A rectangle is inscribed in a right triangle with legs of length 5 and 12. The sides of the rectangle are parallel to the legs of he triangle. Find the dimensions of the rectangle that has the largest area?
no ideas on any of these?
....but the last one I did was fun :)
To find the dimensions of the rectangle with the largest area that is inscribed in the given right triangle, we need to determine the relationship between the sides of the right triangle and the rectangle.
Let's assume the length of the rectangle is x and the width is y.
Since the sides of the rectangle are parallel to the legs of the triangle, we can determine that the width of the rectangle is x, and the length is y.
To maximize the area of the rectangle, we need to find the values of x and y that satisfy the condition of being inscribed in the triangle and maximize the area at the same time.
Given that the legs of the right triangle have lengths of 5 and 12, we can use these values to set up an equation for the relationship between the sides of the triangle and the rectangle.
The area of a rectangle is given by the formula: A = length * width.
We know that the length of the rectangle is y and the width is x.
Therefore, the area of the rectangle is A = xy.
The right triangle's legs of length 5 and 12 form a right angle, so we can use Pythagoras' theorem to find the length of the hypotenuse (z).
Pythagoras' theorem states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.
In this case, we have:
z^2 = 5^2 + 12^2
z^2 = 25 + 144
z^2 = 169
Taking the square root of both sides, we find:
z = 13
Now, since the sides of the rectangle are parallel to the sides of the right triangle, the diagonal of the rectangle is equal to the hypotenuse of the triangle (z).
Therefore, we have:
x^2 + y^2 = z^2
x^2 + y^2 = 13^2
x^2 + y^2 = 169
Now, we need to maximize the area of the rectangle, A = xy, which means we need to maximize the product of x and y.
To do that, we can rewrite the equation above as:
y = sqrt(169 - x^2)
Substituting this expression for y in the equation A = xy, we have:
A = x * sqrt(169 - x^2)
To find the dimensions of the rectangle with the largest area, we need to find the maximum value of A, which occurs when the derivative of A with respect to x is zero.
To find the derivative of A = x * sqrt(169 - x^2), we can use the chain rule:
dA/dx = sqrt(169 - x^2) + x * (1/2) * (169 - x^2)^(-1/2) * (-2x)
= sqrt(169 - x^2) - x^2/sqrt(169 - x^2)
Setting the derivative equal to zero, we have:
sqrt(169 - x^2) - x^2/sqrt(169 - x^2) = 0
To simplify this equation, we can multiply both sides by sqrt(169 - x^2):
(169 - x^2) - x^2 = 0
169 - 2x^2 = 0
2x^2 = 169
x^2 = 169/2
x^2 = 84.5
Taking the square root of both sides, we find:
x = sqrt(84.5)
Therefore, the width of the rectangle is approximately x = 9.20.
Substituting this value of x into the equation y = sqrt(169 - x^2), we can find the length of the rectangle:
y = sqrt(169 - (sqrt(84.5))^2)
= sqrt(169 - 84.5)
= sqrt(84.5)
Therefore, the dimensions of the rectangle with the largest area that is inscribed in the given right triangle are approximately width = 9.20 units and length = sqrt(84.5) units.