Question

janets school is due west of her house and due south of her friend luthers house. the distance between the school and luthers house is 4.8 miles and the straight-line distance between janets house and luthers house is 9 miles. how far is janets house from school? if necessary, round to the nearest tenth. miles submit

Explanation:

Step1: Apply Pythagorean theorem

Let \(a\) be the distance between Janet's house and school, \(b = 4.8\) miles (distance between school and Luther's house), \(c=9\) miles (distance between Janet's house and Luther's house). By Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), so \(a^{2}=c^{2}-b^{2}\).

Step2: Substitute values

Substitute \(b = 4.8\) and \(c = 9\) into \(a^{2}=c^{2}-b^{2}\). Then \(a^{2}=9^{2}-4.8^{2}=81 - 23.04=57.96\).

Step3: Solve for \(a\)

Take square root of \(a^{2}\), \(a=\sqrt{57.96}\approx7.6\)

Answer:

\(7.6\)