Find the area of the trapezoid Important Geometry skills explained
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- Опубликовано: 11 фев 2025
- Find the area of the trapezoid Important Geometry skills explained #geometryskills #mathpuzzles
We are required to find the area of the blue triangle.
This will be so much appreciated.
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“a” and “b” can be determined without the assumption that they are integers. Actually, solving the equations 4a*b+a^2=17 and 4a*b+b^2=32 simultaneously gives us a=1 and b=4 where we use the relation h^2=(a+b)^2-(b-a)^2=4a*b.
how did you figure out that 4a*b+a^2=17 and 4a*b+b^2=32?
We can easily see that h^2+a^2=17 and h^2+b^2=32 where we use the relation h^2=(a+b)^2-(b-a)^2=4a*b to arrive to the equations 4a*b+a^2=17 and 4a*b+b^2=32.
also, nowhere in the problem statement does it say that a and b are integers.
why you consider that a and b are integer numbers maybe they are not integers in b^2-a^2=15 ?
Yes you are right “a” and “b” can be determined without the assumption that they are integers. Actually, solving the equations 4a*b+a^2=17 and 4a*b+b^2=32 simultaneously gives us a=1 and b=4.
my own perplexity
@@Okkk517 which way is the most convenient? I try by sostitution... (b=17-a²/4a)
The diagonal 4 sqrt2 actually demands that b=h=4 given that b and h are perpendicular to each other. Right? Then a=1 based on the sqrt 17 and h=4. No ambiguity.
Since the figure included a circle, I was initially bound and determined to use inscribed circle/triangle relationships. I didn’t, but it made me think in terms of having a large, inscribing triangle made by adding a smaller, similar triangle to the top of the trapezoid. The small triangle is 1/4 of the big, for any length dimension, therefore 1/16 the area. Therefore the trapezoid is 15/16 of the big triangle. Big triangle has base 8, sides 20/3, 20/3, and height 16/3. Area 64/3. Multiplying by 15/16 gives trapezoid area of 20. (Or just calculate the trapezoid directly from a,b,h. D’oh!🤪).
4√2=,( 16 )^2(1) √(17)^2(1) =196 +(1)=289 (288+196)=484 (484-100)=√384 2√^19√ 2^2 √2^√1 1^2 √1 1^2 (x+1x-2)
La figura es simétrica respecto al diámetro vertical, cuya longitud es 2r.
Longitud base =2b; longitud lado superior =2a
El segmento de longitud 4sqr2 sugiere que b=2r=4》r=2 》 Si 17=a^2 +4r^2=a^2+16》a=1 》La hipótesis de partida se ajusta a las premisas del trazado geométrico propuesto 》》Área del trapecio =(4+1)4=20
Gracias y saludos.
Que sea b=2r=4 tambièn se confirma haciendo:
(a+b)² = (2r)² + (a-b)²
(1+4)² = 4² + (4-1)²
Nice solution!!
Yes!
(4\/2)*2-в*2=(\/17)*2-а*2 , 32-в*2=17-а*2 , в*2-а*2=15 , (\/17)*2-а*2=(в+а)*2-(в-а)*2 , 17-а*2=4ва , ( h*2=4ва ) , а*2+4ва=17 . (в*2-а*2=15) х 17 , (а*2+4ва=17) х 15 , (17в*2-17а*2=255) - (15а*2+60ва=255) , 17в*2-32а*2-60ва=0 , 17в*2/а*2-32а*2/а*2-60ва/а*2 =0 , 17в*2/а*2-32-60в/а=0 , 17в*2/а*2-60в/а-32=0 , в/а=у , 17у*2-60у-32=0 , у1,2=(60+-\/3600+4х17х32)/2х17=(60+-\/5776)/34 , = (60+-76)/34 , у1=(60+76)/34=4 , у2=(60-76)/34=-8/17 - негатив , у=в/а =4 , в=4а , в*2-а*2=15 , (4а)*2-а*2=15 , 16а*2-а*2=15 , а = 1, в=4а=4х1=4 , h=2\/ва=2\/4х1=2х2=4 , А=((2а+2в)/2)хh =(а+в)хh =(1+4)х4=20 .