Question

kevins house is due west of kingwood and due south of summerfield. kingwood is 56 kilometers from kevins house and 65 kilometers from summerfield. how far is summerfield from kevins house, measured in a straight line?
□ kilometers
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Explanation:

Step1: Identify the triangle type

We have a right triangle here, where Kevin's house (K), Kingwood (W), and Summerfield (S) form a right triangle with the right angle at Kevin's house. So, \( \triangle KWS \) is right - angled at \( K \). Let \( KW = 56 \) km (distance from Kevin's house to Kingwood), \( WS=65 \) km (distance from Kingwood to Summerfield), and we need to find \( KS \) (distance from Kevin's house to Summerfield). By the Pythagorean theorem, for a right - triangle with hypotenuse \( c \) and legs \( a \) and \( b \), \( c^{2}=a^{2}+b^{2} \), or \( b^{2}=c^{2}-a^{2} \) (if we want to find one of the legs). Here, \( WS \) is the hypotenuse (\( c = 65 \)), \( KW \) is one leg (\( a = 56 \)), and \( KS \) is the other leg (\( b \)).

Step2: Apply the Pythagorean theorem

We use the formula \( b=\sqrt{c^{2}-a^{2}} \). Substitute \( c = 65 \) and \( a = 56 \) into the formula:
First, calculate \( c^{2}-a^{2}=65^{2}-56^{2} \).
We know that \( 65^{2}=65\times65 = 4225 \) and \( 56^{2}=56\times56=3136 \).
Then \( 65^{2}-56^{2}=4225 - 3136=1089 \).
Now, find the square root of \( 1089 \), \( \sqrt{1089}=33 \).

Answer:

33