I don't understand these 2 problems

[dcowboys107]

I have the file right here. Thanks for your help! I'd appreciate an explanation on how you solve the problems.

[ecoli]

hint for question 13, do you know anything about 3,4,5 triangles?

[The Thing]

Hint for 14 then:

If two triangles are similar, then:

One side of triangle 1 / the corresponding side of triangle 2 = another side of traingle 1/ the corresponding side of triangle 2.

[Garfield]

13- similar triangles, find k.

14- stick triangel CDB into ACD, similar triangles, find k.

[s pepperchin]

I have the file right here. Thanks for your help! I'd appreciate an explanation on how you solve the problems.

 

For the first problem you can use the fact that the two triangles share an angle and that means the ratio of two sides of the triangle are equal for any triangle

 

for the second one use the properties of angles and perpendicular lines.

 

That should help but I don't think I should give you the answer. I would however like to see your conclusion when you get it.

[EvoN1020v]

For #14, you can use algerba to find the length of x and y. Simple as that.

[s pepperchin]

for the first one the ratio of the top left side to the top right side is the same for both triangles, so:

 

[math]16/12 = 12/x[/math]

 

[math]16x = 144[/math]

 

[math]x = 9[/math]

 

same idea for y:

 

[math]12/15 = 4/y[/math]

 

[math]12y = 60[/math]

 

[math]y = 5[/math]

 

for the second one

 

triangles ACD and CBD are similar to ABC

 

so the ratios of sides are the same like the previous problem.

[andie]

#13 is just using c^2 = a^2 + b^2 and solving for x and y !

Very simple!

[andie]

for #14 you can take the cosine and sine of 45degrees and then solve for x and y . Very simple once again

 

what level of study is this if you dont mind me asking... im guessing highschool???