MHB Is the perimeter of a right triangle equal to its area? (Part 2)

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In the discussion, a right triangle is defined with legs u and v, and hypotenuse w. The values of u, v, and w are expressed in terms of variables m and n. The objective is to demonstrate that the area of the triangle, calculated as (1/2)(uv), equals the perimeter, represented by u + v + w. The participants confirm that the calculations involve multiplying u and v by 1/2 and then adding the lengths to verify the equality. The exercise aims to illustrate that the perimeter and area of this specific right triangle are numerically equal.
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A right triangle is given. One leg is u units and the other leg is v units. The hypotenuse is given to be w units.

If u = [2(m + n)]/n, v = 4m/(m - n), and
w = [2(m^2 + n^2)/(m - n)n, show that

(1/2)(uv) = u + v + w

I must multiply u times v times (1/2), right? I then must add u + v + w. The right side must equal the left, right?

This exercise will show that the perimeter is numerically equal to the area.
 
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RTCNTC said:
I must multiply u times v times (1/2), right? I then must add u + v + w. The right side must equal the left, right?
Yes.
 
Evgeny.Makarov said:
Yes.

Cool.
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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