MATH QUESTION!!

Question

Incline mats, or triangle mats, are offered with different levels of incline to help gymnasts learn basic moves. As the name may suggest, two sides of the mat are right triangles. If the height of the mat is  14 inches shorter than the length of the mat and the hypotenuse is 2  inches longer than the length of the mat, what is the length of the mat?

Answer 1

Solution:

h = L -14 in, hyp = L + 2in , also hypotenuse = √(L² + h²) = L + 2in take square of both sides

L²+ h² = ( L + 2)² , but h = L-14 substitute h and expand

L² + (L -14)² = L² + 4L + 4, L² + L² - 28L + 196 = L² + 4L + 4, simplify and rearrange

L² - 32 L + 192= 0 Solve by completing the square

L²-32 L = -192 , b=32 and (b/2) = -16, ( b/2)² = 256 add to both sides

L² -32 L + 256 = -192 + 256 = 64

(L-16)² = 64 take square root of both sides, L -16 = ± 8 , L -16 = 8. L = 24 in, L -16 = -8. L = 8 in

Try L = 24 in, h = 24 -14 = 10in . hyp = √(24)²+(10)² = √(576 + 100) = 26 in = L + 2 = 24 + 2 = 26 in,

Try L = 8in, h = 8-14 =-6 -ve value and this option is not possible therefore L = 24 in

Answer 2

a= l-14 a=height, l= length

h=2+l h=the hypotenuse

h^2 = a^2 + l^2

(2+l)^2 = (l-14)^2 + l^2

l^2 + 4l + 4 = l^2 -28l + 196 + l^2

l^2 -32l +192 =0

(l-24)(l-8) = 0

l = 8 or 24 inches long (ignore 8, as that would make height a negative number)

a =24-14 = 10 inches high

hypotenuse = 26 inches

26^2 = 24^2 + 10^2

676 = 576 +100