The area of triangle BPQ is 1/2*BP*BQ*sinθ=1/2*BP*BQ*1/√10=5, hence BP*BQ=10√10. When we flip triangle BQC onto triangle ABP according to AB with AC, we get a triangle with a vertical angle of 90-θ, hence its area is 1/2*BP*BQ*sin(90-θ)=1/2*10√10*3/√10=15.
Yes, I forgot to write sin θ, but I wrote it later. This rule is A=1/2*a*b*sinA, where a and b are the lengths of the two sides and A is the angle between them.
Suppose triangle with given area of 5 has its two sides c and d and angle in between is theta where we are given tan (theta) =1/3.. Area of triangle given in terms of c, d, Theta is: 5 = 0.5 .c . d .sin(Theta) Now duplicate the square, rotate 90 deg clockwise and add to the bottom of existing square. In new joined figure, you will get those two shaded areas merged into a single triangleular area with its two sides same as the given one (c, d) but angle between those sides is now 90deg- Theta.. Therefore shaded area A = 0.5 .c .d . sin(90 deg - Theta) A = 0.5 .c . d . cos(Theta) Therefore the ratio of given area triangle and new merged triangle (shaded area) is simply tan(Theta) = 1/3.. The shaded area threfore 15 unit square.
Thank you for this puzzle. I have put together all that I can see here , but, as pupils say, "stuck!" My thoughts might help others to feel that they have good insights into geometry. t = tan(theta/2) with half-angle formula: 1/3 = 2t / (1 - t^2) Cross multiplying 1- t^2 = 6t which is t^2 +6t -1 = 0 t = (-6 +/- (36 +4)^½ )/2 and 0 < t < 1 t = 10^½ -3 Constructing BM perpendicular to PQ with M on PQ we have area BPQ = 5 = ½ BM × (PM+MQ) Now if BM bisects angle PBQ we might proceed but it is not a given. It is one case of where the triangle BPQ might be. However if we fit the square retrospectively to the triangle it is only a square in the case where BMD is a straight line. To prove that this is true, we should show that ABCD would be just a rectangle. A cut out scale figure of the triangle to place on a drawing of various rectangles is not good enough because neither of the lengths of BP and BQ are given , nor are they given as equals. So this is a good case for enjoying the work of Math Booster
My first husband was an engineer. I was chief cook and bottlewasher. For my grandchildren I try to preserve my ageing brain. I hope you will have a good career as an engineer. I would choose a job where the interview could show whether I could program or use a sheetsheet. These problems do not use the trial and improvement techniques that are common in engineering.
You may need to draw the diagram for this proof. Sin theta = 1/rt10 and sin (90-theta) = 3/rt10. Let BP = a and BQ =b. In triangle BPQ, area = 5 = 1/2a.b.sin theta = ab/2rt10, so ab = 10rt10. Rotate BCQ 90 degrees anticlockwise to BAQ' so that BC lands on BA, and angle PBQ' = 90 - theta and BQ' = b. Then area BPQ' = (1/2)absin(90-theta) = (1/2)10(rt10).3/rt10 = 15, and this is the area wanted.
I do not think the video solutions are too complicated. They are clear and thorough with each line of the algebra and use as few textbook results as possible. If people can do it without the solution, that is good for them. They can do it with the sound turned down and peeping every three minutes to see if they are winning! This is also fun. This puzzle was not the hardest one and it was not the easiest one. The puzzles arrrive often. It is a brilliant pick-what-you-will selection.
@@kateknowles8055 Puzzle?? You watch vídeo just for fun ??? Suppose this same exercise, in your next job interview, and answer as you commented !!! or solve it too complicated !! How will you get that hypothetical job?, with my solution?, or with video solution? or talking about puzzles ?? What kind of engineer will your employer be looking for ??
The area of triangle BPQ is 1/2*BP*BQ*sinθ=1/2*BP*BQ*1/√10=5, hence BP*BQ=10√10. When we flip triangle BQC onto triangle ABP according to AB with AC, we get a triangle with a vertical angle of 90-θ, hence its area is 1/2*BP*BQ*sin(90-θ)=1/2*10√10*3/√10=15.
The area of ∆BPQ is not BP×PQ/2, only if BP were the height of the triangle.
Yes, I forgot to write sin θ, but I wrote it later. This rule is A=1/2*a*b*sinA, where a and b are the lengths of the two sides and A is the angle between them.
@@imetroangola17 There is an extra factor sin((90-θ)) = cos((θ)=3/√10, so it works. Faster than my method!
Now it's done! The fastest and simplest method!
@@pwmiles56It's the method of the friend who posted.
Suppose triangle with given area of 5 has its two sides c and d and angle in between is theta where we are given tan (theta) =1/3..
Area of triangle given in terms of c, d, Theta is:
5 = 0.5 .c . d .sin(Theta)
Now duplicate the square, rotate 90 deg clockwise and add to the bottom of existing square. In new joined figure, you will get those two shaded areas merged into a single triangleular area with its two sides same as the given one (c, d) but angle between those sides is now 90deg- Theta.. Therefore shaded area
A = 0.5 .c .d . sin(90 deg - Theta)
A = 0.5 .c . d . cos(Theta)
Therefore the ratio of given area triangle and new merged triangle (shaded area) is simply tan(Theta) = 1/3..
The shaded area threfore 15 unit square.
A = A₁ / tan θ
A = 5 / (1/3)
A = 15 cm² ( Solved √ )
Trigonometry applied to areas !!!
Thank you for this puzzle. I have put together all that I can see here , but, as pupils say, "stuck!"
My thoughts might help others to feel that they have good insights into geometry.
t = tan(theta/2) with half-angle formula: 1/3 = 2t / (1 - t^2) Cross multiplying 1- t^2 = 6t which is t^2 +6t -1 = 0
t = (-6 +/- (36 +4)^½ )/2 and 0 < t < 1 t = 10^½ -3 Constructing BM perpendicular to PQ with M on PQ we have area BPQ = 5 = ½ BM × (PM+MQ)
Now if BM bisects angle PBQ we might proceed but it is not a given.
It is one case of where the triangle BPQ might be. However if we fit the square retrospectively to the triangle it is only a square in the case where BMD is a straight line.
To prove that this is true, we should show that ABCD would be just a rectangle.
A cut out scale figure of the triangle to place on a drawing of various rectangles is not good enough because neither of the lengths of BP and BQ are given , nor are they given as equals.
So this is a good case for enjoying the work of Math Booster
My first husband was an engineer. I was chief cook and bottlewasher. For my grandchildren I try to preserve my ageing brain. I hope you will have a good career as an engineer.
I would choose a job where the interview could show whether I could program or use a sheetsheet. These problems do not use the trial and improvement techniques that are common in engineering.
You may need to draw the diagram for this proof.
Sin theta = 1/rt10 and sin (90-theta) = 3/rt10.
Let BP = a and BQ =b. In triangle BPQ, area = 5 = 1/2a.b.sin theta = ab/2rt10, so ab = 10rt10.
Rotate BCQ 90 degrees anticlockwise to BAQ' so that BC lands on BA, and angle PBQ' = 90 - theta and BQ' = b. Then area BPQ' = (1/2)absin(90-theta) = (1/2)10(rt10).3/rt10 = 15, and this is the area wanted.
(5)^2={25 *1}/3=25/3=8.1 {90°A+90°B+90°C+90°D}=360°ABCD/8.1=45.ABCD 3^15.1 3^3^5.1 1^3^5.1 3^5(ABCD ➖ 5ABCD+3).
A inteligência artificial não funciona para tradução matemática
tanα = 1/3 --> α = 18,435°
A₁ = ½.a.b.sinα = 5 cm² (given)
a.b = 2.A₁/ sinα = 31,623 cm²
A = ½ a.b sin(90-α)
A = ½*31,623*sin(71,565°)
A = 15 cm² ( Solved √ )
Too complicated vídeo solution !!
A = A₁ / tan θ
A = 5 / (1/3)
A = 15 cm² ( Solved √ )
Trigonometry applied to areas !!!
I do not think the video solutions are too complicated. They are clear and thorough with each line of the algebra and use as few textbook results as possible.
If people can do it without the solution, that is good for them. They can do it with the sound turned down and peeping every three minutes to see if they are winning!
This is also fun. This puzzle was not the hardest one and it was not the easiest one. The puzzles arrrive often. It is a brilliant pick-what-you-will selection.
@@kateknowles8055
Puzzle?? You watch vídeo just for fun ???
Suppose this same exercise, in your next job interview, and answer as you commented !!! or solve it too complicated !!
How will you get that hypothetical job?, with my solution?, or with video solution? or talking about puzzles ??
What kind of engineer will your employer be looking for ??
Magnificent
tan theta = 1/3
tan^2 theta = 1/9
sec^2 theta - 1 = 1/9
sec^2 theta = 10/9
cos^2 theta = 9/10
sin^2 theta = 1/10
cos theta = 3/sqrt(10) [1]
sin theta = 1/sqrt(10 [2]
Let BP=a, BQ=b
5 = 1/2 ab sin theta = ab/ 2sqrt(10)
Use [2]
ab = 10 sqrt(10) [3]
Let AP=x, cQ=y, AB=AD=c
x^2 + c^2 = a^2 [4]
y^2 + c^2 = y^2 [5]
From cosine rule on PQ and right triangle DPQ
(c-x)^2 + (c-y)^2 = a^2 + b^2 - 2ab cos theta
Use [1] and [3]
2c^2 - 2c(x+y) + x^2 + y^2 = a^2 + b^2 - 60 [6]
[4]+[5]-[6]
2c(x+y) = 60
Shaded area = 1/2 cx + 1/2 cy = 15
Very very very long solve…