Question

If it is 90 feet between bases, how far is it from home base to 2nd base?

90^2+b^2=90^2; 127.3 ft
90^2+90^2=c^2; 127.3 ft
90^2+90^2=c^2; 180 ft
90^2+90^2=c^2; 19 ft

Answer 1

To find the distance from home base to second base in a baseball diamond, which is a square with each side measuring 90 feet, we can use the Pythagorean theorem.

The distance from home base to second base is the hypotenuse of a right triangle formed by two sides (the distances from home to first base and from first base to second base), both of which are 90 feet.

Using the Pythagorean theorem:

\[
c^2 = a^2 + b^2
\]

where:
- \( c \) is the distance from home base to second base,
- \( a \) (home to first base) is 90 feet,
- \( b \) (first to second base) is 90 feet.

Plugging in the values:

\[
c^2 = 90^2 + 90^2
\]
\[
c^2 = 8100 + 8100
\]
\[
c^2 = 16200
\]
\[
c = \sqrt{16200}
\]
\[
c = 90\sqrt{2} \approx 127.3 \text{ feet}
\]

Therefore, the distance from home base to second base is approximately **127.3 feet**.