Find the dimensions and area of the largest rectangle that can be inscribed in a right triangle whose sides are 9cm, 12cm, 15cm.
Use quadratic functions to solve.
Put the points of the triangle at (0,0), (0,9) and (12,0)
Let the insribed rectangle have corners at:
(0,0), (x,0), (x,y'), and (0,y')
y' is on the hypotenuse of the triangle, so that
y = 12 - (4/3)x
So y' = 12 - (4/3)x
The rectangle area is
A(x) = x*y' = 12 x - (4/3)x^2
There is a maximum when dA/dx = 0
12 = (8/3) x
x = (3/8)*12 = 4.5
y' = 12 - (4/3)(9/2) = 5
The maximum rectangle area is xy'= 22 and the side lengths are 4.5 and 5.
This is normally a challenging calculus problem, however it can be worked with similar triangles, or algebra.
First, note the triangle is a right triangle. Lay it out so the 12 dimension is on the x axis, and the 9 is on the y axis.
Note the equation of the hypotenuse is
y=-9/12 x + 9
let one side of the rectangle going upward as h, and the side along the x axis as W. So the intersection with the hypotenuse is w,-9w/12 + 9
So the drill is to find the max area.
h= -9/12 w + 9
Area= hw= -9/12 w^2+9w
so when is Area max?
I calculated the point on the hypotenuse incorrectly. It is
y' = 12 - (4/3)(9/2) = 6
The rectangle side lengths are 4.5 and 6, and the area is 27. That equals half the area of the triangle, (1/2)*9*12 = 54. BobPursley tells me that Euclid proved this in many different ways, without calculus.
Thanks to Bob for pointing this out
A package with square ends has a combined length and girth (girth is the perimeter of a cross section) of 120 in. The surface area of the entire package is 3600 sq. in.
Determine the dimensions of thr package. s in.*s in.*l in.
one solution is: 11.08in*11.08in*75.68in.
find the other solution
hint: if 4S + length=120, then the length=120-4S
Round your answer to 2 decimals
Enter the 3 dimensions separated by comas
To find the dimensions and area of the largest rectangle that can be inscribed in the right triangle, we need to maximize the area of the rectangle.
Let's start by drawing the right triangle with sides measuring 9 cm, 12 cm, and 15 cm.
To inscribe a rectangle in the right triangle, we need to find the dimensions of the rectangle such that its sides are parallel to the legs of the triangle.
Let's assume the dimensions of the rectangle are x cm and y cm. We'll place the shorter side of the rectangle along the shorter side of the triangle. So, x will be the width of the rectangle and y will be its height.
Now, if we draw a diagram, we see that the length of the rectangle (y) forms the hypotenuse of a smaller right triangle with sides x and 9 cm. Similarly, the width of the rectangle (x) forms the hypotenuse of another smaller right triangle with sides y and 12 cm.
Using the Pythagorean theorem, we can relate the sides of the smaller right triangles to find the relationship between x and y:
For the first smaller right triangle:
x^2 + 9^2 = y^2
For the second smaller right triangle:
y^2 + 12^2 = x^2
Rearranging the equations, we get:
x^2 - y^2 = 81 .....(1)
y^2 - x^2 = 144 .....(2)
Now, we have a system of two equations. To solve this system, we'll use substitution or elimination.
Let's solve the system using elimination. Adding equations (1) and (2), we get:
2y^2 = 225
y^2 = 225/2
y = √(225/2)
y = √112.5
y ≈ 10.61 cm
Substituting this value of y into equation (1), we get:
x^2 - (√112.5)^2 = 81
x^2 - 112.5 = 81
x^2 = 112.5 + 81
x^2 = 193.5
x = √193.5
x ≈ 13.92 cm
So, the dimensions of the largest rectangle inscribed in the right triangle are approximately 13.92 cm (width) and 10.61 cm (height).
To find the area of this rectangle, we multiply the width and height:
Area = width × height
Area ≈ 13.92 cm × 10.61 cm
Area ≈ 147.49 cm^2
Therefore, the dimensions of the largest rectangle that can be inscribed in the right triangle are approximately 13.92 cm (width) and 10.61 cm (height), and its area is approximately 147.49 square centimeters.