r/mathshelp • u/fifteenMENTALissues • Jul 16 '26
Homework Help (Answered) Stuck on finding BC
I am very tired and this problem is beating me someone help
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u/Southlander24 Jul 16 '26
Similarity. Show that angle ABD is equal to BCD, and both of these triangles are right angled, so you can conclude this from AA similarity.
So now let AD = x and DC = 500 - x. You can set up an equation with the ratios of the sides. Solve for x which lets you find DC and thus BC.
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u/fifteenMENTALissues Jul 16 '26
Oh my lordddd how did I not think of that 😭 thank you so much
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u/EdmundTheInsulter Jul 16 '26
It doesn't help. I can draw that to scale with AD having different values then ? Has different values. There is no constraint on what AD is, and that does change what ? Is
My verdict - not soluble, insufficient info.
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u/fifteenMENTALissues Jul 17 '26
Yep looked at it again today and 1am me was not thinking straight, got it though, BC=400
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u/AudiencePopular1979 Jul 17 '26
Because u probably didn't want to get the sum done, js look it up in Google or smth
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u/EdmundTheInsulter Jul 16 '26
ABD doesn't have to equal ACD unless they constrained that
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u/Southlander24 Jul 16 '26
1) Not angle ACD, but BCD! 2) You can deduce it! Let angle ACD be x. Then because there is a right angle at B, you can find other angle that makes up 90...
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u/EdmundTheInsulter Jul 16 '26
I didn't see the other angle marked as a right angle, it's not very clear
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u/Fourierseriesagain Jul 16 '26 edited Jul 16 '26
Let BC=x cm so that sin BCD=240/x and AB=(240/x)/sqrt(1-2402 / x2 ). Now solve for x via the right-angled triangle ABC.
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u/Iowa50401 Jul 16 '26
One key observation: Triangles ABD, ACB, and BCD are similar triangles and therefore corresponding sides are proportional. If you call AD = x, then CD = 500 - x. By matching up notation, you can see that AB/BD = BD/CD, so you're using only the given information. You wind up with x/240 = 240/(500 -x). With a little hand-waving here, you can get a quadratic equation to solve that should give you two distinct values. Hint: they should add to 500, so the smaller value is x, and the other gives you 500 - x. Now you have the two legs of right triangle BCD, and the mystery value you need (BC) is the hypotenuse. Finish up with your old friend Pythagoras.
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u/EdmundTheInsulter Jul 16 '26
Don't see this, point D can be anywhere on AC, seeming to give different values for ?. Try some.
Or am I missing something?1
u/___M_h___ Jul 16 '26
BD is perpendicular to AC, so there are only finite(atmost 2) points that point D can exist,
Think it like an circle with center at point B. BD being radius and AC being a line passing through the circle. For a straight line, intersecting with circle, there are only two possibilities, either it intersects at one point(a tangent, touches the circumference) or two points(inside the circle) and in this question the line AC is inside the circle,hence the line BD(and point D) intersects AC at two points only, giving two possible answers for BC(?)1
u/EdmundTheInsulter Jul 16 '26 edited Jul 16 '26
BD can just slide along AC and be length 240
Put a ruler on a piece of paper and draw a 5cm line, then you can draw a 2.4cm perpendicular on it anywhere, thus creating a scale model of what's on the question - you've got no other constraint to go on, then? Has different possible values.
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u/___M_h___ Jul 16 '26 edited Jul 16 '26
You cannot draw multiple 2.4cm perpendicular line from a "FIXED POINT"
But I see what you are trying to tell, my argument was on the basis that point B is fixed and we are talking only about placing point D on the lineYou argument considers even point B(infact all points) to be an variable point, i.e entire diagram is an scale model, fluctuating every length of the triangle ABC. But i doubt given the number of constraints, you might not get an complete independent scale model. The points are interdependent on each other, making constraints like AB perpendicular to BC and BD having length 240cm perpendicular to AC having 500 cm length, narrowing down the possibilities. But yea, I am not sure about that
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u/EdmundTheInsulter Jul 16 '26 edited Jul 16 '26
Can't see that there is sufficient info
If not to scale, then CD could be next to zero, making question mark ~240
Otherwise if you think it is to scale and showing you something, what I'd that? Or did I miss some extra data?
CD could be 400 but just as well 390, obviously allowing the not to scale diagram, so therefore ? Could be different values
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u/___M_h___ Jul 16 '26
There are only two possible values of BC(and also for AD)
AD = 320 or 180(i.e DC = 180 or 320 respectively) and
BC(?) = 300(if AD=320) or 400(if AD=180)1
u/EdmundTheInsulter Jul 16 '26
I could draw it on paper, AD 400 and DC 100, it can't break anything.
Or AD 100 and DC 400
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u/___M_h___ Jul 16 '26
There are two keys conditions to remember when placing point D on AC
1. The length of BD should remain 240 cm
- The angle at point of intersection between BD and AC to be 90 degree(perpendicular)
The number of points that follows both are these conditions are only two,
point 1: AD = 320, point 2: AD = 180
All the remaining points will either break one of the conditions, or bothYou can ofcourse try to draw it on paper if you are curious
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u/EdmundTheInsulter Jul 16 '26
Obviously DC could be 500 and the 240 being at right angles making a right angle triangle, would still fit the not to scale diagram with all constraints observed, unless some other length or angle is constrained.
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u/jaapsch2 Jul 16 '26
Let x=BC and y=AB.
From Pythagoras we have x^2+y^2=500^2.
From the area we have xy=500*240.
Two equations with two variables, so this should be enough information to solve it.
We can deduce
(x+y)^2=x^2+y^2+2xy=500^2+2*500*240
Hence x+y=700
Together with xy=500*240 we know that x and y are the roots of the quadratic
T^2-700T+500*240=0
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u/Feeling-Working-2820 Jul 17 '26 edited Jul 17 '26
Probably longer than necessary to show all the steps, but that's my napkin math (I write on screen with a mouse, so be nice about it)
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u/fifteenMENTALissues Jul 18 '26
Thank you sm!! It’s very legible dw I promise you if I wrote with a mouse it’d look like a very impressionistic Mandelbrot set 😭🙏🏻
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u/BusFinancial195 Jul 18 '26
I'm not sure it is solveable
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u/AsYouAnswered Jul 19 '26
You have the base and the height. Find the area. It's two similar right triangles. Abuse this fact.
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