Question

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 1

what he said

Explanation:

Answer 2

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.