Find the lengths of the sides of the triangle.
2007-08-22 05:44:14 · 5 answers · asked by Raelene G 1 in Science & Mathematics ➔ Mathematics
so from a^2+b^2=c^2 we have a^2+(a+1.6)^2=2.7^2
so we get 2a^2+3.2a+1.6^2=2.7^2
and then 2a^2+3.2a+(1.6^2-2.7)^2=0
now use the quadratic formula to solve for a.
2007-08-22 05:49:12 · answer #1 · answered by Saul L 2 · 1⤊ 0⤋
a^2 + b^2 = c^2
c = 2.7
a - b = 1.6
a = b + 1.6
c^2 = (b + 1.6)^2 + b^2
(2.7)^2 = b^2 + 3.2b + (1.6)^2 + b^2
7.29 = 2b^2 + 3.2b + 2.56
2b^2 + 3.2b - 4.73 = 0
use quadratic:
b = -2.533
a = -2.533 + 1.6 = -0.933
lengths must be positive so take the absolute value of each to get:
a = 0.933
b = 2.533
check:
(0.933)^2 + (2.533)^2 = 0.87 + 6.41 = 7.28
sqrt(7.28) = 2.7 = c
2007-08-22 06:02:32 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Triangle sides a^2+b^2 = C^2 (C is the hyp)
hyp = 2.7
short side = x
long side = x+1.6
so
x^2 + (x+1.6)^2 = 2.7
x^2 + (x^2+3.2x+2.56) = 2.7
solve for x
2007-08-22 05:55:02 · answer #3 · answered by b_ney26 3 · 0⤊ 0⤋
(x+1.6)² + x² = 2.7²
x² + 3.2x + 2.56 + x² = 7.29
2x² + 3.2x - 4.73 = 0
Apply quadratic.
2007-08-22 05:49:28 · answer #4 · answered by gebobs 6 · 0⤊ 1⤋
Sorry your question is not that clear to me
2007-08-22 05:48:01 · answer #5 · answered by King 2 · 0⤊ 0⤋