Question

devon likes to skate at a movie theater that is due south of her school and due west of her favorite game store. if the movie theater is 4.2 miles from her school and the straight - line distance between the school and the game store is 4.6 miles, how far is the movie theater from the game store? if necessary, round to the nearest tenth.

Explanation:

Step1: Identify the right - angled triangle

The distance from school to movie - theater, distance from movie - theater to game store and distance from school to game store form a right - angled triangle. Let the distance from school to movie - theater be $a = 4.2$ miles, the distance from movie - theater to game store be $b$, and the distance from school to game store be $c = 4.6$ miles. By the Pythagorean theorem $a^{2}+b^{2}=c^{2}$.

Step2: Rearrange the Pythagorean theorem to solve for $b$

We can rewrite the Pythagorean theorem as $b=\sqrt{c^{2}-a^{2}}$.

Step3: Substitute the values of $a$ and $c$

Substitute $a = 4.2$ and $c = 4.6$ into the formula: $b=\sqrt{4.6^{2}-4.2^{2}}=\sqrt{(4.6 + 4.2)(4.6 - 4.2)}$ (using the difference - of - squares formula $x^{2}-y^{2}=(x + y)(x - y)$). First, calculate $(4.6 + 4.2)=8.8$ and $(4.6 - 4.2)=0.4$. Then $b=\sqrt{8.8\times0.4}=\sqrt{3.52}$.

Step4: Calculate the value of $b$

$b=\sqrt{3.52}\approx1.9$ miles.

Answer:

$1.9$