Can you solve this impossible math question?

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Brain Station

Күн бұрын

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@brain_station_videos 29 күн бұрын

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@MarieAnne. Ай бұрын

Information is incomplete.. Not once do you mention that hypotenuse of blue triangle passes through top left vertex of rightmost square. Sure it seems to do so, but if you look closely at a large image of diagram, it looks like the hypotenuse passes slightly to the right of this vertex. So please be mindful of such details in the future.

@vincechittenden Ай бұрын

Agree completely. I paused the video at the beginning and decided that it can't be done because the hypotenuse of blue triangle clearly (I was watching on a43" TV) does not pass through the top left corner of the right hand square. So I played the video and realised I'd been mislead. You could call it click bait. The problem is relatively simple given that missing piece of information.

@colinslant Ай бұрын

Exactly. If the diagram is to scale, it doesn't pass through the vertex. If it's not to scale, it's not specified in the question that it passes through the vertex. So the proposed solution is wrong.

@MarieAnne. Ай бұрын

Let θ = angle at bottom right vertex of blue triangle. Let α = angle at bottom right vertex of white triangle below blue triangle. Since hypotenuse of blue triangle forms diagonal of rightmost square, then α + θ = 45° → θ = 45° - α tan α = 2/6 = 1/3 tan θ = tan(45°−α) = (tan 45° − tan α) / (1 + tan 45° tan α) = (1 − 1/3) / (1 + 1/3) = 1/2 Let b = base of blue triangle (hypotenuse of white triangle) b² = 2² + 6² = 40 b = 2√10 Let h = height of blue triangle h/b = tan θ h/(2√10) = 1/2 h = √10 Area of blue triangle = 1/2 * b * h = 1/2 * 2√10 * √10 = 10

@joeydifranco0422 Ай бұрын

Beat me to it.

@harikatragadda Ай бұрын

Without trigonometry or coordinate geometry. Let the Blue triangle vertices be A, B, C clockwise from the top, and intersecting the rectangle at D. Draw a perpendicular CE on AB. ∆CED is an Isosceles Right Triangle, with CD = 4, and CE= DE= 2√2. EB= 2√2+2√2=4√2 CB=√40 ∆AEC is Similar to ∆CEB. AC/CE=BC/EB AC= √10 Blue Area= ½*AC*CB=10

@saswatabanerjee2949 Ай бұрын

Could you elaborate on how you are proving that CED is isosceles?

@harikatragadda Ай бұрын

@@saswatabanerjee2949 If the upper right corner of the rectangle is F, then ∠CED=∠BDF =45°. Hence, ∠DCE =45°

@harikatragadda Ай бұрын

@saswatabanerjee2949 Notice ∠ADC=45°

@saswatabanerjee2949 Ай бұрын

@@harikatragadda Thank you so much

@RyanLewis-Johnson-wq6xs Ай бұрын

Sqrt[6^2+2^2]=2Sqrt[10]

@fire_x9 Ай бұрын

Thanks for the video. It was very interesting how you managed to find the area. Congrats! 😊

@tenhocaradellamaehnaosouum7347 Ай бұрын

We can complete the figure as a rectangle, where the base is 6, and the height is x. The blue triangle is inscribed within this rectangle, with each vertex touching a side of the rectangle. The top side of the rectangle, measuring 6, is divided into x and 6 - x, as the hypotenuse of the blue triangle intersects the rectangle's sides to form a right isosceles triangle. This happens because the hypotenuse of the blue triangle forms a 45° angle with the smaller square, meaning it also forms 45° with one side of the larger rectangle. Additionally, one of the legs of the blue triangle, the shorter one, is the hypotenuse of a smaller right triangle (the rectangle is divided into four triangles in total). Since the height of the rectangle is x, and the smaller right triangle extends upward to the rectangle's top edge, one leg of this smaller triangle is (x - 2). Therefore, the hypotenuse of this smaller triangle is sqrt(2x^2 - 16x + 40), which corresponds to one of the legs of the blue triangle. In the right isosceles triangle we constructed, the hypotenuse is x×sqrt(2), due to the relationship between the sides in an isosceles right triangle 2x^2. Since the hypotenuse of the isosceles triangle is also the hypotenuse of the blue triangle, we equate the areas of both triangles. This gives us the equation: 2x^2 = (2x^2 - 16x + 40) + 40, which simplifies to x = 5. With x = 5, the legs of the blue triangle are sqrt(10) and sqrt(40). Width that, se obtain sqrt(400) = 20, and the area of the blue triangle is 10.

@raghvendrasingh1289 Ай бұрын

👍 tan ( 45° - alpha ) = {1-(1/3)}/{1+(1/3)} = 1/2 x = (2√10)(1/2) = √10 area = 10

@RyanLewis-Johnson-wq6xs Ай бұрын

Sqrt(1-0)^2+(5-2)^2)=Sqrt[10]

@RyanLewis-Johnson-wq6xs Ай бұрын

(Sqrt[40]xSqrt[10])/2=10

@RyanLewis-Johnson-wq6xs Ай бұрын

Amazing video.

@brain_station_videos Ай бұрын

Glad you liked it! 👍

@阿中阿中 Ай бұрын

The method you used is too complicated.Actually, it's not necessary. You can draw two auxiliary lines and use the area ratio of similar triangles to get the result.