Find the length x #thinkoutsidethebox #geometryskills #mathpuzzles

Find the length x #geometryskills #mathpuzzles #thinkoutsidethebox

This will be so much appreciated.

[GrayYeonMath]

Hi there to all, my solution is quite long.
Here is another solution for the problem above.


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[Ramkabharosa]

By Appolonius's theorem, in any triangle ABC, if AD is the median,
then 2.(|AB|²+|AC|²) = |BC|² + 4.|AD|². So 2(8²+10²) = (2x)² + 4(6²).
Hence x² = (328 - 144)/4 = 184/4 = 46. Thus x = √46 & |BC| = 2√46.

[bsmith]

Any median meets the bisected base forming a pair of supplementary angles. Let's call one of then theta, then the other will be 180-theta.
Now apply the law of cosines to both smaller triangles, on the supplementary angles.
64 = 36 + x^2 - 12x*cos(theta)
100 = 36 + x^2 - 12x*cos(180-theta)
We know that cos(180-t) = -cos(t), so the second equation reduces to 100 = 36 - x^2 + 12x*cos(theta).
This is really useful since now it is easy to cancel out the cos() terms by adding the two equations together. Then we get 164 = 72 + 2x^2.
Very simple to solve to get an answer of x=+/-sqrt(46). Since we are working with plane figures the negative root must be discarded leaving us with a final answer of x=sqrt(46).

[じーちゃんねる-vn]

Multiply by 1/2 and consider the position vector with the top vertex as the origin.
|a|=4 |b|=5 |(a+b)/2|=3 ∴ (16+25+2a･b)/4=9 ∴a･b=-5/2 ∴X=2x=2√(23/2)=√46

[gelbkehlchen]

Solution:
a = 10, b = 8, e/2 = 6, e = 12.
Side 6 is the bisector of the opposite side and you can double this bisector and you will then get a parallelogram with the sides a = 10 and b = 8 and the diagonals f = 2*6 = 12 and e = 2x. You can apply the parallelogram equation to this parallelogram, which says that:
e²+f² = 2*(a²+b²) ⟹
(2x)²+12² = 2*(10²+8²) ⟹
4x²+144 = 328 |-144 ⟹
4x² = 184 |/4 ⟹
x² = 46 |√() ⟹
x = √46 ≈ 6.7823

[solomou]

Καλημέρα σας. Όταν πήγαινα σχολείο είχα συναντήσει την παρακάτω άσκηση η οποία έχει εξαιρετικά δύσκολη λύση. Αν θέλετε προσπαθήστε να την λύσετε και να την δημοσιεύσετε. Υποδείξτε μου ένα τρόπο να σας στείλω κι εγώ τη λύση μου γιατί δεν γνωρίζω πως να το κάνω. (το σχήμα είναι απαραίτητο). "Εάν οι διχοτόμοι δύο γωνιών ενός τριγώνου είναι ίσες να αποδειχθεί ότι το τρίγωνο είναι ισοσκελές".

[solomou]

Καλησπέρα σας από Ελλάδα. Η λύση προκύπτει πολύ απλά και καθαρά Γεωμετρικά, εφαρμόζοντας το θεώρημα των διαμέσων για την διάμεσο ΑΜ=6.