Question 152428
a.) Same as GEOMETRY PROBLEM #151709
SEE ANSWER# {{{111525}}} -------> DONE.
b.)The {{{Length=L}}} is 100ft longer than the {{{width=W}}} --->{{{L=W+100ft}}}
The {{{diagonal=D}}} side given = 500ft, by PYTHAGOREAN THEOREM:
{{{D^2=L^2+W^2}}}, but {{{L=W+100ft}}}, so
{{{D^2=(W+100)^2+W^2}}}
{{{500^2=W^2+200W+10000+W^2}}}
{{{0=2W^2+200W+10000-250000}}} ---> {{{2W^2+200W-240000=0}}}, Divide the whole eqn by 2:
{{{cross(2)W+cross(200)100W-cross(240000)120000=0}}}{{{(1/cross(2))}}}
{{{W+100W-120000=0}}} ---> {{{a=1}}},{{{b=100}}}, & {{{c=-120000}}}
By Pyth. theorem:
{{{w=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{w=(-100sqrt(100^2-4*1*(-120000)))/(2*1)}}}
{{{w=(-100+-sqrt(10000+480000))/2}}}
{{{w=(-100+-sqrt(490000))/2}}} ------> {{{w=(-100+-700)/2}}}
2 values:
{{{w=(-100+700)/2=600/2=300ft}}} ----------> TO BE USED ----> {{{WIDTH=300ft}}}
{{{w=(-100-700)/2=-800/2=-400ft}}} --------> DO NOT USE
So, {{{L=300+100=400ft}}}
To check, go back PYTH Theorem: {{{500^2=400^2+300^2=160000+9000}}}
{{{25000=25000}}}, good!
.
3.) Let {{{X}}} = 1st odd integer
{{{X+2}}}= 2nd odd integer
{{{X+4}}}= 3rd integer
condition: square of 3rd is 264 more than suqrae of 2nd: to show,
{{{(x+4)^2=264+(x+2)^2}}}
{{{cross(x^2)+8x+16=264+cross(x^2)+4x+4}}}
{{{8x-4x=264+4-16}}} --> {{{cross(4)x/cross(4)=cross(252)63/cross(4)}}}
{{{x=63}}} -----------------------> 1st odd integer
{{{63+2=65}}} --------------------> 2nd odd integer
{{{63+4=67}}} --------------------> 3rd odd integer
in doubt? go back to the condition:
{{{(67^2)=264+(65^2)}}}
{{{4489=264+4225}}}
{{{4489=4489}}}
</pre><font size=4><b>Thank you,
Jojo</pre>