The computations of the areas at the beginning of video (about 4 and an half minutes) could be skipped. Once length r has been found to be 6, we have the lengths of all 3 sides of each of the 2 triangles. We can then use Heron's formula to compute the areas of each triangle and add the areas. To avoid the constructions, at 6:18 we have the equation 96 = r² + 20rcos(x) from ΔCDE. Applying the cosine rule to ΔABD, we get a second equation, (14)² = r² + (r + 10)² - 2r(r + 10)cos(x). We should be able to substitute (96 - r²)/20r for cos(x) in the second equation and solve for r. (Without doing the algebra, we note that r = 6 and x = 60° balances both equations.)
Area of shaded isósceles triangle: Heron's formula: s = ½[14+14+(10+10+6)]= 27 cm A² =s(s-a)(s-b)(s-c) A² = 27(27-14)(27-14)(27-26) A = 67,55 cm² = 39√3 cm²
In my solution there's no need to introduce that equilateral triangle or that parallelogram, and there's no need to calculate angle "x" neither angle "y"
@@lwels49 You wrote : h² = ¾r² = ¾6² --> h = 3√3cm How did you get that ratio "3/4" ??? I wrote : h² = 14²-(½26)² --> h = 3√3cm (Pytagorean theorem) or what is the same: h² = r² - (½26-10)² = 6² - 3²
@marioalb9726 The height of the isosceles triangle is a side of a right triangle, the other sides being 14 and 10 + r/2. This height is also one of the sides of a small right triangle . The other sides are hypotenuse r and r/2. Pythagoras gives height squared = r squared- (r/2) squared = 3/4 of r squared. I hope this helps and good luck with future puzzles.
Interesting problem, I think it was presented some time ago but now it's solved by a different method. If we can't solve it by construction, then the formula helps. I will look into it to see if I get a different idea for solving it.
Are you sure this is it, because I can only remember solving the Question only 3days ago, Are you referring to the other Question whith a square which you said you did the solving but had a slight error in the arithmetic? Perhaps you are right, I may have created the Question without realizing that I made one similar some times back, let me check
@@MathandEngineering I checked all videos and this problem was NOT presented before, but a similar one was presented, with the same title as this video, presented 6 months ago. That video is about an X-Y region with areas of 54, 14, and X for different sections (2 have X).
The computations of the areas at the beginning of video (about 4 and an half minutes) could be skipped. Once length r has been found to be 6, we have the lengths of all 3 sides of each of the 2 triangles. We can then use Heron's formula to compute the areas of each triangle and add the areas.
To avoid the constructions, at 6:18 we have the equation 96 = r² + 20rcos(x) from ΔCDE. Applying the cosine rule to ΔABD, we get a second equation, (14)² = r² + (r + 10)² - 2r(r + 10)cos(x). We should be able to substitute (96 - r²)/20r for cos(x) in the second equation and solve for r. (Without doing the algebra, we note that r = 6 and x = 60° balances both equations.)
Shaded area can be converted into an isosceles triangle, sides 14, 14 and 10+10+r , therefore :
Angle BAD = Angle DCE = α
Cosine rule, triangle CDE :
r²=14²+10²-2*14*10*cosα
cosα = (296-r²)/280
Cosine rule, triangle ABD :
r²=14²+(10+r)²-2*14*(10+r)*cosα
r²=14²+(10²+20r+r²)-(280+28r)cosα
cosα=(296+20r)/(280+28r)
Equalling:
(296-r²)/280= (296+20r)/(280+28r)
(296-r²)*(280+28r)=(296+20r)*280
82880-280r²+8288r-28r³=82880+5600r
28r³+280r²-2688r=0
r² + 10r -96= 0 --> r=6 cm
b = 10+10+r = 26 cm
Blue shaded area, isósceles triangle:
h² = 14²-(b/2)² --> h = 3√3cm
A = ½.b.h = ½*26*3√3 = 39√3
A = 67,55 cm² ( Solved √ )
Area of shaded isósceles triangle:
Heron's formula:
s = ½[14+14+(10+10+6)]= 27 cm
A² =s(s-a)(s-b)(s-c)
A² = 27(27-14)(27-14)(27-26)
A = 67,55 cm² = 39√3 cm²
In my solution there's no need to introduce that equilateral triangle or that parallelogram, and there's no need to calculate angle "x" neither angle "y"
The height of this isosceles triangle turns out to be the square root of 3/4 of r squared. Get a simple quadratic in r squared using Pythagoras
@@lwels49
You wrote :
h² = ¾r² = ¾6² --> h = 3√3cm
How did you get that ratio "3/4" ???
I wrote :
h² = 14²-(½26)² --> h = 3√3cm
(Pytagorean theorem)
or what is the same:
h² = r² - (½26-10)² = 6² - 3²
@marioalb9726 The height of the isosceles triangle is a side of a right triangle, the other sides being 14 and 10 + r/2. This height is also one of the sides of a small right triangle . The other sides are hypotenuse r and r/2. Pythagoras gives height squared = r squared- (r/2) squared = 3/4 of r squared. I hope this helps and good luck with future puzzles.
Impressive video as usual (except sometimes when it's mafia)
My method was :
Cosine Rule n°1 (al-kashi) in triangle ABD : AB^2=AD^2+BD^2-2*AB*BD*cos(x)
Cosine Rule n°2 (al-kashi) in triangle CDE : CE^2=CD^2+DE^2-2*CD*DE*cos(180°-x) (same than the video at 5:26)
Cosine Rule n°1 (al-kashi) : 14^2=(10+r)^2+r^2-2*(10+r)*r*cos(x)
Cosine Rule n°2 (al-kashi) : 14^2=10^2+r^2-2*10*r*cos(180°-x)
14^2=10^2+r^2+2*10*r*cos(x)
Equation 1 - Equation 2 :
14^2-14^2=(10+r)^2+r^2-2*(10+r)*r*cos(x)-(10^2+r^2+2*10*r*cos(x))
0=2*10*r+r^2-2*(2*10+r)*r*cos(x)
2*10*r+r^2=2*(2*10+r)*r*cos(x)
1=2*cos(x)
cos(x)=1/2
x=60°
Equation 2 :
14^2=10^2+r^2+2*10*r*(1/2)
r^2+10*r+(10^2-14^2)=0
r^2+10*r-96=0
(r-6)*(r+16)=0
r=6
Triangle ABD : AB=14 ; BD=r=6 ; AD=6+10=16
Triangle CDE : CD=10 ; DE=r=6 ; CE=14
Heron's formula :
A(blue)=A|ABD|+A|CDE|
A(blue)=1/4*sqrt((6+16+14)*(-6+16+14)*(6-16+14)*(6+16-14))+1/4*sqrt((6+10+14)*(-6+10+14)*(6-10+14)*(6+10-14))
A(blue)=1/4*sqrt(3*96^2)+1/4*sqrt(3*60^2)
A(blue)=96/4*sqrt(3)+60/4*sqrt(3)
A(blue)=39*sqrt(3)
@@matthieudutriaux 😊, friend thank you for sharing this amazing method, the first sentence in the comment was funny though. Thanks
Good solution using no advanced fornulas. Congrats.
Thanks sir 👍, your comment means a lot to me. I appreciate
Amazing
Thanks
Interesting problem, I think it was presented some time ago but now it's solved by a different method. If we can't solve it by construction, then the formula helps. I will look into it to see if I get a different idea for solving it.
Are you sure this is it, because I can only remember solving the Question only 3days ago,
Are you referring to the other Question whith a square which you said you did the solving but had a slight error in the arithmetic?
Perhaps you are right, I may have created the Question without realizing that I made one similar some times back, let me check
@@MathandEngineering I checked all videos and this problem was NOT presented before, but a similar one was presented, with the same title as this video, presented 6 months ago. That video is about an X-Y region with areas of 54, 14, and X for different sections (2 have X).