Is the answer 1? I got it wrong? ?

A rectangle is inscribed in a right isosceles triangle with a hypotenuse of length 1 units.

 What is the largest area the rectangle can have?

Area=_____units^2

3 Answers

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  • 4 weeks ago
    Favourite answer

    if length of hypotenuse of the right isosceles triangle is 1, the the remain sides are ½√2 each.

    constructive the right isosceles triangle as bounded by line y = x

    if height of rectangle = y then length of rectangle = ( ½√2 - x )

    Area of rectangle inscribed, A:

    A = ( ½√2 - x )( y )

    A = ( ½√2 - x )( x ) since y = x

    -A = x² - ½x√2 

    let’s complete the square

    -A = ( x - ¼√2 )² - ( ¼√2 )²

    A = ⅛ - ( x - ¼√2 )²

    A is maximum if the term ( x - ¼√2 )² is 0;

    ∴ 

    A = ⅛ - ( x - ¼√2 )²

    A = ⅛ - 0

    A = ⅛ square units

  • Amy
    Lv 7
    4 weeks ago

    Hypotenuse of 1 means the sides are 1/√2

    Suppose the rectangle is drawn in the right-angled corner. The opposite corner touches the hypotenuse.

    |\

    | \

    | _\

    | |\

    |__|_\

     x

    The small triangles have the same angles as the big triangle, so they similarly have height = width. This makes the height of the rectangle (1/√2-x).

    Area = (1/√2 - x)x =  x/√2 - x^2

    Set the derivative equal to zero.

    1/√2 - 2x = 0

    x = 1 / 2√2

    Area = 1/8

    And the second derivative is -2 everywhere, so this is a maximum.

    We could instead draw the rectangle with its side along the hypotenuse.

    |\

    | \

    |/ \ x

    |\ . \

    |_\/_\

    If this rectangle has length x along the hypotenuse, then after some geometry you'll find its area is: 

    A = (1-x)x/2

    Which also has a maximum of 1/8

  • rotchm
    Lv 7
    4 weeks ago

    If the hypo is 1 then the base and height are each √2/2. Hence the area of the whole triangle is .... 1/4. Convince yourself of this. So the inscribed rectangle must have an area smaller than that! So, try again....

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