right triangle

[logistic_guy]

here is the question

Find the values of $\displaystyle x$ and $\displaystyle y$ in the triangle.




my attemb
i say it's impossible to find $\displaystyle x$ and $\displaystyle y$
proof of my claim
1. $\displaystyle x^2 + y^2 = h^2$
can't be solved with $\displaystyle 2$ sides unknown

2. $\displaystyle \sin \theta$ and $\displaystyle \cos \theta$ and $\displaystyle \tan \theta$
can't be used because angles not given

3. $\displaystyle x + y + h =$ perimeter
can't be used because i've $\displaystyle 3$ unknown

4. $\displaystyle A = \frac{1}{2}xy$
even area not given mean $\displaystyle 3$ unknown

my think tell me drawing a line from $\displaystyle B$ to $\displaystyle D$ won't change anything
so i think this question aim is only to measure the student reaction

am i correct?

 

[topsquark]

here is the question

Find the values of $\displaystyle x$ and $\displaystyle y$ in the triangle.

View attachment 38954


my attemb
i say it's impossible to find $\displaystyle x$ and $\displaystyle y$
proof of my claim
1. $\displaystyle x^2 + y^2 = h^2$
can't be solved with $\displaystyle 2$ sides unknown

2. $\displaystyle \sin \theta$ and $\displaystyle \cos \theta$ and $\displaystyle \tan \theta$
can't be used because angles not given

3. $\displaystyle x + y + h =$ perimeter
can't be used because i've $\displaystyle 3$ unknown

4. $\displaystyle A = \frac{1}{2}xy$
even area not given mean $\displaystyle 3$ unknown

my think tell me drawing a line from $\displaystyle B$ to $\displaystyle D$ won't change anything
so i think this question aim is only to measure the student reaction

am i correct?

Nope. Do you see any similar triangles in there?

-Dan

 

[pka]

here is the question

Find the values of $\displaystyle x$ and $\displaystyle y$ in the triangle.

View attachment 38954

Question: Have you studied the mean proportional?
If you have then does it follow that $||BD||=6$ (the length)?

 

[logistic_guy]

thank

Do you see any similar triangles in there?

yes

i foget to write the $\displaystyle 5$th idea which is similar triangles
that's the idea of the line $\displaystyle BD$. as i said it won't help

5. $\displaystyle \frac{a}{b} = \frac{c}{d}$
take any ratio of the similar triangles, you'll always have equation with $\displaystyle 2$ unknown

i'll show you one try
$\displaystyle \frac{y}{4} = \frac{x}{BD}$

Question: Have you studied the mean proportional?
If you have then does it follow that $||BD||=6$ (the length)?

thank pka
i don't see your reply before i post

no i don't study the mean
do you mean the length of $\displaystyle BD = \lfloor\frac{9 + 4}{2}\rfloor = \lfloor \frac{13}{2}\rfloor = \lfloor 6.5\rfloor = 6$, always?

 

[pka]

thank pka
i don't see your reply before i post

no i don't study the mean
do you mean the length of $\displaystyle BD = \lfloor\frac{9 + 4}{2}\rfloor = \lfloor \frac{13}{2}\rfloor = \lfloor 6.5\rfloor = 6$, always?

NO! this has nothing to do with the floor function. Do a web search for the mean proportional.

 

[logistic_guy]

NO! this has nothing to do with the floor function. Do a web search for the mean proportional.

i read about the mean proportional
it say the mean proportional between two numbers $\displaystyle \alpha$ and $\displaystyle \beta$ is the number $\displaystyle \gamma$ such that
$\displaystyle \frac{\alpha}{\gamma} = \frac{\gamma}{\beta}$
$\displaystyle \gamma^2 = \alpha\beta$
$\displaystyle \gamma = \sqrt{\alpha\beta}$

let's assumed i agree to what it say

then it say in right triangle If an altitude is drawn from the right angle to the hypotenuse, it divides the hypotenuse into two segments, say $\displaystyle a$ and $\displaystyle b$
then the altitude say $\displaystyle h$ is the mean proportional between these two segments
$\displaystyle h^2 = ab$
$\displaystyle h = \sqrt{ab}$

what i understand from this
$\displaystyle BD^2 = 4 \times 9$
it don't make sense

 

[topsquark]

thank


yes

i foget to write the $\displaystyle 5$th idea which is similar triangles
that's the idea of the line $\displaystyle BD$. as i said it won't help

5. $\displaystyle \frac{a}{b} = \frac{c}{d}$
take any ratio of the similar triangles, you'll always have equation with $\displaystyle 2$ unknown

i'll show you one try
$\displaystyle \frac{y}{4} = \frac{x}{BD}$

You are forgetting the Pythagorean Theorem. Two equations, two unknowns.

-Dan

 

[logistic_guy]

You are forgetting the Pythagorean Theorem. Two equations, two unknowns.

-Dan

you mean two equations, three unknown
$\displaystyle x$
$\displaystyle y$
$\displaystyle BD$

 

[Aion]

here is the question

Find the values of $\displaystyle x$ and $\displaystyle y$ in the triangle.

View attachment 38954


my attemb
i say it's impossible to find $\displaystyle x$ and $\displaystyle y$
proof of my claim
1. $\displaystyle x^2 + y^2 = h^2$
can't be solved with $\displaystyle 2$ sides unknown

2. $\displaystyle \sin \theta$ and $\displaystyle \cos \theta$ and $\displaystyle \tan \theta$
can't be used because angles not given

3. $\displaystyle x + y + h =$ perimeter
can't be used because i've $\displaystyle 3$ unknown

4. $\displaystyle A = \frac{1}{2}xy$
even area not given mean $\displaystyle 3$ unknown

my think tell me drawing a line from $\displaystyle B$ to $\displaystyle D$ won't change anything
so i think this question aim is only to measure the student reaction

am i correct?

From similarity $\Delta ADB \sim \Delta ABC$ the ratios of corresponding sides are equal.

$$\frac{AD}{AB}=\frac{AB}{AC}$$
Now $AB=y, AC=13, AD=4$ thus

$$\frac{4}{y}=\frac{y}{13}$$

 

[logistic_guy]

From similarity $\Delta ADB \sim \Delta ABC$ the ratios of corresponding sides are equal.

$$\frac{AD}{AB}=\frac{AB}{AC}$$
Now $AB=y, AC=13, AD=4$ thus

$$\frac{4}{y}=\frac{y}{13}$$

i think you've a mistake

 

[Aion]

i think you've a mistake

What part do you think is incorrect?

$\Delta ADB \sim \Delta ABC$, because they both share $\angle A$ and $\angle ADB=\angle ABC =90^{\circ}$.

 

[logistic_guy]

What part do you think is incorrect?

$\Delta ADB \sim \Delta ABC$, because they both share $\angle A$ and $\angle ADB=\angle ABC =90^{\circ}$.

the line $\displaystyle BD$ divide the big triangle into two small similar triangles
the big triangle isn't similar to them so you can't compare it with anyone of them

the correct way is to say $\displaystyle \Delta ABD \sim \Delta BCD$

 

[Aion]

the line $\displaystyle BD$ divide the big triangle into two small similar triangles
the big triangle isn't similar to them so you can't compare it with anyone of them

the correct way is to say $\displaystyle \Delta ABD \sim \Delta BCD$

No, all three triangles are similar to each other. $\Delta ADB \sim \Delta BDC \sim \Delta ABC$. The reason is that $\angle ADB=\angle BDC =\angle ABC= 90^{\circ}$, and since $\angle BAD=\angle CBD$, each of the three triangles have two angles in common, which is sufficient to prove that they are similar.

 

[logistic_guy]

No, all three triangles are similar to each other. $\Delta ADB \sim \Delta BDC \sim \Delta ABC$. The reason is that $\angle ADB=\angle BDC =\angle ABC= 90^{\circ}$, and since $\angle BAD=\angle CBD$, each of the three triangles have two angles in common, which is sufficient to prove that they are similar.

thank teacher very much
this i don't know before

by this new information, what pka say in post $\displaystyle 3$ now make sense
so $\displaystyle BD = 6$

$\displaystyle x^2 = 6^2 + 9^2 = 36 + 81 =117$
$\displaystyle x = \sqrt{117} = 3\sqrt{13}$

$\displaystyle y^2 = 6^2 + 4^2 = 36 + 16 = 52$
$\displaystyle y = \sqrt{52} = 2\sqrt{13}$

test my answer in the big triangle
$\displaystyle x^2 + y^2 = (3\sqrt{13})^2 + (2\sqrt{13})^2 = 9(13) + 4(13) = 117 + 52 = 169$
$\displaystyle (4 + 9)^2 = 13^2 = 169$

they're equal

 

[Agent Smith]

$x^2 = z^2 + 81$ and $y^2 = z^2 + 16$

$x^2 - 81 = y^2 - 16 \implies x^2 - y^2 = 65$ (equation 1)

$(x + y)(x -y) = 65$

$x + y = 13 \text{ and } x - y = 5$

$2x = 18 \implies x = 9 \text{ and } y = 4$

I think

PS: Obviously wrong, but I don't know where I made a mistake.

It seems that $x^2 + y^2 = 169$ (equation 2)

$2x^2 = 234 \implies x^2 = 117 \implies x = \sqrt {117}$

$y = \sqrt{52}$

 

[logistic_guy]

View attachment 38956

$x^2 = z^2 + 81$ and $y^2 = z^2 + 16$

$x^2 - 81 = y^2 - 16 \implies x^2 - y^2 = 65$ (equation 1)

$(x + y)(x -y) = 65$

$x + y = 13 \text{ and } x - y = 5$

$2x = 18 \implies x = 9 \text{ and } y = 4$

I think

PS: Obviously wrong, but I don't know where I made a mistake.

It seems that $x^2 + y^2 = 169$ (equation 2)

$2x^2 = 234 \implies x^2 = 117 \implies x = \sqrt {117}$

$y = \sqrt{52}$

you give me three equations with three unknown
$\displaystyle x^2 = z^2 + 9^2$
$\displaystyle y^2 = z^2 + 4^2$
$\displaystyle x^2 + y^2 = (4 + 9)^2$

that's possible to solve the questionwithout trouble

thank smith very much

 

[Dr.Peterson]

$x^2 = z^2 + 81$ and $y^2 = z^2 + 16$
$x^2 - 81 = y^2 - 16 \implies x^2 - y^2 = 65$ (equation 1)
$(x + y)(x -y) = 65$
$x + y = 13 \text{ and } x - y = 5$ --- NO!
$2x = 18 \implies x = 9 \text{ and } y = 4$

I think

PS: Obviously wrong, but I don't know where I made a mistake.

You can't arbitrarily factor 65 and assign one factor to x+y and the other to x-y.

This is a result of teaching students to expect everything to be integers!

While 9 and 4 satisfy this one equation, they don't satisfy the other, which you ignored.

Given the correct solution, you can see that $x+y=3\sqrt{13}+2\sqrt{13}=5\sqrt{13}$, and $x-y=3\sqrt{13}-2\sqrt{13}=\sqrt{13}$. Their product is 65, but they are not integers.

 

[Agent Smith]

You can't arbitrarily factor 65 and assign one factor to x+y and the other to x-y.

This is a result of teaching students to expect everything to be integers!

While 9 and 4 satisfy this one equation, they don't satisfy the other, which you ignored.

Given the correct solution, you can see that $x+y=3\sqrt{13}+2\sqrt{13}=5\sqrt{13}$, and $x-y=3\sqrt{13}-2\sqrt{13}=\sqrt{13}$. Their product is 65, but they are not integers.

Thanks a million. I didn't understand the question.