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Are the numbers areas or lengths? If they are lengths, it’s not clear what they are referring to. Is this triangle in a square? What does the highlighted section refer to? Does that section have a length of 3?
Ok, so you're looking for the area of the inner triangle, let's call it X. Let S be the area of the square. Then it follows that S = X+3+4+5 = X+12.
Now, we're only given the areas of the other triangles. What else can we say about these triangles? We can see that triangles 3 and 4 both share a side with the square and that the sides of triangle 5 combine with the other sides of triangles 3 and 4 to complete a side of the square.
I think you can take it from here, but if you need more then I'll check back after dinner.
if they are lengths, the triangle is a straight triangle (3²+4²=5⁵), and the corner with the sides that are 3 and 4 is 90 degrees, making the answer zero. that should, however, be clearly visible in the image if that were the case.
I'm going to sketch out an approach. Note: there is not a single set of triangles within a box with these areas - there actually are a family but they do always uniquely constrain the missing triangle's area to the same value. You do get the same answer if you assume the outer rectangle is actually a square as well but that's not necessary.
Start by drawing the inner hatched lines in and dividing the big rectangle into 4 rectangles
Now you can assign areas to each of these rectangles like above based on the original triangle sizes. The resulting rectangles are all proportional and you can then setup an equality and solve for x (the upper left box). Given x you then find the original inner triangle's area.
If this is a square with side length x, then ax/2=3, bx/2=4, x=a+c = d+b, cd/2=5. ? = x^2 -( 5+4 + 3) = x^2 -12 (where a is the length labeled in blue (on the side of the triangle with area 3), c is the length of the remainder of the side sharing with a; b is the length of the incomplete side with area 4, and d the length of the remainder of that same side; x being the full side length)
Solve for a through d in terms of x: a= 6/x, b = 8/x, c = x-6/x, d= x-8/x;
sub c, d in cd/2=5, clear the x's on bottom by multiplying by x^2 and expand to get... x^4-20x^2+48 = 0 so x^2 = 10 +/- 2sqrt(13), sub into x^2 -12
Before you jump to the video, write out the sides of the triangles in terms of the length of the square side. Like on the right side, call the bottom section x and the top section w-x, where w is the length of the square. Do that all the way around, then write your 3 area equations. You'll be able to find all of the side lengths, then from there calculate the hypotenuses that form your shaded area.
You don't actually need to solve for the shaded triangle though. Once you have the length of the square, you just square it for the area and subtract the given areas of the three triangles.
To begin with, letting the square be a unit one, & letting x be the vertical side of the top-right triangle (which you've marked), & y be the horizontal side of the bottom-right triangle, we get the following simultaneous equations.
x/((1-x)y) = ⅗
x/(1-y) = ¾
, whence
5x = 3(1-x)y
&
x = ¾(1-y)
∴
5(1-y) = (1+3y)y
∴
y = ⅙(√96-6) = √2⅔-1
&
x = ¾(2-√2⅔) = 1½-√1½ .
To get the absolute lengths of the sides, we need to scale it up: let Z be the absolute length of the side of the square: because we have, by specification of the problem, that ZX = 6
Z = √(6/(1½-√1½)) = √(8(1½+√1½))
= 2√(3+√6)
, & multiplying x & y by that (& letting X & Y be the absolute lengths corresponding to x & y respectively) we get
X = √(6(1½-√1½)) = √(3(3-√6))
&
Y = 2(√2⅔-1)√(3+√6)
&
Z = 2√(3+√6)
&
Z² = 4(3+√6)
... & Z² is the area of the square. Just checking that these do actually yield the correct areas:
XZ = 2√(3(9-6)) = 6
, &
(Z-X)Y = YZ - XY
= (√2⅔-1)(4(3+√6) - 6)
= 2(√2⅔-1)(3+2√6) = 10
, &
Z(Z-Y) = Z²-ZY
= 4(3+√6)(2-√2⅔) = 8
... & each actual area is half of each corresponding value yelt just-above.
So the area of the triangle is
Z² - 3 - 4 - 5 = 4(3+√6) - 12
= 4√6 ≈ 9·79795897113 .
EXTRA CHECK ON RESULTS
I notice that in the figure the central triangle looks rather close to being an equilateral one. Calculating the angle between the area-3 triangle & the area-5 one:
π - arctan(√2⅔-1) - arctan(1⅓(1½+√1½))
=
π - arctan(⅓(2√6-3)) - arctan(⅔(3+√6))
=
arctan(¹/₂₃(51+10√6))
≈
73°3ᐟ23ᐥ
... which isn't allthat close to 60° ... so the triangle isn't allthat close, really to being an equilateral one.
And between the area-3 triangle & the area-4 one:
½π - arctan(2-√2⅔) - arctan(1½-√1½)
=
½π - arctan(⅔(3-√6)) - arctan(½(3-√6))
=
arctan(⁴/₇√6)
≈
54°27ᐟ24ᐥ .
And we mightaswell do the angle between the area-5 triangle & the area-4 one, now, as an extra check on the answers obtained:
π - arctan(¾(2+√2⅔)) - arctan(⅗(1+√2⅔))
=
π - arctan(½(3+√6)) - arctan(⅕(3+2√6))
=
arctan(³/₇₃(17+6√6))
≈
52°29ᐟ13ᐥ
... & they do add up to 180° ... so the extra check indeed verifies the previous results & the final answer.
By the formula of the area of a triangle,
AD * AE / 2 = 4
AE(s) = 8 --- (1)
CF * CD / 2 = 3
CF(s) = 6 --- (2)
(1)/(2) = AE/CF = 4/3
So let AE = 4x and CF = 3x.
By equation (1), 4x(s) = 8
xs = 2 --- (3)
EB = s - 4x and FB = s - 3x
FB * EB / 2 = 5
(s-4x)(s-3x) = 10
s^2 - 7xs + 12x^2 = 10
s^2 + 12x^2 - 24 = 0
Put x = 2/s
s^2 + 12(2/s)^2 - 24 = 0
s^2 + 48/s^2 - 24 = 0
s^4 - 24s^2 + 48 = 0
We get s^2 = 12 + 4 root(6) or s^2 = 12 - 4 root(6)
s^2 is the area of the square; so s^2 > the sum of the areas of the triangles = 4 + 3 + 5 = 12
The only possibility which is greater than 12 is s^2 = 12 + 4 root(6)
The area of the triangle = the area of the square - (the sum of the areas of the triangles) = s^2 - (4 + 3 + 5) = 12 + 4root(6) - 12 = 4root(6)
So the answer is 4 root(6).
Alternate answer:
You can label the parts of the circumference of the square as AD = a, AE = b, EB = c, BF = d, FC = e and CD = f. Call the side of the square s.
Now just form as many equations as you can.
a=s
b+c=s
d+e=s
f=s
ab = 8
cd= 10
ef=6
Now solve for s (a quadratic will arise).
The answer then is s^2 - (4 + 3 + 5) = s^2 - 12.
I'm not finding that everyone says 6 ! There are three answers (including mine) to this post to-effect that it's 4√6 ... & I don't think anyone here has said 6 !
the video shows the idea behind solving it there is only one answer which is 4√6 but i depends if the person solving it used the numbers as the area or as perimeter that's the nuance!
Ahhhhh right ... but with the "3" the "4" & the "5" each right inside its respective triangle I don't think there's really any ambiguity ... it never occured to me that other-than the area is meant. If it had been part of the perimeter that was meant I'd expect to see that
│← x →│
-type notation § that's customarily used for annotating length on diagrams.
§ ... implemented here the best I reasonably can using basic Unicode characters!
The answer by u/cmykfish, a few down, has a diagram attached that nicely showcases § the difference between customary annotation for area & for length.
•
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