Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 4 cm and 6 cm if two sides of the rectangle lie along the legs.
let the base of the rectangle be x
let the height of the rectangle be y
If you make a sketch you can see three similar triangles,
by ratios:
(6-x)/y = 6/4
6y = 24 - 4x
y = (12 - 2x)/3
area = xy
= x((12-2x)/3 = 4x - (2/3)x^2
d(area)/dx = 4 - (4/3)x
= 0 for a max of area
(4/3)x = 4
4x = 12
x = 3, then y = (12-6)/3 = 2
So the largest rectangle has an area of 6 units^2
6 on bottom
4 up
w horizontal
h up from 6 side
then
w/6 = (4-h)/4
or
h/4 = (6-w)/6 (same thing by similar triangles)
Area = (1/2) w h = A
A =(1/2) w (4 - 2 w/3) = 2 w - w^2/3
dA/dw = 0 at max/min = 2 - 2 w/3
w = 3
then h = 4 - 2(3)/3 = 2
To find the largest possible area of a rectangle inscribed in a right triangle, we need to consider the relationship between the sides of the rectangle and the legs of the triangle.
Let's label the legs of the right triangle as A, B, and C, where A and B are the legs and C is the hypotenuse.
Here, A = 4 cm and B = 6 cm.
The area of a rectangle is given by A = length × width. In this case, we want to maximize the area.
Let's assume the length of the rectangle is x cm and the width is y cm.
To start finding the relationship between x, y, A, and B, we can draw the right triangle and rectangle together.
The larger side of the rectangle will be equal to the length of the triangle's leg, and the smaller side of the rectangle will be equal to the width of the triangle's leg.
Now, if we consider the larger side of the rectangle to be x cm, it will be equal to the base of the triangle, B. Similarly, the smaller side of the rectangle, y cm, will be equal to the height of the triangle, A.
Therefore, x = B and y = A.
Now, we can rewrite the area of the rectangle in terms of x and y:
A = length × width = x × y.
Substituting the values of x and y, we have:
A = B × A = 6 cm × 4 cm = 24 cm².
So, the largest possible area of the rectangle that can be inscribed in the right triangle, with legs of lengths 4 cm and 6 cm, is 24 cm².
To find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 4 cm and 6 cm, we need to understand the geometric properties involved.
Let's label the right triangle ABC, with side AB measuring 6 cm and side BC measuring 4 cm. The right angle is at vertex B.
For the rectangle to be inscribed in this triangle, two sides of the rectangle must lie along the legs of the right triangle. Let's call these sides of the rectangle CD and DE. CD lies along side AB, and DE lies along side BC.
To find the area of the largest rectangle, we need to determine the dimensions of the rectangle — specifically, the length of CD and DE. Once we have these dimensions, we can multiply them to obtain the area.
Let's examine the situation more closely. The length of CD corresponds to the width of the rectangle, and the length of DE corresponds to the height of the rectangle.
Since CD lies along the longer side of the right triangle, which is AB measuring 6 cm, CD cannot be longer than 6 cm. Similarly, DE cannot be longer than 4 cm because it lies along the shorter side BC.
Now, let's consider the specific lengths of CD and DE that maximize the area of the rectangle. To do this, we use the fact that the area of a rectangle is given by multiplying its length and width.
We have already established that CD cannot be longer than 6 cm, so let's consider the case when CD equals 6 cm. In this case, DE would be 4 cm since it can't exceed the length of BC. Thus, the area of the rectangle would be 6 cm × 4 cm = 24 cm².
However, we also need to consider the scenario where the width and height are swapped. In other words, CD is 4 cm and DE is 6 cm. This also gives us an area of 4 cm × 6 cm = 24 cm².
Therefore, in both cases, we obtain the same area of 24 cm². This means that the maximum area of the rectangle that can be inscribed in the right triangle with legs of lengths 4 cm and 6 cm is 24 cm².