Let a = width of blue triangle It is given that the hypotenuse of the blue triangle is perpendicular to the hypotenuse of the green triangle. As such, we know that the gradients (slopes) of the two lines are negative reciprocals, so they multiply to -1: (-6/a)*(4/(11-a))=-1 -24/(11a-a²)=-1 24/(11a-a²)=1 24=11a-a² a²-11a+24=0 Then we solve for a value of a. Of the two values, 8 and 3, We choose the bigger one. We then use the Pythagorean triples (3,4,5) and (6,8,10) (a multiple of 345) to find the two sides of the rectangle, 5 and 10. Hence the area is 50.
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Let a = width of blue triangle
It is given that the hypotenuse of the blue triangle is perpendicular to the hypotenuse of the green triangle. As such, we know that the gradients (slopes) of the two lines are negative reciprocals, so they multiply to -1:
(-6/a)*(4/(11-a))=-1
-24/(11a-a²)=-1
24/(11a-a²)=1
24=11a-a²
a²-11a+24=0
Then we solve for a value of a. Of the two values, 8 and 3, We choose the bigger one. We then use the Pythagorean triples (3,4,5) and (6,8,10) (a multiple of 345) to find the two sides of the rectangle, 5 and 10. Hence the area is 50.
Great explanation!