Calculate the distance X in the quadrilateral | Important Geometry skills explained

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Calculate the distance X in the quadrilateral | Important Geometry skills explained

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Күн бұрын

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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Пікірлер: 54

@garypaulson5202 Жыл бұрын

Very interesting! Figuring out which equations to combine is the trick

@PreMath Жыл бұрын

Glad you think so! Thank you! Cheers! 😀

@rabotaakk-nw9nm Жыл бұрын

Профессор @PreMath блестяще доказал одно из известных свойств ортодиагонального четырёхугольника и существование известной пифагорейской тройки 20,21,29 😂

@ghhdcdvv5069 Жыл бұрын

تمرين جيد جميل. رسم واضح مرتب. شرح واصح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا. تحياتنا لكم من غزة فلسطين

@ybodoN Жыл бұрын

From PreMath's beautiful solution, we can deduce that AB² = AD² + BC², which is simply the Pythagorean equation.

@honestadministrator Жыл бұрын

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

@PreMath Жыл бұрын

Thanks for sharing! Cheers!

@klementhajrullaj1222 Жыл бұрын

👍👌 10*

@georgebliss964 Жыл бұрын

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.

@PreMath Жыл бұрын

Thank you! Cheers! 😀

@ybodoN Жыл бұрын

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😎

@mohanramachandran4550 Жыл бұрын

Unbelievable solution

@jimlocke9320 Жыл бұрын

Note that ΔBMC and ΔCAD are similar (both are right triangles and have equal interior angles

@mhtsosmart 9 ай бұрын

This is the simplest, fastest and more "euclidean" solution. It took me just a few minutes.

@nehronghamil4352 Жыл бұрын

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

@ybodoN Жыл бұрын

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

@PreMath Жыл бұрын

Thank you! Cheers! 😀

@jonchester9033 10 ай бұрын

Great problem. Of course, the question is, what was your thought process in deciding how to combine equations (1) through (5).

@KAvi_YA666 Жыл бұрын

Thanks for video.Good luck sir!!!!!!!!!!!

@devondevon4366 Жыл бұрын

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

@rishudubey1533 Жыл бұрын

thankyou v.much dear ❤

@PreMath Жыл бұрын

You're so welcome! So nice of you. Thank you, Rishu! Cheers! 😀

@amitsinghbhadoriya6318 Жыл бұрын

Thanks

@PreMath Жыл бұрын

You are very welcome! So nice of you. Thank you! Cheers! 😀

@MarieAnne. Жыл бұрын

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

@PreMath Жыл бұрын

Excellent! Thank you! Cheers! 😀

@rangaswamyks8287 Жыл бұрын

you r great sir i love you such an excellent problem.. blless me sir

@PreMath Жыл бұрын

So nice of you, dear. Thank you! Cheers! 😀 Always stay blessed 🙏

@narendrarawat307 11 ай бұрын

Can we have different values of *y* for this quadrilateral?

@theoyanto Жыл бұрын

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.

@Istaphobic Жыл бұрын

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.

@hookahsupplier.5155 Жыл бұрын

Can you please elaborate how b + c = 90 in the embarkment of your solution?

@user-bd6ut2gs5r Жыл бұрын

x^(4)-x+1=0 find x^(10) can you help with that pls. I did not find any solution in the internet.

@misterenter-iz7rz Жыл бұрын

A quadrilateral with one right angle, may be or not may be a rectangle ?

@ajbonmg Жыл бұрын

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.

@kennethstevenson976 Жыл бұрын

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.

@vidyadharjoshi5714 Жыл бұрын

Xsq = AEsq + EBsq. AEsq + MEsq = AMsq = 105sq +DMsq; EBsq = 100sq - ECsq; ECsq = MCsq - MEsq. So Xsq = 100sq + 105sq. X = 145

@PreMath Жыл бұрын

Thank you! Cheers! 😀

@misterenter-iz7rz Жыл бұрын

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)=.....😅

@PreMath Жыл бұрын

Thanks for sharing! Cheers!

@AdityaKumar-pj3gp Жыл бұрын

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 👍

@misterenter-iz7rz Жыл бұрын

@@AdityaKumar-pj3gp 2x2=4

@alancs85 10 ай бұрын

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.