A Very Nice Geometry Problem | You should be able to solve this!

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[@marioalb9726]

Labelling 'z' = DC = BD and 'y' = BE = EC (Isósceles triangle BCD) We have a Pytagorean Triple right triangle 7-24-25, so z=25. Next, sides 24 and 7+z=32, is other Pytagorean Triple 3k-4k-5k, that is 24-32-40, so 2y = 40 cm Next, sides x-y-z is other Pytagorean Triple 3k-4k-5k, that is 15-20-25, so: x=15 cm ( Solved √ ) Solved mentally whithout any calculador, and any paper.

[@marioalb9726]

Labelling DC = BD = 'z' Pytagorean theorem: z² = 24² + 7² z = 25 cm Similarity of triangles: x/z = 24 / √[24²+(7+z)²] x / 25 = 24/ √[24²+(7+25)²] x/25 = 24/40 x = 15 cm ( Solved √ )

[@JINSEOKSONG]

The equation 𝑥/𝑧 = 24/(7+𝑧) is incorrect. In fact, calculating 𝑥/25 = 24/(7+25) gives 𝑥=18.75 The correct equation is 𝑥/𝑧 = 24/𝐵𝐶

[@marioalb9726]

x/z = 24 / √[24²+(7+z)²] x / 25 = 24/ √[24²+(7+25)²] x/25 = 24/40 x = 15 cm ( solved √ ) This is the correct Thanks I wrote wrongly: """"""' Similarity of triangles: x / z = 24 / (7 + z) x / 25 = 24 / (7+25) x = 15 cm """""""

[@santiagoarosam430]

DC=DB=√(7²+24²)=25=5*5 ---> AC=7+25=32 ---> BC=√(24²+32²)=40 ---> BE=EC=40/2=20=5*4 ---> DE=5*3=X. Gracias y saludos.

[@devondevon4366]

x=15 Can use a similar triangle to solve this problem, but I will do some guesswork, then test later to see if it adds up First, draw a line from the top of the quadrilateral to the bottom to form two triangles, one with bases 7 and 24 and the other right with base x nd ". The line drawn is the hypotenuse of the triangle to the left sqrt (7^2 + 24^2) = 25 Its length is 25. If it is a 3-4-5 right triangle scaled up by 5, then x corresponds to the 3, and if scaled up by 5, then x = 3*5 =15 Therefore, the hypotenuse of the large triangle is 40 (20 + 20 or + "). He ce, it is a 3-4-5 right triangle scaled up by 8 (8*5) =40 the 24 corresponds to the (8*3=4), and hence the other side, 32 32-7 = 25 and sqrt ( x^2 + 20^2) = sqrt ( 15^2 + 20^2) =15 Again showing that x =15 Notice that 7^2 + 24^2 = 625 and 15^2 + 20^2 = 625

[@michaeldoerr5810]

The answer is 15. And I did that in my head. I now learn that if there is a right tringle inacribed in a right triangle, that means the Pythagorean Theorem must be applied thrice. Also I got 15 from thia calculation: 25[25-16]. You get 25(9) take the square root and you get 5(3). You get 15 after applying the Pythagorean Theorem twice or thrice. And that is one of the *easier* problems!!!

[@ludmilaivanova1603]

a very interesting problem. I solved it differently: I used two parallel to AB line segments drawn from E ans D. The triangles I've got are similar. DO=a, I found a=9, then x=15.

[@user-ll5wl9gq4x]

Отличное задание! РЕСПЕКТ! В принципе, зная лишь ДВЕ из ПИФАГОРОВЫХ ТРИАД : 7 - 24 -25 и 3 - 4 - 5, можно ответ получить и устно! Обязательно предложу своим онлайн ученикам в качестве хорошей тренировки!😊🖕

[@Tmwyl]

I got this one too! As soon as I realized to draw an Isosceles triangle with the leg as the hypotenuse the rest was gravy. Pythagoras is trending. Haha.

[@quigonkenny]

Draw BD. Triangle ∆DAB: DA² + AB² = BD² 7² + 24² = BD² BD² = 49 + 576 = 625 BD = √625 = 25 As BE = EC = a, ∠BED = ∠DEC = 90°, and DE is common, ∆BED and ∆DEC are congruent triangles, and thus CD = BD = 25. As AB = 24 = 3(8) and CA = 25+7 = 32 = 4(8), then ∆CAB is an 8:1 ratio 3-4-5 Pythagorean triple right triangle and BC = 5(8) = 40. As BC = 2a, 2a = 40 ==> a = 20. As ∠CAB = ∠DEC = 90° and ∠C is common, then ∆CAB and ∆DEC are similar triangles. Triangle ∆DEC: DE/EC = AB/CA x/20 = 24/32 = 3/4 x = (3/4)20 = 15 units

[@user-sw9lb2zs6e]

Three times Pythagoras, thats it. An easy one.

[@prossvay8744]

X=√25^2-20^2=15

[@RealQinnMalloryu4]

(24)^2 =576H/O/ASino° (7)^2=49H/O/BTano° {576H/O/ASino°+49H/O/BTano°}=626H/O/ASino°BTano° 180°/625H/O/ASino°BCTano° = 3 .85H/O/ASino°BTano° 3^1.5^17 3^1.5^1^1 3^1.5^1^1^1 3^1.1^1^1^1 3.1 H/O/ASino°BTano° (H/OASino°BTano° ➖ 3H/O/ASino°BTano°+1)

[@tes3272]

x =15

[@user-gu6dc7yu1m]

15

[@AmirgabYT2185]

15