sqrt 1024/17 or 7.7611 BEF is a 3-4-5 scaled up by 2 In the right triangle BEF, let angle B=33 degrees, then E=57 degrees In the right triangle CBF, let B= 17 degrees, then angle B in triangle ABE = 40, (notice that 40+ 33+ 17=90) then angle F in BCF= 73 degrees. Hence, BCF would be a 90-73-17 right triangle (it is likely not, to see why it doesn't matter) then angle F in the right triangle CDF would = 17, D= 90 and C=73. Hence, it would ALSO be a 90-73- 17 right triangle. Hence, triangle BCF and triangle CDF are similar. It doesn't matter what degree I let angle B equal; it will show that BCF and CDF are similar triangles. Since line EF = 6 (Pythagorean, also recall BEF is a 3-4-5 scaled up by 2, hence 3*2 =6) and BCF is similar to CDF, then line DF/6 = x/8 DF = 6/8 x DF= 3/4 x If DF= 3/4x, then CF = 1/4 x (1/4 + 3/4 =1 or x) Hence, for a triangle BCF, the bases in terms of x are 'x' and '1/4 x) Using Pythagorean Theorem x^2 + (1/4x)^2 = 8^2 x^2 + 1/16 x^2 = 64 16/16 x ^2 + 1/16 x^2 = 64 17/16 x^2 = 64 x^2 = 64 * 16/17 x^2 = 1,024/17 x = sqrt (1,024/17) or 7.7611
In triangle ∆EFB, as the hypotenuse BE = 10 and the longer leg FB = 8, ∆EFB is a 2:1 ratio 3:4:5 Pythagorean triple right triangle and the shorter leg EF must equal 6. This can also be determined directly from the Pythagorean Theorem as 10²-8² = 6². Let ∠DFE = α and ∠FED = β, where α and β are complementary angles that sum to 90°. As DC is a straight line and ∠EFB = 90°, ∠BFC = 180°-(90°+α) = 90°-α = β, and as ∠BFC = β, ∠CBF = α. Thus ∆EDF and ∆FCB are similar triangles. DF/FE = CB/BF DF/6 = x/8 DF = 6x/8 = 3x/4 DC = DF + FC x = 3x/4 + FC FC = x - 3x/4 = x/4 Triangle ∆FCB: FC² + CB² = BF² (x/4)² + x² = 8² x²/16 + x² = 64 17x²/16 = 64 x² = 64(16/17) = 1024/17 x = √(1024/17) = 32/√17 x = (32√17)/17 ≈ 7.761 units
Razón de semejanza entre los triángulos rectángulos EDF y FCB =s=6/8=3/4→ DF=s*X=3X/4→ FC=X-(3X/4)=X/4→ X²+(X/4)²=8²→ X=32/√17 =7,76114....ud. Gracias y un saludo cordial.
A questão é muito bonita. Eu a fiz de duas maneiras distintas, sendo uma delas igual à sua primeira maneira. Parabéns pela escolha! Brasil Setembro 2024. The question is very beautiful. I did it in two different ways, one of which is the same as your first way. Congratulations on your choice! Brazil September 2024.
The answer is 1/17[32*sqrt(17)]. This is similar to some other problems in this channel. I got the answer. Also owe you an apology. I got the wrong answer yesterday. I screwed up the calculation and over-calculated it. For yesterday's problem, it was 5/8[sqrt(65)]. I believe that I passed my "sanity check" for today's problem!!!
sqrt 1024/17 or 7.7611
BEF is a 3-4-5 scaled up by 2
In the right triangle BEF, let angle B=33 degrees, then E=57 degrees
In the right triangle CBF, let B= 17 degrees, then angle B in triangle ABE = 40, (notice that 40+ 33+ 17=90)
then angle F in BCF= 73 degrees. Hence, BCF would be a 90-73-17 right triangle (it is likely not, to see why it doesn't matter)
then angle F in the right triangle CDF would = 17, D= 90 and C=73. Hence, it would ALSO be a 90-73- 17 right triangle.
Hence, triangle BCF and triangle CDF are similar. It doesn't matter what degree I let angle B equal; it will show that
BCF and CDF are similar triangles.
Since line EF = 6 (Pythagorean, also recall BEF is a 3-4-5 scaled up by 2, hence 3*2 =6)
and BCF is similar to CDF, then line DF/6 = x/8
DF = 6/8 x
DF= 3/4 x
If DF= 3/4x, then CF = 1/4 x (1/4 + 3/4 =1 or x)
Hence, for a triangle BCF, the bases in terms of x are 'x' and '1/4 x)
Using Pythagorean Theorem x^2 + (1/4x)^2 = 8^2
x^2 + 1/16 x^2 = 64
16/16 x ^2 + 1/16 x^2 = 64
17/16 x^2 = 64
x^2 = 64 * 16/17
x^2 = 1,024/17
x = sqrt (1,024/17) or 7.7611
In triangle ∆EFB, as the hypotenuse BE = 10 and the longer leg FB = 8, ∆EFB is a 2:1 ratio 3:4:5 Pythagorean triple right triangle and the shorter leg EF must equal 6. This can also be determined directly from the Pythagorean Theorem as 10²-8² = 6².
Let ∠DFE = α and ∠FED = β, where α and β are complementary angles that sum to 90°. As DC is a straight line and ∠EFB = 90°, ∠BFC = 180°-(90°+α) = 90°-α = β, and as ∠BFC = β, ∠CBF = α. Thus ∆EDF and ∆FCB are similar triangles.
DF/FE = CB/BF
DF/6 = x/8
DF = 6x/8 = 3x/4
DC = DF + FC
x = 3x/4 + FC
FC = x - 3x/4 = x/4
Triangle ∆FCB:
FC² + CB² = BF²
(x/4)² + x² = 8²
x²/16 + x² = 64
17x²/16 = 64
x² = 64(16/17) = 1024/17
x = √(1024/17) = 32/√17
x = (32√17)/17 ≈ 7.761 units
Applying cosine-projection gives 6.cos(b)+8.cos(a)=8.cos(b)=x gives cos(b)=4/sqrt(17)
X=32/√17
EF=6,
BC/EF=8/6=4X/3X,
FC=2 , 64-16X²=0,
ВС=√(64-4)≈7.75.
Sqrt (1024/17) or 7.611
Razón de semejanza entre los triángulos rectángulos EDF y FCB =s=6/8=3/4→ DF=s*X=3X/4→ FC=X-(3X/4)=X/4→ X²+(X/4)²=8²→ X=32/√17 =7,76114....ud.
Gracias y un saludo cordial.
arccos(x/8)+arccos(8/10)+arccos(x/10)=90..arccos(x/8)+arccos(4/5)=90-arccos(x/10)...applico sin..√(1-x^2/64)(4/5)+(x/8)(3/5)=(x/10)...1-x^2/64=((1/40)x(5/4))^2=x^2/1024...1024=17x^2...x=32/√17
A questão é muito bonita. Eu a fiz de duas maneiras distintas, sendo uma delas igual à sua primeira maneira. Parabéns pela escolha! Brasil Setembro 2024. The question is very beautiful. I did it in two different ways, one of which is the same as your first way. Congratulations on your choice! Brazil September 2024.
Congrats. I like the solutions obtained without trigonometrics. They are are more criative.
Thank you 🙂
EF=√[10²-8²]=6 ⊿EDF∞⊿FCB 6 : 8 = 3 : 4
DC=3x/4 FC=x/4
x²+(x/4)²=8² 17x²/16=64 x²=1024/17 x=32/√17
(10^2=100 (8)^2=64,{100+64}= 164 360°ABCD/164 2.32ABCD 2.2^16 1.1^1^4^4 2^2^2^2 1^1^1^2 (ABCD ➖ 2ABCD+1).
can't we skip the derivation of EF (& similar) now?
most of us at this level can recognise a 3-4-5 triangle, no?
Essa eu acertei ! Fiz exatamente igual ao professor !
X=7.76
The answer is 1/17[32*sqrt(17)]. This is similar to some other problems in this channel. I got the answer. Also owe you an apology. I got the wrong answer yesterday. I screwed up the calculation and over-calculated it. For yesterday's problem, it was 5/8[sqrt(65)]. I believe that I passed my "sanity check" for today's problem!!!
asnwer=9 cm isit
asnwer=32/17 isit
7:23 samjhe nhi