Calculate the distance X in the quadrilateral | Important Geometry skills explained
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- Опубликовано: 29 сен 2024
- Learn how to find the distance X in the quadrilateral. Important Geometry and Trigonometry skills are also explained: Pythagorean Theorem. Step-by-step tutorial by PreMath.com
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Calculate the distance X in the quadrilateral | Important Geometry skills explained
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Very interesting! Figuring out which equations to combine is the trick
Glad you think so!
Thank you! Cheers! 😀
Профессор @PreMath блестяще доказал одно из известных свойств ортодиагонального четырёхугольника и существование известной пифагорейской тройки 20,21,29 😂
تمرين جيد جميل. رسم واضح مرتب. شرح واصح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا. تحياتنا لكم من غزة فلسطين
From PreMath's beautiful solution, we can deduce that AB² = AD² + BC², which is simply the Pythagorean equation.
AB^2 + MC^2
= AE^2 + BE^2 + EM^2 + EC^2
= BE^2 + CE^2 + AE^2 + EM^2
= BC^2 + AM^2
= BC^2 + DM^2 + AD^2
Herein DM = MC
Hereby x^2 = AB^2 = AD^2 + BC^2
Thanks for sharing! Cheers!
👍👌 10*
I solved this using coordinate geometry:
D = (0, 0), A = (0, −105), M = (a, 0), C = (2a, 0), B = (b, −c)
Since AC ⊥ MB, then
Slope AC × Slope MB = −1
105/(2a) × (−c)/(b−a) = −1
*105c = 2a(b−a)*
BC = 100
(2a−b)² + (0+c)² = 100²
4a² − 4ab + b² + c² = 10000
−2 *(2a)(b−a)* + b² + c² = 10000
−2 *(105c)* + b² + c² = 10000
b² + c² − 2(105c) + 105² = 10000 + 105²
b² + (c−105)² = 10000 + 11025
x = AB
x² = (b−0)² + (−c+105)²
x² = b² + (c−105)² = 10000 + 11025 = 21025
x = 145
Excellent!
Thank you! Cheers! 😀
Alternate Solution:
if you assume MCB is a right angle, which apparently it is then:
triangle CDA is similar to triangle CEM is similar to triangle BCM
and 2y / 105 = b / c
and 100 / y = b / c
2y^2 = 100 * 105
y = ((100 * 105)/2)^.5
x^2 = [(2y)^2+(105-100)*2]^.5
x=145
Length DC is not defined.
For simplicity, triangle ACD can be selected as a 3,4.5 ratio such that DC = 140 and AC = 175.
Triangles ACD and ECM are similar.
Therefore, 140/175 = EC/70.
EC = 56.
In triangle ECB, Cosine ECB = 56/100 = 0.56.
Cosine rule applied to triangle ABC.
X^2 = 175^2 + 100^2 - (2 x 175 x 100 x Cos ECB).
X^2 = 30625 +10000 - (35000 x 0.56)
X^2 = 21025.
X = 145.
Thank you! Cheers! 😀
In a similar approach, I solved another "simple" case where ∠DCB was 90°.
Although this is not a proof that AB is constant, x also turned out to be 145.
PremMath's solution boils down to x = √(AD² + BC²), solving all the cases😎
Draw in AM. You now have a quadrilateral, ABCM, in which the diagonals are perpendicular. There is a theorem which states that "in a quadrilateral with perpendicular diagonals, the sum of the squares of opposite sides are equal."
apfstatic.s3.ap-south-1.amazonaws.com/s3fs-public/5_quadrilaterals-with-perpendicular-diagonals.pdf
In other words, AM² + BC² = AB² + MC².
By Pythagoras on ΔADM, we get AM² = 105² + (x/2)²
Hence, 105² + (x/2)² + 100² = x² + (x/2)²
So 105² + 100² = x²
.... giving x = 145.
Trying to solve by similar triangles is a trap. There are only two similar triangles: ACD and CEM. And there's nothing useful that we can do with'em. If you assume that CBE is similar to them, which is wrong, you'll fall in the trap, doing hard and endless calculations to have a wrong solution.
Great problem. Of course, the question is, what was your thought process in deciding how to combine equations (1) through (5).
thankyou v.much dear ❤
You're so welcome!
So nice of you.
Thank you, Rishu! Cheers! 😀
x^(4)-x+1=0 find x^(10) can you help with that pls. I did not find any solution in the internet.
Thanks
You are very welcome!
So nice of you.
Thank you! Cheers! 😀
When ∠DCB is 90° (which is not necessarily the case) then we have DC / 105 = 100 / (DC / 2) ⇒ DC = √21000 ⇒ AC = √32025
AD ∥ BC ⇒ ∠DAC = ∠ACB = θ ⇒ cos θ = 105 / √32025. Then x² = 100² + 32025 − 2 ⋅ 100 ⋅ √32025 ⋅ (105 / √32025) ⇒ x = 145
Thank you! Cheers! 😀
Can we have different values of *y* for this quadrilateral?
I am afraid my method is rather cumbersome, let DM be m, so AC^2=105^2+4m^2, as triangles ADC and CEM are similar, CE/CM=CE/m=CD/AC=2m/sqrt(105^2+4m^2), thus CE=2m^2/sqrt(105^2+4m^2), thus EB^2=100^2-4m^4/(105^2+4m^2), AE=sqrt(105^2+4m^2)-2m^2/sqrt(105^2+4m^2)=1/sqrt(105^2+4m^2)x(105^2+2m^2), there the answer is the square root of EB^2+AE^2=100^2-4m^2/(105^2+4m^2)+(105^2+2m^2)^2/(105^2+4m^2)=100^2+1/(105^2+4m^2)x(105^2+2m^2)^2-(2m)^2)=.....😅
Thanks for sharing! Cheers!
Why ac^2=105^2+4m^2 if you take dm as m then then you should thke 2m instead of 4m . If any problem to my understanding please ignore or if you made it please correct thanks 👍
@@AdityaKumar-pj3gp 2x2=4
you r great sir
i love you
such an excellent problem.. blless me sir
So nice of you, dear.
Thank you! Cheers! 😀
Always stay blessed 🙏
Note that ΔBMC and ΔCAD are similar (both are right triangles and have equal interior angles
This is the simplest, fastest and more "euclidean" solution. It took me just a few minutes.
Xsq = AEsq + EBsq. AEsq + MEsq = AMsq = 105sq +DMsq; EBsq = 100sq - ECsq; ECsq = MCsq - MEsq.
So Xsq = 100sq + 105sq. X = 145
Thank you! Cheers! 😀
Un sistem de 5 ecuatii cu 7 necunoscute ...
Clearly not a rectangle as unequal opposite sides, 😅 also AB and CD cannot be equal.
Good point!
However, ABCD could be a right trapezoid🤔
145
Let the length of the red line that meets the diagonal = m, and let the line perpendicular to it = p
Let the remaining diagonal = s and the remaining red line = r
Let to equal line = y and y
A line from the center of y and y is drawn to the left of the quadrilateral, meeting the diagonal.
Hence the following equations using Pythagorean Theorem:
p^2 + r^2 = 100^2 [equation 1]
m^2 + p^2 =y^2 [equation 2]
m^2 + s^2 =105^2 + y^2 or 0= -y^2 +m^2 + s^2 -105^2 [equation 3]
p^2 + r^2 = 100^2 or r^2 =100^2 -p^2 [equation 4]
r^2 + s^2 =x^2 or r^2 = x^2 - s^2 [equation 5]
Hence 100^2 -p^2 = x^2 - s^2
Hence p^2 = s^2 -x^2 + 100^2 but
p^2 also = y^2 -m^2 [ see equation 2)
Hence s^2 -x^2 + 100^2 = y^2 -m^2 [equation 6]
Hence - x^2 = y^2 - m^2- s^2- 100^2 [equation 7
0 = -y^2 + m^2 + s^2 -105^2 [see equation 2]
-x^2 = 0 + 0+0 -100^2 - 105^2 [ adding equation 2 and equation 7]
-x^2 = -100^2 - 105^2
x^2 = 100^2 + 105^2
x^2 = 10,000 +11,025
x^2 = 21,025
x = sqrt 21,025 or 145 Answer
I got 105/2y = y/100 incorrectly through similar triangles and compensating errors with a result of 144.9 units which was close to the real answer. Then I remembered that close only counts in Horseshoes, Hugging and Hand grenades.
Wow, I love these types problem with system of equations... I can't do most of them (yet 😕) but still love em.
Often indistinguishable from Magic.
Unbelievable solution
Just used similar triangles for a quick solution.
Let MC = a (therefore, DC = 2a).
Let angle(MBC) = b and angle(BMC) = c with b + c = 90 degrees.Therefore angle(ACD) = b (as angle(CEM) = 90) and angle(ACB) = c (as angle(BEC) = 90). Thus, angle BCD is also 90 degrees.
As angle(ADC) is also 90 (given) and angle(ACD) = b, therefore angle(CAD) = c. Therefore triangles MBC and ACD are similar (AAA).
Therefore, as MC = a, a/100 = 105/2a and a = sqrt(5250). Thefefore, CD = 2a = 2 x sqrt(5250). Drawing a line from B that meets AD orthogonally to what we can call point F gives rectangle BCDF with resulting triangle ABF with sides 2a, 5 and x.
Now, sqr(2a) + sqr(5) = sqr(x) using Pythagoras' Theorem; and
4 x 5250 + 25 = sqr(x)
Therefore, sqr(x) = 21025; and
x = 145.
Can you please elaborate how b + c = 90 in the embarkment of your solution?
A quadrilateral with one right angle, may be or not may be a rectangle ?
YOM!
Thanks for video.Good luck sir!!!!!!!!!!!
100
Hi
Hi dear
❤cool❤
Thanks