{A[1] =60 * side}}} ............. (2)



Set (1)  = (2)      and solve for the side:


{{{60 * side  =  65 * height}}}

{{{side  = (65/60) * height}}}

{{{side  =  (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  -  h^2 )   =  (1/2) base}}}

Substitute side

{{{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}}}

 {{{(5h / 12)  =  (1/2) base}}}

 
since {{{(1/2) base=130/2=65}}}

{{{(5/12)h  =  65}}}

{{{h = 65/(5/12)}}}

 {{{h = (65*12)/5}}}

{{{h  = 13 * 12}}}

 {{{h =   156 m}}}

 
So, the area is:

{{{A=(1/2)base * height}}}

{{{A=(1/2)130m * 156m}}}

{{{A=65m * 156m}}}  

{{{A=10140m^2}}}

 
so, your answer is: d) {{{10 140}}}