many triangles

[logistic_guy]

here is the question

In each figure, find the value of \(\displaystyle x\).

Figure \(\displaystyle 1\)


Figure \(\displaystyle 2\)



my attemb
i'll start with figure \(\displaystyle 1\)
from the first triangle i've \(\displaystyle 13^2 = x^2 + a^2\)
from the second triangle i've \(\displaystyle 15^2 = x^2 + b^2\)
rule of thumb say if i've three unknown, i need three equations
the problem is the big triangle isn't right triangle, so i think it's wrong to say \(\displaystyle 15^2 = 14^2 + 12^2\)
so what formula should i use to use the big triangle? i don't have something in mind

 

[Dr.Peterson]

here is the question

In each figure, find the value of \(\displaystyle x\).

Figure \(\displaystyle 1\)
View attachment 38988

Figure \(\displaystyle 2\)
View attachment 38989


my attemb
i'll start with figure \(\displaystyle 1\)
from the first triangle i've \(\displaystyle 13^2 = x^2 + a^2\)
from the second triangle i've \(\displaystyle 15^2 = x^2 + b^2\)
rule of thumb say if i've three unknown, i need three equations
the problem is the big triangle isn't right triangle, so i think it's wrong to say \(\displaystyle 15^2 = 14^2 + 12^2\)
so what formula should i use to use the big triangle? i don't have something in mind

You've written two equations, \(\displaystyle 13^2 = x^2 + a^2\) and \(\displaystyle 15^2 = x^2 + b^2\).

There's a third equation in only a and b, which is so obvious you missed it. (If you'd added a and b to the picture, you would have seen it.)

Now solve that system of equations. I'd eliminate x first and find a and b, then use them to find x; but there are other ways.

 

[logistic_guy]

There's a third equation in only a and b, which is so obvious you missed it.

i don't see this

\(\displaystyle a + b = 14\)
\(\displaystyle a = 14 - b\)
\(\displaystyle 13^2 = x^2 + (14 - b)^2\)
\(\displaystyle 13^2 - (14 - b)^2 = x^2\)
\(\displaystyle 15^2 = 13^2 - (14 - b)^2 + b^2\)
\(\displaystyle 225 = 169 - 196 + 28b - b^2 + b^2\)
\(\displaystyle 225 = -27 + 28b\)
\(\displaystyle 225 + 27 = 28b\)
\(\displaystyle 252 = 28b\)
\(\displaystyle b = \frac{252}{28} = 9\)
if my caclulation correct it mean \(\displaystyle a = 5\)

i'll solve this equation
\(\displaystyle 13^2 = x^2 + a^2\)
\(\displaystyle 13^2 = x^2 + 5^2\)
\(\displaystyle 169 = x^2 + 25\)
\(\displaystyle 169 - 25 = x^2\)
\(\displaystyle 144 = x^2\)
\(\displaystyle x = \sqrt{144} = 12\)

is my analize correct?

 

[Dr.Peterson]

You can, of course, check your answer in all three equations:

\(\displaystyle 13^2 = x^2 + a^2\Rightarrow13^2 = 12^2 + 5^2\)
\(\displaystyle 15^2 = x^2 + b^2\Rightarrow15^2 = 12^2 + 9^2\)
\(\displaystyle a+b=14\Rightarrow4+9=14\)

So it's correct.

My method was to subtract the first two equations, giving [imath]15^2-13^2=b^2-a^2[/imath], so [imath]b^2-a^2=56[/imath], which factors as [imath](b+a)(b-a)=56[/imath], so [imath]b-a=56/14=4[/imath]. Then solve this with the third equation to find a and b.

 

[khansaheb]

You can, of course, check your answer in all three equations:

\(\displaystyle 13^2 = x^2 + a^2\Rightarrow13^2 = 12^2 + 5^2\)
\(\displaystyle 15^2 = x^2 + b^2\Rightarrow15^2 = 12^2 + 9^2\)
\(\displaystyle a+b=14\Rightarrow4+9=14\)................................................ typo

So it's correct.

My method was to subtract the first two equations, giving [imath]15^2-13^2=b^2-a^2[/imath], so [imath]b^2-a^2=56[/imath], which factors as [imath](b+a)(b-a)=56[/imath], so [imath]b-a=56/14=4[/imath]. Then solve this with the third equation to find a and b.

Small typo:
5 + 9 = 14

 

[logistic_guy]

You can, of course, check your answer in all three equations:

\(\displaystyle 13^2 = x^2 + a^2\Rightarrow13^2 = 12^2 + 5^2\)
\(\displaystyle 15^2 = x^2 + b^2\Rightarrow15^2 = 12^2 + 9^2\)
\(\displaystyle a+b=14\Rightarrow4+9=14\)

So it's correct.

My method was to subtract the first two equations, giving [imath]15^2-13^2=b^2-a^2[/imath], so [imath]b^2-a^2=56[/imath], which factors as [imath](b+a)(b-a)=56[/imath], so [imath]b-a=56/14=4[/imath]. Then solve this with the third equation to find a and b.

i like this idea

so \(\displaystyle b - a = 4\) and \(\displaystyle (a + b) + (b - a) = 14 + 4 = 18 = a + b + b - a = 2b\)
\(\displaystyle b = \frac{18}{4} = 4.5\) which is wrong

i know if solve
\(\displaystyle a + b = 14\)
\(\displaystyle b - a = 4\)
i'll get a correct answer

but in the same time i know \(\displaystyle (a + b) + (b - a) = 18 = 2b\) is also true but it give worng answerwhy?

 

[Dr.Peterson]

i like this idea

so \(\displaystyle b - a = 4\) and \(\displaystyle (a + b) + (b - a) = 14 + 4 = 18 = a + b + b - a = 2b\)
\(\displaystyle b = \frac{18}{4} = 4.5\) which is wrong

No, if 2b = 18, then b = 9.

I clearly read my typo as what I meant it to be, 5. I don't know why you divided by 4 instead of 2 ...

i know if solve
\(\displaystyle a + b = 14\)
\(\displaystyle b - a = 4\)
i'll get a correct answer

but in the same time i know \(\displaystyle (a + b) + (b - a) = 18 = 2b\) is also true but it give worng answerwhy?

Because it does not give the wrong answer ...

 

[logistic_guy]

No, if 2b = 18, then b = 9.

I clearly read my typo as what I meant it to be, 5. I don't know why you divided by 4 instead of 2 ...

i don't know why i divide by \(\displaystyle 4\)

Because it does not give the wrong answer ...

i see it. i appreciate that

for figure \(\displaystyle 2\). what's the plan there? is the leg \(\displaystyle 36\) connected to the leg \(\displaystyle x\) make a straight segment? it appear to me not straight

 

[Dr.Peterson]

for figure \(\displaystyle 2\). what's the plan there? is the leg \(\displaystyle 36\) connected to the leg \(\displaystyle x\) make a straight segment? it appear to me not straight

Why bother to assume anything until you know whether it's needed? And why assume anything that is not stated, at all, and doesn't even look true? Just use what is clearly true.



This one is very straightforward, and all you need is the series of right triangles. Start with one where two sides are known, and repeat.

Just do it!

 

[The Highlander]

Just do it!

He hardly ever does anything... it's just question after question!

Using Pythagoras' Theorem (throughout)

Find the lengths of: c
then (a + b)
then b
then a (not Pythagoras, just simple subtraction!)
then x
and show us your working and result for x, please.