Interview quesitons

A right triangle has a hypotenuse equal to 10 and an altitude to the hypotenuse equal to 6. Find the area of the triangle.

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math.stackexchange.com/questions/1594740/v-i-arnold-says-russian-students-cant-solve-this-problem-but-american-student
en.m.wikipedia.org/wiki/Geometric_mean_theorem
i.stack.imgur.com/JSZHV.png
twitter.com/AnonBabble

Im not doing your homework faggot.

What're you interviewing for a middle school math teacher position?

a lot

Can I go home?

Base*height/2

go fucking kill yourself

>hypotenuse
base

>altitude to the hypotenuse
height

>area
bh/2

No one has given a correct answer yet.

I thought you guys were good at math.

You drew and labeled your shit wrong.

You must be over 18 to post on Sup Forums.

Literally learned this shit in middle school and again in 9th grade

well, you never said altitude is from the vertex, hence area can not be calculated

You all failed at basic geometry.

Nigger wtf u on about.
The inner trianble is a 45 45 90 so it should be ez finding thr sides. Then for the base just adjust Pythagorean to fit the missing variable.

49.3?

that doesn't even make any sense

40.8

/thread

math.stackexchange.com/questions/1594740/v-i-arnold-says-russian-students-cant-solve-this-problem-but-american-student

10 * 6 / 2 = 30

Been a while since college, but can someone confirm if i'm guessing how to do it right?

you use two pythagorean formulas to find the length and then the width use W x L / 2 to find the area of the triangle?

That's pretty simple. So we know that the angle at C is the same angle as the angle at B (in the small triangle). Then we know that the tan of that angle is 6/10 and (looking at the small triangle) p/6 (p is the small part of A->C. The drawing is badly captioned). Then we get that p = 3.6. The area is then ( 2.6 * 6 + 6 * 10 ) / 2 = 40.8.

the cake is a lie

College? this is middle school work. GJ you're not a total retard

>The circumsphere of a circle is not pi in non-euclidean geometry
well no shit

You just proved that curve -> length isn't a continuous function. Well done user.

It's an impossible triangle. That's cute. I tried to find the sides and ended up with impossible numbers.

does the triangle have to exist in order to solve it?

length of BC = sqrt(6^2 + 10^2) = sqrt(36 + 100) = 11.662
length of AC = 10 + R
length of AB = sqrt((10 + R)^2 - 11.662^2)
length of AB = 100 + 10R + 10R + R^2 - 136
length of AB = 100 + 20R + R^2 - 136

R = sqrt((length of AB)^2 - 6) = sqrt( (100 + 20R + R^2)^2 - 6)
R^2 = (100 + 20R + R^2)^2 - 6
R^2 + 6 = (100 + 20R + R^2)^2
R^2 + 6 = 10000 + 4000R + 600R2 + 40R3 + R4

I'm not getting anywhere with this. I guess trigonometry is needed.

6cm * 10cm / 2
square -----> 60cm / 2 = 30cm

You haven't gave us enough information

i wonder why 4! didn't trigger you too

Nah, you don't really need any trigonometric function for this. Here is how it's done:

You orient the triangle like pic related. Then you can describe the edges of the triangle with the functions provided in the picture, each with t in [0,1]. The big triangle is trivial. For the edge of the small one you simply use the orthogonality condition. Then you just set t * 10 + 6 = 0 and get t = 0.6 for the point of intersection with the x-axis. The exact point is then (-3.6, 0). The area is then easily determined by ( 3.6 * 6 + 6 * 10 ) / 2 = 40.8

Forgot the pic.

MATH = mental abuse to humans

Only to retarded humans

>In a right triangle the perpendicular of the hypotenuse c cannot be longer than c/2.

Faggit, the area of the triangle is the bit within the lines, it's not that hard to find dumb ass

>This is not a right triangle.

How do I measure the area of something that can't exist?

Where did Sup Forums learn trig. School appears to be wasted on most of you. The smaller triangles are similar triangles to the large triangle. A property of similar triangles is that all sides on one triangle are proportional to the sides on the other. So calculate k which works out to be .6 , and use that value to get the rest of the sides.

Can I assume, that we are in a non-euclidean geometry?

The triangle doesn't exist

didnt divide the 2 out of the 2i sorry

Kill yourself. Why did you post this here? This fucking type of bait is meant for /sci/.
Why does this even have over 40 replies? Why do the fucking mods have to do this to me? THIS IS NOT FUCKING TECH RELATEEEED
IT'S GOING TO FUCKING REACH BUMP LIMIT JUST LIKE THAT FUCKING PONY PORN THREAD DID LAST NIGHT BECAUSE THE MODS ARE SO INCOMPETENT.
FUCK YOU MOOK YOU FUCKING PIECE OF SHIT. YOU RUINED Sup Forums FOREVER. I FUCKING HATE ALL OF YOU.

BC = (10*10 + 6*6)^0.5
AB = 10*10 - BC*BC
Area = AB * BC / 2

Took me about 40 seconds.

Let us make a category Tri, where objects are triangles and arrows are affine transformations of triangles. We see that we can make a category STri, where objects are the sets of all triangles with equal areas, and arrows the sets of transformations. Defining a corresponding functor F would be left as an exercise to the reader (hint - rigid transformations would be sent to identity).
One can notice that STri is isomorphic to R. Define a functor A: STri -> R which takes a triangle to its area.

So, to find an area of a triangle, just apply (A * F) to it. Simple as that.

So wrong it's not even funny

>10*10
>^0.5
Why would you write ^2 as a product, but then write sqrt as a rational power? That makes absolutely no sense.

That is some high intensity retardation you have there.

I dont understand why you aren't just using
Area = 1/2(b*h)

Whats the base and whats the height?

not him but just turn the image
might want to look at though

a right triangle is impossible with the given dimensions

at first glance I assumed it was 10 and 6 respectively

half right. 10 is not the base though. Only a portion of it, and the remaining portion is impossible.

is 10 the length of AB or the length of B to the intersecting line? is ABC the triangle we are trying to find the area for?

Read the problem. The hypotenuse is stated as 10. If it were just a portion then it would be the base of the smaller triangle. It really is just the hypotenuse. The triangle doesn't exist, though, as I already said see

Yep it works out (roughly) at 30.85

Its Isoceles, We can see the height bisects the 90 degrees through ABC, so we must have angles 90, 45, 45. Thus, BAC must be 45 too.

AC = 10
AB = ACsin45
AB = 5sqrt(2)

Area of a right angle triangle = 1/2 * b * h
= 1/2 * 10 * 5 * sqrt(2) = 42.4 u^2

sorry, 35.4u^2

altitude = h = 6
hypotenuse A-C = two segments with lengths, x, y

x + y = 10
x = 10 - y

h^2 = x * y
36 = y(10 - y)
36 = 10y - y^2
0 = -y^2 + 10y -36

...

oh

You find things that do exist and aproach it. Then you take the limit of those things.
Same way you analyse algorithm complexity.
Geese Sup Forums ...

this is great, especially the unquestioning arrogance

>Technology

>Its Isoceles
Where the fuck did you get that? It clearly isn't.

Lmao I can't believe Sup Forums is retarded half of the 18 replies are wrong and the other I didn't look at. Use a proportion answers .5(b*h) with h of 6 and b of 10 + (36/10) cause 6/10 = x/6

...

you dont learn simultaneous equations in middle school, which is how you would solve a question like that minus the troll aspect to OP

here's a real maths question.
You want to cut a circular cake, diameter d, into x pieces of equal area.
The traditional way of doing this would be to cut normal slices every 360/x deg around the circle.
However you are an autist from Sup Forums and need each slice to have the exact same area - you also only have a ruler to hand.
Therefore devise an equation/ set of equations to cut the cake in parallel slices into slices of exactly equal area simply in terms of x and d.

Nigger

This is the answer.

6^2 = 10x

x = 3.6

((3.6 + 10)*6)/2 = 40.8

harder than it looks ehhh

en.m.wikipedia.org/wiki/Geometric_mean_theorem

Everybody who saw this and was too pussy to attempt it, you are the true novices of this world

Fuck I went back in time to junior high school.

This actually is a good question. It looks complex, but the answer is actually very simple.

The area is equal to 0.5*base*height.
1/2 * 10 * 6 = 30 units^2

Now try which looks simple but is actually very complex

I'm an autist and tried solving it.

It is impossible

Maybe I misunderstood the question but it seems simple. After cutting the cake twice laterally (into four slices of equal area) just start cutting it longitudinally (one slice across the middle gives you eight slices, then two cuts into those portions yields 16 slices, then four cuts into those sections yields 64 slices, etc). Repeat ad infinitum

why does the triangle have to fit within the circle?
Why not outside it?

Then it wouldn't be a right triangle

>same area
Cut the cake horizontally

Area = 6*10/2
you really have serious metal problems

It's 6 times the square root of 10.
Which is exactly 10 times the square root of 6.
Which is 10 * 3 = 30.

Too easy.

>terms of x and d.

lol look at all this shit

6/10 = 0.6

(6*10)/2 + (3.6*6)/2 = 40.8

Those two triangles shown do not have the same area.
The area of the triangle in the OP is 30 a.u

i.stack.imgur.com/JSZHV.png

>find area of triangle
what triangle?

right, left or both

wth Sup Forumsuys, problem still not solved?

It's not four, because no matter how many times you repeat that, there are still going to be more squares. It's never going to turn into a curve. It will always be 4 and it will NEVER match the true circumference of the circle.

Why is this on Sup Forums, by the way?
Fuck off to OP. They'll gladly help you with your homework. No shame in not understanding it.

But 4! is 24?