#24 inscribing and weird triangles

Kristy

Junior Member
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Jun 26, 2007
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Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3cm and 4cm if two sides of the rectangle lie along the legs??

Huh? I don’t even know what this is talking about.
I do know that the hypotenuse of the triangle must be 5 because
5^2 = 3^2 + 4^2
25 = 9 + 16

I looked up inscribed:

in•scribe ( n-skr b )
tr.v. in•scribed, in•scrib•ing, in•scribes
1.
a. To write, print, carve, or engrave (words or letters) on or in a surface.
b. To mark or engrave (a surface) with words or letters.
2. To enter (a name) on a list or in a register.
3.
a. To sign one's name or write a brief message in or on (a gift book or photograph, for example).
b. To dedicate to someone.
4. Mathematics To draw (one figure) within another figure so that every vertex of the enclosed figure touches the outer figure.
________________________________________
[Latin nscr bere : in-, in, on; see in-2 + scr bere, to write; see skr bh- in Indo-European roots.]


So I guess this means that there is a rectangle inside the triangle? And the one pointy end touches the hypotenuse or something?

I would guess that the rectangle's sides have to be littler than 3 and 4? Do they have to be exactly half as little (area of a triangle is half the area of a rectangle so just guessing?)
 
let the right triangle vertex be at the origin.
let the other two vertices be at (0,4) and (3,0)

pick any point that lies on the hypotenuse ... this will be the upper-right corner of a rectangle inscribed in the triangle.

the hypotenuse lies on the line y = (-4/3)x + 4

the area of the rectangle A = (base)(height) = xy = x[(-4/3)x + 4]

so ... find the value of x that will maximize A = (-4/3)x<sup>2</sup> + 4x
 
Kristy said:
I would guess that the rectangle's sides have to be littler than 3 and 4? Do they have to be exactly half as little (area of a triangle is half the area of a rectangle so just guessing?)

After you work out the arithmatic - following skeeter's suggestion - you'll find that your "intuition" was correct.
 
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