why area of triangle changes when measured as components of triangles? [closed] If we measure an area of triangle directly using a formula I got 90 square unit. But if we measure by components like 2 triangles and one rectangle and take some it counts 90.5

Why 0.5 square unit difference occur?

Any Help will be appreciated

closed as off-topic by user21820, Lord Shark the Unknown, Eevee Trainer, Leucippus, Xander HendersonJan 15 at 4:57

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• Is this meant as a riddle? The union of the colored areas is not a triangle. – Martin R Jan 14 at 9:31
• yes.I wanted to know why such thing is happening – MathLover Jan 14 at 9:33
• – Andrew T. Jan 14 at 16:19
• See the addendum to THIS ANSWER – steven gregory Jan 14 at 21:38

Because the joint point of the two segments is not on the point of $$(11,5)$$. As the equation of the line is $$y = \frac{9}{20} x$$, hence the coordinate of the joint point is $$(\frac{20\times 5}{9} = \frac{100}{9}, 5)$$ which is not exactly $$(11,5)$$.

This picture is basically what happens here.

The difference is that your picture tries to make the slope of the orange and the blue triangles almost the same as the big one, hence giving the illusion of them being sub-triangles. The points A$$(0,0)$$ , P$$(11,5)$$ and B$$(20,9)$$ are not on the same straight line. It forms a very small triangle (APB). So, the half area of the rectangle is not the sum of the colored pieces. $$a=AP=\sqrt{11^2+5^2}\quad;\quad b=PB=\sqrt{11^2+4^2}\quad;\quad c=AB=\sqrt{9^2+20^2}$$

The calculus of the area of triangle (APB) is easy thanks to the Heron's formula :

$$\frac14\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}=0.5$$

Similar fake problem with solution on page 23 in https://fr.scribd.com/doc/15493868/Pastiches-Paradoxes-Sophismes