SOLUTION: A rectangle is inscribed in an isosceles triangle with one of the sides of the rectangle on the base of the triangle. Prove that the rectangle of greatest area occupies half the ar

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A rectangle is inscribed in an isosceles triangle with one of the sides of the rectangle on the base of the triangle. Prove that the rectangle of greatest area occupies half the ar      Log On


   

Question 1152184: A rectangle is inscribed in an isosceles triangle with one of the sides of the rectangle on the base of the triangle. Prove that the rectangle of greatest area occupies half the area of the triangle,
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

rectangle-inscibed-in-isoscales-triangle.png

the area of the triangle:
A%5Bt%5D=%281%2F2%292b%2Ah
A%5Bt%5D=b%2Ah
the area of the rectangle:
A%5Br%5D=b%2A%28h%2F2%29
A%5Br%5D=bh%2F2
=>proven that A%5Br%5D=A%5Bt%5D%2F2

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let the isosceles triangle have the center of the base at (0,0); let the vertices be at (-b,0), (0,a), and (b,0). For help with the problem later on, draw the altitude from (0,0) to (0,a).

Let one vertex of the rectangle be at (x,0), with 0 < x < b. Draw that rectangle.

The equation of the "right" side of the triangle, with endpoints at (0,a) and (b,0), is

y+=+%28-a%2Fb%29x%2Ba

Then the dimensions of the rectangle are 2x and (-a/b)x+a. And the area is then

A+=+%282x%29%28%28-a%2Fb%29x%2Ba%29+=+%28-2a%2Fb%29x%5E2%2B2ax

To find the maximum area, find the condition that makes the derivative of the area function equal to 0:

dA%2Fdx+=+%28-4a%2Fb%29x%2B2a

%28-4a%2Fb%29x%2B2a+=+0
2a+=+%284a%2Fb%29x
x+=+b%2F2

So the maximum area of the rectangle is obtained when one vertex is halfway between the center of the base and one endpoint of the base.

When a sketch is made of that condition, it should be obvious that the area of the rectangle is equal to the area of the four small triangles; and that means the area of the rectangle is half the area of the original triangle.