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From her home, Brittany would have to walk due north to get to her friend Christine's house and due east to get to her friend Tisha's house. It is $4$ miles from Brittany's house to Tisha's house and a straight-line distance of $5$ miles from Christine's house to Tisha's house. How far is Brittany's house from Christine's house? $\newline$ $\_\_\_\_\_\_\_\_\_$ miles

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Pythagorean theorem: word problems

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Q. From her home, Brittany would have to walk due north to get to her friend Christine's house and due east to get to her friend Tisha's house. It is

4

miles from Brittany's house to Tisha's house and a straight-line distance of

5

miles from Christine's house to Tisha's house. How far is Brittany's house from Christine's house?

\newline

\_\_\_\_\_\_\_\_\_

miles

Identify Triangle Formed: Identify the triangle formed by Brittany's, Christine's, and Tisha's houses. Brittany to Tisha is one leg ( $4$ miles), and Christine to Tisha is the hypotenuse ( $5$ miles). We need to find the other leg, which is the distance from Brittany's house to Christine's house.
Apply Pythagorean Theorem: Apply the Pythagorean Theorem: $a^2 + b^2 = c^2$ , where $c$ is the hypotenuse. Here, $a$ is the distance from Brittany to Christine, $b$ is $4$ miles, and $c$ is $5$ miles. Set up the equation: $a^2 + 4^2 = 5^2$ .
Calculate Squares: Calculate the squares: $4^2 = 16$ , $5^2 = 25$ . Plug these into the equation: $a^2 + 16 = 25$ .
Solve for $a^2$ : Solve for $a^2$ : $a^2 = 25 - 16$ . $a^2 = 9$ .
Find $a$ : Find $a$ by taking the square root of $9$ . $a = \sqrt{9}$ , so $a = 3$ .

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