Right triangle problem

[lev888]

Take the ratio of the corresponding parts and equate them. So we get:
6/9 = EF/DE
2/3 = EF/(3 - EF)
Cross multiplying, we get:
6 -2EF = 3EF
6 = 5EF
EF = 1.2
DE = 1.8

Using the pythagorean theorem, AE = Square root[36 + (1.2)^2]
= Sq. rt(37.44)
= 6.12 rounded
Similarly, EC = 9.18
Adding the two we get AC = 15.3
x^2 = 15.3^2 - 49
= 185.09
x = 13.60 cm

[math]AE = \sqrt(37.44)[/math][math]CE = \sqrt(84.24)[/math][math]AC = \sqrt(37.44) + \sqrt(84.24) [/math][math]AC^2 = 37.44 + 2*\sqrt(37.44*84.24) + 84.24 = 37.44 + 2*56.16 + 84.24 = 234[/math][math]AC = \sqrt(234)[/math]
Edited: replaced the first * with + in the 4th line.

 

[mathnerd2021]

[math]AE = \sqrt(37.44)[/math][math]CE = \sqrt(84.24)[/math][math]AC = \sqrt(37.44) + \sqrt(84.24) [/math][math]AC^2 = 37.44 * 2*\sqrt(37.44*84.24) + 84.24 = 37.44 + 2*56.16 + 84.24 = 234[/math][math]AC = \sqrt(234)[/math]

Thanks a ton!

 

[Cubist]

How did you get from the first step to step two?

AC = AE + CE, use eqn (1) and (2)

[imath]=\sqrt{6^2+\left(3-\frac{9}{5}\right)^2} + \sqrt{9^2+\left(\frac{9}{5}\right)^2}[/imath]

[imath]=\sqrt{6^2+\left(\frac{\cancel{3\times 5 - 9}\,6}{5}\right)^2} + \frac{9}{5}\sqrt{\left(\frac{5}{9}\right)^29^2 + \left(\frac{5}{9}\right)^2\left(\frac{9}{5}\right)^2}[/imath]

[imath]=\frac{6}{5}\sqrt{\left(\frac{5}{6}\right)^2 6^2 + \left(\frac{5}{6}\right)^2\left(\frac{6}{5}\right)^2} + \frac{9}{5}\sqrt{5^2 + 1}[/imath]


[imath]=\frac{6}{5}\sqrt{5^2+1}[/imath] + [imath]\frac{9}{5}\sqrt{26}[/imath]

[imath]=3\sqrt{26}=\sqrt{234}[/imath]

The @Dr.Peterson method is MUCH more awesome!

BTW: If you're giving this question to students, since you teach math, then I'd recommend giving them a less distorted picture than the original post. However, if it's homework or a "fun test" (is there such a thing?) then the distorted diagram is a good lesson that a reasonable quality sketch helps to solve a problem, and a bad (misleading) sketch can hinder progress.

The question asks for an exact answer. Are my calculation steps fine for getting an exact answer? But I have rounded in between, so is this incorrect If I want to get exact answers?

Once you've rounded, then you've abandoned all hope of an exact answer (IMO !)

 

[mathnerd2021]

AC = AE + CE, use eqn (1) and (2)

[imath]=\sqrt{6^2+\left(3-\frac{9}{5}\right)^2} + \sqrt{9^2+\left(\frac{9}{5}\right)^2}[/imath]

[imath]=\sqrt{6^2+\left(\frac{\cancel{3\times 5 - 9}\,6}{5}\right)^2} + \frac{9}{5}\sqrt{\left(\frac{5}{9}\right)^29^2 + \left(\frac{5}{9}\right)^2\left(\frac{9}{5}\right)^2}[/imath]

[imath]=\frac{6}{5}\sqrt{\left(\frac{5}{6}\right)^2 6^2 + \left(\frac{5}{6}\right)^2\left(\frac{6}{5}\right)^2} + \frac{9}{5}\sqrt{5^2 + 1}[/imath]


[imath]=\frac{6}{5}\sqrt{5^2+1}[/imath] + [imath]\frac{9}{5}\sqrt{26}[/imath]

[imath]=3\sqrt{26}=\sqrt{234}[/imath]

The @Dr.Peterson method is MUCH more awesome!

BTW: If you're giving this question to students, since you teach math, then I'd recommend giving them a less distorted picture than the original post. However, if it's homework or a "fun test" (is there such a thing?) then the distorted diagram is a good lesson that a reasonable quality sketch helps to solve a problem, and a bad (misleading) sketch can hinder progress.


Once you've rounded, then you've abandoned all hope of an exact answer (IMO !)

Thank you!