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Question
from her home, eva would have to walk due north to get to her friend nellies house and due east to get to her friend jims house. it is 2.7 miles from evas house to jims house and a straight - line distance of 5.5 miles from nellies house to jims house. how far is evas house from nellies house? if necessary, round to the nearest tenth.
Step1: Identify the right - angled triangle
The path from Eva's house to Nellie's house, from Eva's house to Jim's house, and from Nellie's house to Jim's house form a right - angled triangle. Let the distance from Eva's house to Nellie's house be $a$, the distance from Eva's house to Jim's house be $b = 2.7$ miles, and the straight - line distance from Nellie's house to Jim's house be $c = 5.5$ miles. By the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, and we want to find $a$.
Step2: Rearrange the Pythagorean theorem
We can rewrite the Pythagorean theorem to solve for $a$: $a=\sqrt{c^{2}-b^{2}}$.
Step3: Substitute the values
Substitute $b = 2.7$ and $c = 5.5$ into the formula: $a=\sqrt{5.5^{2}-2.7^{2}}=\sqrt{(5.5 + 2.7)(5.5 - 2.7)}$ (using the difference of squares formula $x^{2}-y^{2}=(x + y)(x - y)$). First, calculate $(5.5 + 2.7)(5.5 - 2.7)=(8.2)\times(2.8)=22.96$. Then $a=\sqrt{22.96}\approx4.8$ miles.
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