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How can we prove that a triangle with a base shared with a rectangle and the last point touching the opposite side to the base on the rectangle always exactly has 1/2 of the rectangle's area?

 Jan 26, 2022
 #1
avatar+717 
+4

Properties of Rectangle say that the opposite sides are congruent, and all angles are 90 degrees. The diagonal is congruent to itself so by HL congruency theorem, we can say that the two triangles are congruent, because they are congruent they add up to the area of the rectangle, so that is why one of them has half the area of the rectangle. 

 

smiley

 Jan 26, 2022
edited by proyaop  Jan 26, 2022
 #2
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last point? Can you give a picture?

but if it looks like a right triangle, then since the top left point lines with the bottm left, there is exactly enough room for another right triangle in the rectangle

 Jan 26, 2022
 #3
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nice! i didn't see it until now proyaop!

 Jan 26, 2022
 #4
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I believe the triangle only shares 1 side to the rectangle, not necessarily 2. Here is my way of proving it. First, though, let's assign some names to the different sides of the rectangle.

Here, h is the height of the rectangle, and b is the base of the rectangle. As you can see, the red triangle shares both the base and height of the rectangle: therefore, the red triangle should have exactly 1/2 of its area, since the triangle's area will be bh/2, while the rectangle's area is bh. Since you know the two small triangles make up for the rest, you can deduce that they cover 1 - 1/2 = 1/2 of the area of the rectangle. If you want to prove it using mathematical notation, here's how I would approach it: you know that the small triangles' heights are the same ones as the rectangle, and you also know that the two triangles' bases add up to the rectangle's base (b1+b2 = b). Therefore, you know that their total area is: (b1+b2)h/2, and since b1+b2 = b, then their area is bh/2, which is half of bh, that is, the rectangle's total area. 

 Feb 4, 2022

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