Respuesta :
The area of the largest rectangle that can be inscribed in the right triangle is calculated to be 6 cm^2
As we know that for a rectangle of width W and length L the area is given as;
A = L × W
Now let's say that L is along the 8 cm side and W is along the 3 cm side, let's find a relation between these two.
When we inscribe the rectangle in the larger triangle the quotient between the catheti of the right triangle must be equal to the quotient of the catheti of the right triangle that is generated.
Therefore that new triangle will have legs equal to the;
8cm - L and W.
Then we have that:
8cm - L / W = 8 cm / 3 cm
L = 8 cm - 8/3 × W
Now we can write L in terms of W so that we can replace that in the area equation to get
A = (8cm - (8/3)×W)×W = 8cm×W - (8/3)×W^2
This is the equation we have to maximize, also this is a quadratic polynomial of negative leading coefficient, which means that the maximum is at the vertex.
As we know the vertex of the general quadratic polynomial;
a × x^2 + b × x + c
is at:
x = -b / 2a
Then, in this scenario, the vertex is at;
W = (-8 cm) / (2 × 8/3) = 1.5 cm
Then the length is given by:
L = 8cm - (8/3) × 1.5cm = 4 cm
Therefore the area of the largest rectangle that can be inscribed in the given right triangle is;
A = 1.5 cm × 4 cm = 6 cm^2
To learn more about the area; click here:
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