+35 What is the area of the largest rectangle that can be inscribed in an isosceles triangle with side lengths 8, sqrt{80}, and sqrt{80}? Assume that one side of the rectangle lies along the base of the triangle.
THX!
+23246 Start by drawing this triangle on a sheet of graph paper.
Place the base on the x-axis, centered at the origin [point O].
Call the left-hand corner point A(-4,0) and the right-hand corner point B(4,0).
Place the vertex of the triangle [point C] on the positive y-axis.
The first problem is to find the coordinates of C.
The distance from A to O is 4; the distance from A to C is sqrt(80).
Use the Pythagorean Theorem to find the distance from ) to C; it iis 8; so the coordinates of C are (0,8).
Find the equation of line CB.
Its slope is (8 - 0) / (0 - 4) = -2.
The equation is: y = -2x + 8.
Draw a rectangle inscribed in the triangle with its base on the x-axis.
Let the right-hand corner point of this base be (x,0) and the left-hand corner point of this base be (-x,0).
To find the height of the rectangle: the height will be the y-value -- and at point (x,0), the height is -2x + 8.
The area of the rectangle will be the length of its base (from point (-x,0) to (x,0) which is equal to 2x) times its height (which is -2x + 8).
Area = (2x)(-2x + 8) ---> Area = -4x2 + 16x
I'm assuming that you are taking calculus -- find the first derivative: Area' = -8x + 16
To find a maximum (or a minimum, but this time it's a maximum), set the first derivative equal to zero and solve:
---> -8x + 16 = 0 ---> -8x = -16 ---> x = 2
So, the total length of the base is 4.
The height is -2x + 8 = -2(2) + 8 = -4 + 8 = 4
So, the area is: 4 x 4 = 16 square units.
+23246 Start by drawing this triangle on a sheet of graph paper.
Place the base on the x-axis, centered at the origin [point O].
Call the left-hand corner point A(-4,0) and the right-hand corner point B(4,0).
Place the vertex of the triangle [point C] on the positive y-axis.
The first problem is to find the coordinates of C.
The distance from A to O is 4; the distance from A to C is sqrt(80).
Use the Pythagorean Theorem to find the distance from ) to C; it iis 8; so the coordinates of C are (0,8).
Find the equation of line CB.
Its slope is (8 - 0) / (0 - 4) = -2.
The equation is: y = -2x + 8.
Draw a rectangle inscribed in the triangle with its base on the x-axis.
Let the right-hand corner point of this base be (x,0) and the left-hand corner point of this base be (-x,0).
To find the height of the rectangle: the height will be the y-value -- and at point (x,0), the height is -2x + 8.
The area of the rectangle will be the length of its base (from point (-x,0) to (x,0) which is equal to 2x) times its height (which is -2x + 8).
Area = (2x)(-2x + 8) ---> Area = -4x2 + 16x
I'm assuming that you are taking calculus -- find the first derivative: Area' = -8x + 16
To find a maximum (or a minimum, but this time it's a maximum), set the first derivative equal to zero and solve:
---> -8x + 16 = 0 ---> -8x = -16 ---> x = 2
So, the total length of the base is 4.
The height is -2x + 8 = -2(2) + 8 = -4 + 8 = 4
So, the area is: 4 x 4 = 16 square units.
+128599 Building on geno's answer, here is another approach.......
Whenever we have an isosceles triangle that has a base = height, any inscribed rectangle will always have the same perimeter.......and this perimeter = twice the height of the triangle = twice the base of the triangle
Proof :
Let the base of any inscribed rectangle = 2x and [per geno's answer], let the height be -2x + 8
Then......the perimeter for any x in the domain will be = 2 [ length of rectangle base + height of rectangle ] =
2 [ 2x + (-2x) + 8] = 2 [ 8] = 16 units
Note that this is twice the base/height of the triangle......and it can be proved [ by using calculus ] that the area of any rectangle with a fixed perimeter, P, will be maximized when each side = P/4
So....the area of the inscribed rectangle = [P/4]^2 = [16/4]^2 = 4^2 = 16 units^2
