Yes, it is usually a good idea to check to see if the problem was constructed from Pythagorean triples. As you noted, the triple 8, 15, 17 produces a hypotenuse of 17, which is one of the givens in the problem. AB = 8 and BD = 15 - 9 = 6. So, ΔABD has sides 6 - 8 -10 which is the 3 - 4 -5 Pythagorean triple scaled up by a factor of 2. ΔABD is a right triangle with hypotenuse AD, so x = 6 produces a valid solution to the problem. We haven't proved that there is only one solution. Math Booster's general solutions do prove that there is only one solution. Using part of method #2, we have enough information to prove that there can be only one acute value of Math Booster's Θ, without solving for Θ. Θ must be acute because it is an angle of a right triangle and not the right angle. With only one possible valid value for Θ, there can be only one solution.
I did it the other way around, since 8:15:17 isn't one of the ones I have readily memorized. The fact that we already have two integers at DC and CA make it that much more likely that we're dealing with Pythagorean triples, after all.
Everybody calls it the Law of Cosines or Cosine Rule, nobody calls it the Al Kashi theorem. In case of Pythagoras, the theorem itself is called the Pythagoras Theorem. And the reason is that Cosine Rule was developed by many mathematicians over centuries but expressed differently deoending on the era, starting with Euclid, then Heron, al Biruni, de Muris, al Khwarizmi, Neelakantha, and then al Kashi. Then it was further simplified by Francois Viette. French mathematicians do call it Theorem de Al-Kashi sometimes.
1. Let the length of segment AB be 'y'. 2. By Pythagoras, we can write the two equations: a) y² + (x + 9)² = 289 b) y² + x² = 100 3. Subtract b) from a), losing the y² term in the process, to get: c) (x + 9)² - x² = 189 4. Expand and simplify c) as follows: x² + 18x + 81 - x² = 189, 18x = 108, x = 6 Q.E.D.
3rd method: use Heron's formula to find area of triangle ADC = 36. Height of triangle ADC = 36 * 2/9 = 8 Knowing Y = 8 you can solve for X = 6 using Pythagorean formula.
Use Heron's rule as to triangle ADC. Get area, then find internal angles . Angle ADC is 126.87, which makes angle ADB 53.13. Cosine of Angle ADB = .60, thus when applied to line AD, BD = 6.
Seeing 10 as the hypotenuse for a triangle (in ∆ABD) immediately makes me think 6:8:10 Pythagorean triple., so I plug in 6 for x and 8 for y (since the other way around would make BC = 17 amd ∆ABC would be degenerate). 8² + (6+9)² =(?) 17² 64 + 225 =(?) 289 289 = 289 Checks out. x = 6.
Third Method: Use Heron's formula to get area of triangle ACD. Calculate height h of right triangle. Calculate x by Pythagorean Theorem.
Pytagorean theorem:
(x+9)² + y² = 17²
(x²+18x+81) + y² = 289
18x = 289-81-(x² + y²) = 208-10²
18x = 108
x = 6 cm ( Solved √ )
I noticed it right away : )
Knowing the 8, 15, 17 right triangle one can try x = 6 and confirm it with the 6, 8 10 right triangle.
Yes, it is usually a good idea to check to see if the problem was constructed from Pythagorean triples. As you noted, the triple 8, 15, 17 produces a hypotenuse of 17, which is one of the givens in the problem. AB = 8 and BD = 15 - 9 = 6. So, ΔABD has sides 6 - 8 -10 which is the 3 - 4 -5 Pythagorean triple scaled up by a factor of 2. ΔABD is a right triangle with hypotenuse AD, so x = 6 produces a valid solution to the problem.
We haven't proved that there is only one solution. Math Booster's general solutions do prove that there is only one solution. Using part of method #2, we have enough information to prove that there can be only one acute value of Math Booster's Θ, without solving for Θ. Θ must be acute because it is an angle of a right triangle and not the right angle. With only one possible valid value for Θ, there can be only one solution.
I did it the other way around, since 8:15:17 isn't one of the ones I have readily memorized. The fact that we already have two integers at DC and CA make it that much more likely that we're dealing with Pythagorean triples, after all.
AB=h
h^2=10^2-x^2 h^2=17^2-(x+9)^2
100-x^2=208-x^2-18x 18x=108 x=6
As you cited 1:10 *Pythagoras* for the theorem, you will have to cite 2:17 *Al Kashi* for the cosine's theorem
Respect the authors !
Everybody calls it the Law of Cosines or Cosine Rule, nobody calls it the Al Kashi theorem. In case of Pythagoras, the theorem itself is called the Pythagoras Theorem.
And the reason is that Cosine Rule was developed by many mathematicians over centuries but expressed differently deoending on the era, starting with Euclid, then Heron, al Biruni, de Muris, al Khwarizmi, Neelakantha, and then al Kashi. Then it was further simplified by Francois Viette.
French mathematicians do call it Theorem de Al-Kashi sometimes.
1. Let the length of segment AB be 'y'.
2. By Pythagoras, we can write the two equations:
a) y² + (x + 9)² = 289
b) y² + x² = 100
3. Subtract b) from a), losing the y² term in the process, to get:
c) (x + 9)² - x² = 189
4. Expand and simplify c) as follows:
x² + 18x + 81 - x² = 189,
18x = 108,
x = 6
Q.E.D.
Cosine rule:
cos(180°-α)= (a²+b²-c²)/2ab
cos α = -(9²+10²-17²)/2.9.10
cos α = 0,6 --> α = 53,13°
x = 10 cosα
x = 6 cm ( Solved √ )
3rd method: use Heron's formula to find area of triangle ADC = 36. Height of triangle ADC = 36 * 2/9 = 8
Knowing Y = 8 you can solve for X = 6 using Pythagorean formula.
Use Heron's rule as to triangle ADC. Get area, then find internal angles . Angle ADC is 126.87, which makes angle ADB 53.13. Cosine of Angle ADB = .60, thus when applied to line AD, BD = 6.
AB^2= 17^2-(9+x)^2;
AB^2= 10^2-x^2;
17^2-(9+x)^2= 10^2-x^2;
18x= 108;
x=108/18= 6.
Треугольник ABD - египетский
Seeing 10 as the hypotenuse for a triangle (in ∆ABD) immediately makes me think 6:8:10 Pythagorean triple., so I plug in 6 for x and 8 for y (since the other way around would make BC = 17 amd ∆ABC would be degenerate).
8² + (6+9)² =(?) 17²
64 + 225 =(?) 289
289 = 289
Checks out. x = 6.
φ = 30°; ∆ ABC → AB = y; BC = BD + CD = x + 9; AC = 17; AD = 10
ADC = δ → BDA = 6φ - δ; ∆ ADC → 289 = 181 - 180 cos(δ) → cos(δ) = -3/5 →
cos(6φ - δ) = -cos(δ) = 3/5 = x/10 → x = 6 → y = 8 →
ABD = pyth. triple = 2(3 - 4 - 5); ABC = pyth. triple = (8 - 15 - 17)
(x+9)^2+(10^2-x^2)=17^2...18x=289-181=108..x=6
Couldn't be easier !
Just another problem easily solved by Stewart's Theorem.
Knew how to solve it right when I saw the thumbnail
17^2=(9+x)^2+(10^2-x^2) → x=6
x=6
The answer is x= 6. Golly the methods that you gave are much better explained than the comments. Then again both methods are REALLY familiar.
(10^2=100 H/A/(17)^2=289a/0/ (9)^2=81H/A {100H/A/+81H/A/+289A/O/}=470H/AA/O =180°ABC/470H/AA/0180°ABC/= 2.110H/AA/O/ABC 2^1.10^10^10 1^1.2^5^2^5^2^5 1^1^1^1^2^1 2^1 (H/AA/OABC ➖ 2H/AA/OABC+1).
Using the Pythagorean theorem is easier.
Yes, but it's good to know and apply different methods.
asnwer=5cm isit
asnwer=6 cm
6
Fiz pelo segundo método !
X=6