la1noxz

2022-10-03

Average area of triangle formed $\frac{1}{8}$ that of square.

A point is taken at random in each of the two adjacent sides of a square. Show that the average area of the triangle formed by joining them is one eighth of the area of the square.

A point is taken at random in each of the two adjacent sides of a square. Show that the average area of the triangle formed by joining them is one eighth of the area of the square.

procjenomuj

Beginner2022-10-04Added 8 answers

Step 1

In compound probability product of two different probabilities $Area=\frac{1}{2}x\cdot y;\text{}\phantom{\rule{1em}{0ex}}{p}_{Area}=\frac{1}{2}\phantom{\rule{thickmathspace}{0ex}}{p}_{x}\phantom{\rule{thickmathspace}{0ex}}{p}_{y}$

Step 2

Here ${p}_{x}={p}_{y}=\frac{1}{2}$ for the full range of side length (0,1) considered

We can directly compare the area of gray triangle area to that of the square $\frac{\frac{1}{2}\cdot (\frac{1}{2}{)}^{2}}{{1}^{2}}=\frac{1}{8}$

In compound probability product of two different probabilities $Area=\frac{1}{2}x\cdot y;\text{}\phantom{\rule{1em}{0ex}}{p}_{Area}=\frac{1}{2}\phantom{\rule{thickmathspace}{0ex}}{p}_{x}\phantom{\rule{thickmathspace}{0ex}}{p}_{y}$

Step 2

Here ${p}_{x}={p}_{y}=\frac{1}{2}$ for the full range of side length (0,1) considered

We can directly compare the area of gray triangle area to that of the square $\frac{\frac{1}{2}\cdot (\frac{1}{2}{)}^{2}}{{1}^{2}}=\frac{1}{8}$

seguitzla

Beginner2022-10-05Added 4 answers

Step 1

Inside your square, draw the quadrilateral with vertices (p,0), (0,q), (p,1), (1,q). The triangles in the corners of the square have total area 1/2. Proof: Draw the dashed lines below. Each triangle is half the area of the rectangle containing it, and those rectangles add up to area 1.

Step 2

Thus, the average area of these 4 triangles is 1/8. By the symmetry of the problem, therefore, the expected area of each of the triangles is 1/8.

Inside your square, draw the quadrilateral with vertices (p,0), (0,q), (p,1), (1,q). The triangles in the corners of the square have total area 1/2. Proof: Draw the dashed lines below. Each triangle is half the area of the rectangle containing it, and those rectangles add up to area 1.

Step 2

Thus, the average area of these 4 triangles is 1/8. By the symmetry of the problem, therefore, the expected area of each of the triangles is 1/8.

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