Right triangle problem

[mathnerd2021]

Hi,

Can someone help me solve this as soon as possible. Many thanks!

 

[mathnerd2021]

I tried extending the lines to figure it out but couldn't get anywhere.

 

[lev888]

Draw the hypotenuse of the right triangle with legs x and 7. It creates 2 triangles with long legs 9 and 6 and short legs whose sum is 3. See them? These 2 triangles are similar. Why? Use this fact to find the hypotenuse, and then x.

 

[Harry_the_cat]

If the angle between the 3cm and 6cm side is a right angle as marked, then the 6cm side and the 9cm side are parallel.

 

[mathnerd2021]

Draw the hypotenuse of the right triangle with legs x and 7. It creates 2 triangles with long legs 9 and 6 and short legs whose sum is 3. See them? These 2 triangles are similar. Why? Use this fact to find the hypotenuse, and then x.

Do you mean this way?

 

[mathnerd2021]

But that line that I drew that extends from the 6 cm line cannot be a straight line.

Do you mean this way? View attachment 28381

 

[mathnerd2021]

If the angle between the 3cm and 6cm side is a right angle as marked, then the 6cm side and the 9cm side are parallel.

Yes I got the same thing. But how do we proceed from there.

 

[lev888]

Do you mean this way? View attachment 28381

No. The 9 and 6 segments are parallel (do you see why?), the lines they belong to can't intersect. The hypotenuse should not be colinear with the 6cm segment.

 

[mathnerd2021]

I was able to add in a few more details.

 

[Cubist]

I think @lev888 means more like this...


EDIT: I had to omit the 3cm measurement for clarity. DF=3cm. But I'd let y=DE, and then EF=(3-y)

 

[mathnerd2021]

No. The 9 and 6 segments are parallel (do you see why?), they can't intersect. The hypotenuse should not be colinear with the 6cm segment.

Oh I get what you mean now.

 

[mathnerd2021]

I think @lev888 means more like this...
View attachment 28383

Thank you.

 

[mathnerd2021]

Awesome. Thank you so much everyone! I got x = 13.6 after rounding.

 

[lev888]

Awesome. Thank you so much everyone! I got x = 13.6 after rounding.

Feel free to post your solution.

 

[mathnerd2021]

Feel free to post your solution.

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

 

[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

Looks good (didn't check all calculations).

 

[Cubist]

You beat me to it! But my work leads to an exact answer ? Let y=DE. Using Pythagoras on △FAE and △DCE...

[imath](AE)^2 = 6^2 + (3-y)^2 \hspace2ex \text{[eqn 1]}[/imath]

[imath](CE)^2 = 9^2 + y^2 \hspace2ex \text{[eqn 2]}[/imath]

By similar triangles △FAE and △DCE...

[imath]\frac{AE}{CE} = \frac{6}{9}[/imath]

[imath]3(AE) = 2(CE) [/imath]

[imath]9(AE)^2 = 4(CE)^2 \hspace2ex[/imath] now use eqn 1 & eqn 2...

[imath]9(6^2 + (3-y)^2) = 4(9^2 + y^2)[/imath]

...leading to [imath]y = 9[/imath] or [imath]\frac{9}{5}[/imath] however [imath]y<3[/imath] therefore [imath]y=\frac{9}{5}[/imath]. Now...

[imath]AC = \sqrt{6^2 + (3-y)^2} + \sqrt{9^2 + y^2}[/imath]

Using [imath]y=\frac{9}{5}[/imath] leads to [imath]AC = \sqrt{234}[/imath]. Then use Pythagoras on △BAC...

[imath]234 = x^2 + 7^2[/imath]

[imath]x = \sqrt{185}[/imath]cm

≈ your answer of 13.6cm

EDIT: Your method to find EF & DE is much better than mine

 

[Dr.Peterson]

For what it's worth, here's my method, where G is the intersection of CD extended and the perpendicular to it from A:



[imath]AC^2 = CG^2 + GA^2 = (9+6)^2 + 3^2 = 234[/imath]

[imath]x^2 = AC^2 - AB^2 = 234 - 7^2 = 185[/imath]

[imath]x = \sqrt{185} \approx 13.6[/imath] cm

 

[mathnerd2021]

For what it's worth, here's my method, where G is the intersection of CD extended and the perpendicular to it from A:

View attachment 28388

[imath]AC^2 = CG^2 + GA^2 = (9+6)^2 + 3^2 = 234[/imath]

[imath]x^2 = AC^2 - AB^2 = 234 - 7^2 = 185[/imath]

[imath]x = \sqrt{185} \approx 13.6[/imath] cm

Awesome!

 

[mathnerd2021]

You beat me to it! But my work leads to an exact answer ? Let y=DE. Using Pythagoras on △FAE and △DCE...

[imath](AE)^2 = 6^2 + (3-y)^2 \hspace2ex \text{[eqn 1]}[/imath]

[imath](CE)^2 = 9^2 + y^2 \hspace2ex \text{[eqn 2]}[/imath]

By similar triangles △FAE and △DCE...

[imath]\frac{AE}{CE} = \frac{6}{9}[/imath]

[imath]3(AE) = 2(CE) [/imath]

[imath]9(AE)^2 = 4(CE)^2 \hspace2ex[/imath] now use eqn 1 & eqn 2...

[imath]9(6^2 + (3-y)^2) = 4(9^2 + y^2)[/imath]

...leading to [imath]y = 9[/imath] or [imath]\frac{9}{5}[/imath] however [imath]y<3[/imath] therefore [imath]y=\frac{9}{5}[/imath]. Now...

[imath]AC = \sqrt{6^2 + (3-y)^2} + \sqrt{9^2 + y^2}[/imath]

Using [imath]y=\frac{9}{5}[/imath] leads to [imath]AC = \sqrt{234}[/imath]. Then use Pythagoras on △BAC...

[imath]234 = x^2 + 7^2[/imath]

[imath]x = \sqrt{185}[/imath]cm

≈ your answer of 13.6cm

EDIT: Your method to find EF & DE is much better than mine

How did you get from the first step to step two? That is, how did you get AC = √234 from the previous step?
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?