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Danville Street and Springtown Avenue intersect. If Danville Street is $8\,\text{meters}$ wide and Springtown Avenue is $6\,\text{meters}$ wide, what is the distance between two opposite corners of the intersection? $\newline$ _________ $\text{meters}$

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Pythagorean theorem: word problems

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Q. Danville Street and Springtown Avenue intersect. If Danville Street is

8\,\text{meters}

wide and Springtown Avenue is

6\,\text{meters}

wide, what is the distance between two opposite corners of the intersection?

\newline

_________

\text{meters}

Identify Intersection Dimensions: Identify the legs and hypotenuse of the intersection. $\newline$ Danville Street is $8$ meters wide, and Springtown Avenue is $6$ meters wide. These are the legs of the right triangle formed by the intersection. The distance between two opposite corners of the intersection will be the hypotenuse. $\newline$ Legs: $8$ meters, $6$ meters $\newline$ Hypotenuse: Let's call it ' $d$ ' meters.
Use Pythagorean Theorem: Use the Pythagorean Theorem to find the hypotenuse. $\newline$ The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides (the legs). This can be written as: $\newline$ $(\text{Leg1})^2 + (\text{Leg2})^2 = (\text{Hypotenuse})^2$ $\newline$ So, we have: $\newline$ $8^2 + 6^2 = d^2$
Calculate Leg Squares: Calculate the squares of the legs. $\newline$ $8^2 = 8 \times 8 = 64$ $\newline$ $6^2 = 6 \times 6 = 36$ $\newline$ Now, add these squares to find the value of $d^2$ . $\newline$ $64 + 36 = d^2$
Add Squares: Add the results to find the value of $d^2$ . $\newline$ $64 + 36 = 100$ $\newline$ So, $d^2 = 100$
Find Hypotenize: Find the value of $d$ by taking the square root of $d^2$ . $\sqrt{100} = \sqrt{d^2}$ $10 = d$
Answer Prompt: Answer the question prompt. $\newline$ The distance between two opposite corners of the intersection is $10$ meters.

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