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First, let’s take the length of the short sides of the rectangle to be 6 to simplify the math. Since the aspect ratio of the rectangle is 4:3, that means the length of the long sides is 8 and the length of the diagonal is 10.
ResponderExcluirThe diagram has a small right triangle in the upper right hand corner of the rectangle on the right. This right triangle is similar to the larger right triangle just below it. So, the smaller right triangle is also a 3:4:5 triangle. But, what are the lengths of the sides? The long leg of the smaller triangle has length equal to the diagonal of the rectangle minus the length of the short side of the rectangle. So, the length of the long leg of the smaller triangle is 10 – 6 = 4, which means that the short leg’s length is 3 and the length of the hypotenuse is 5. Now we notice that the pink quadrilateral is a kite and the short leg of the smaller triangle is one of its short sides. So, the length of the short sides of the kite is 3, and since the short side of the rectangle is one of the kite’s long sides, the length of the kite’s long sides is 6. Finally, since two of the kite’s opposite angles are right angles, that means that their measures sum to 180° and therefore the kite is a cyclic quadrilateral. So, we can use Brahmagupta’s formula to calculate its area:
s = (6 + 6 + 3 + 3)/2 = 9
Area of kite = √[(9 – 6)(9 – 6)(9 – 3)(9 – 3)] = (9 – 6)(9 – 3) = 3*6 = 18
The rectangle has a base of 8 and a height of 6, so its area is 6*8 = 48. Therefore, the fraction of the rectangle that is pink is 18/48, or 3/8.
My answer: 3/8
Yet another nice application of Brahmagupta's formula. :)
ExcluirI calculated it by realizing the sides of the small triangle are 1/2 the sides of the big triangle – therefore, the area of the small triangle is 1/4 the area of the big triangle (which is 1/2 the area of the rectangle).
The area of the pink kite is, therefore, 1/2 – 1/8 of the area of the rectangle: 3/8.
That's actually much simpler. How did I not notice it?
ExcluirYour solution is still very cool. Every possible use of this Brahmagupta's formula is worthy. :)
Excluir