James and Martha are discussing special right triangles. James believes that a 30-60-90 triangle with a hypotenuse of 10 feet has a greater perimeter than a 45-45-90 triangle with a hypotenuse of 10 feet. Martha disagrees. Who is correct?​

Answer:

Step-by-step explanation:

Disclaimer  it is up the student to verify all the geometry algabra and mathematical formulas and calculation .

cosФ = opp/hyp     sinФ = adj/hyp        hyp = 10      perimeter = a + b + c

for the 45° right triangle

a = b     so      2a² = hyp²                       perimeter 45 = a + a + hyp    

P45 = 2a + hyp         a = sin(45) (hyp)  = [1/2√2)] (hyp)        sin(45) = (√2)/2

P45 = 2([1/2√2)] (hyp) + hyp

P45  = hyp (√2 + 1)

for the 30° right triangle

a = hyp×sin(30)     b = hyp×cos(30)           perimeter 30 = a + b + hyp    

            sin(30) = 1/2       cos(30) = (√3)/2

a  = hyp(1/2)           b = hyp(1/2)(√3)

P30 =        a        +      b         + hyp

        =  hyp(1/2)  +  hyp(1/2)(√3) + hyp

        =  hyp[ 1/2  +  (1/2)√3  + 1]

        =  hyp[ 1/2  +  (1/2)√3  + 1]             factored out the hyp

        =  (1/2)hyp[ 1 + √3 + 2]                  factored out a 1/2

P30  =  (1/2)hyp[ 3 + √3]

which is greater  P45 or P30?   vso I set both perimeters equal to each other

              P45           vs             P30

      hyp (√2 + 1)     =    (1/2)hyp[ 3 + √3]            the hyp cancel

               (√2 + 1)     =    (1/2)[ 3 + √3]                    still tough to tell  needed

               1.414 + 1    =    (1/2)[ 3 + 1.732]                to go to the calsulator

                 2.414      vs    1/2( 4.732)

                 2.414      >     2.366        thus P45 > P30    < -----  Final Answer

Disclaimer  it is up the student to verify all the geometry algabra and mathematical calculation.

Answer:

what he said

Step-by-step explanation:

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