Question 1200779
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You are allowed only one question per post. 
I'll do problem 1 to get you started.


x = some positive number, i.e. x > 0
a = x = smaller leg
b = 3x = larger leg
c = unknown hypotenuse


Use the Pythagorean Theorem to determine what c is in terms of x.
{{{a^2+b^2 = c^2}}}


{{{c = sqrt(a^2+b^2)}}}


{{{c = sqrt(x^2+(3x)^2)}}}


{{{c = sqrt(x^2+9x^2)}}}


{{{c = sqrt(10x^2)}}}


{{{c = sqrt(10)*sqrt(x^2)}}}


{{{c = sqrt(10)*x}}} Note that x > 0 so sqrt(x^2) = x


{{{c = x*sqrt(10)}}}


Add up the three sides of the triangle. Set the sum equal to the stated perimeter 45. Solve for x.
{{{a+b+c = perimeter}}}


{{{x+3x+x*sqrt(10) = 45}}}


{{{4x+x*sqrt(10) = 45}}}


{{{(4+sqrt(10))x = 45}}}


{{{x = 45/(4+sqrt(10))}}}


Now let's rationalize the denominator.


{{{x = 45/(4+sqrt(10))}}}


{{{x = (45(4-sqrt(10)))/((4+sqrt(10))(4-sqrt(10)))}}}


{{{x = (180-45sqrt(10))/((4)^2-(sqrt(10))^2)}}}


{{{x = (180-45sqrt(10))/(16-10)}}}


{{{x = (3(60-15*sqrt(10)))/(6)}}}


{{{x = (3(60-15*sqrt(10)))/(3*2)}}}


{{{x = (60-15*sqrt(10))/2}}}


Because each leg of the right triangle is perpendicular to one another, we can treat them as the base and height in either order.


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


{{{area = (1/2)*a*b}}}


{{{area = (1/2)*x*3x}}}


{{{area = (3/2)x^2}}}


Now plug in the expression we found for x.
{{{area = (3/2)x^2}}}


{{{area = (3/2)((60-15*sqrt(10))/2)^2}}}


{{{area = (3/2)(((60-15*sqrt(10))^2)/4)}}}


{{{area = (3/8)(60-15*sqrt(10))^2}}}


{{{area = (3/8)(60^2-2*60*15*sqrt(10)+(15*sqrt(10))^2)}}} Use the formula (m+n)^2 = m^2+2mn+n^2


{{{area = (3/8)(3600-1800*sqrt(10)+2250)}}} 


{{{area = (3/8)(5850-1800*sqrt(10))}}} 


{{{area = (3(5850-1800*sqrt(10)))/8}}}


{{{area = (17550-5400*sqrt(10))/8}}}


{{{area = (2(8775-2700*sqrt(10)))/8}}}


{{{area = (2(8775-2700*sqrt(10)))/(4*2)}}}


{{{area = (8775-2700*sqrt(10))/4}}}


When using a calculator
{{{(8775-2700*sqrt(10))/4 = 59.21258}}} which is approximate to five decimal places.
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