Precalculus word problem

Question

A Major League Baseball diamond has 4 bases, forming a square. The bases are 90 feet apart and the pitchers mound is 60.5 feet away from home plate. (The pitchers mound is not halfway between home plate and second base) 

Find the distance from the pitchers mound to first base. Round to nearest tenth of a foot. 

1. can you show me work as a picture please

Answer 1

Another solution is to use the Law of Cosines

Draw a diagram

The distance from the pitcher's mound to home plate is 60.5 feet.

Draw this segment. It bisects the 90° angle at home plate. (two 45° angles)

The distance from home plate to first base is 90 feet.

Draw a segment from the pitcher's mound to first base. This is the distance that you want.

Use the Law of Cosines.

Call the distance from the pitcher's mound to first base "x".

x^2=60.5^2+90^2-2(60.5)(90)cos45°

x^2=3660.25+8100-10,890(0.70710678)

x^2=11,760.25-7700.392

x^2=4059.858

take the square root of both sides

x=63.717015

x=63.7 feet

Answer 2

Properties of the square:

(1) diagonals are equal and bisect each other

(2) diagonals for 4 congruent right triangles

Iscoceles right triangles have hypotenuse that is sqrt(2) times the leg.

So then the distance from the pitcher's mound to each base is

90/sqrt(2) = 90*sqrt(2)/2 = 45*sqrt(2).

Let M be the point where the diagonals meet.

Let t be the distance from M to the pitcher's mound.

THen t = 45*sqrt(2) - 60.5

From M to first base is 45*sqrt(2).

Finally, from the pitcher's mound to M to first base is a right triangle

with legs t and 45*sqrt(2)

Pythagorean theorem says the distance from the pitcher's mound to first base is:

sqrt [t^2 + (45*sqrt(2))^2 ] =

sqrt[ (45*sqrt(2) - 60.5)(45*sqrt(2) - 60.5) + (45*sqrt(2))^2 ] =

sqrt[ 2025*2 - 2 * 60.5 * 45*sqrt(2) + 60.5^2 + 2025*2 ]

sqrt[ 8100 + 60.5^2 - 121*45*sqrt(2) ] =

sqrt[ 8100 + 3660.25 - 121*45*sqrt(2)) =

sqrt[ 11760.25 - 121*45*sqrt(2)] =

sqrt( 11760.25 - 7,700.3928471215025407251951033218‬]

=63.715046518687385003521786164029‬