"A rectangle is inscribed in a semicircle of radius 2 cm. Find the largest area of such a rectangle".
There is a diagram, but I think the question makes it clear enough what is going on. I'm having problems finding a relationship that I can work with.
draw a radius from the centre to a vertex of the rectangle
Let the base of the rectangle be 2x, making the triangle base x, and let the height of the rectangle by y
Area of rectangle = 2xy
but x^2 + y^2 = 4
y = √(4 - x^2) or (4-x^2)^(1/2)
A = 2x(4-x^2)^(1/2)
take the derivative using the product rule,
set it equal to zero and solve for x
Thank you! Never occurred to me to position the radius like that.
To find the largest area of a rectangle inscribed in a semicircle, we can start by considering the relationship between the rectangle and the semicircle.
Let's label the rectangle as ABCD, with sides AB and BC parallel to the diameter of the semicircle. Let E be the midpoint of AB, and F be the point where the rectangle intersects the semicircle.
Next, let's assign variables to the different lengths involved. Let the length of AB be 2l and the length of BC be 2w. The radius of the semicircle is given as 2 cm.
Using these variables, we can express EF in terms of l and w. Since EF is the diameter of the semicircle, it has a length of 2 times the radius, or 4 cm.
Now, we can see that triangle EBF is a right triangle with a hypotenuse EF and legs 2l and w. From the Pythagorean theorem, we can write:
(2l)^2 + w^2 = (EF)^2
4l^2 + w^2 = 16
To simplify, we can solve for w^2:
w^2 = 16 - 4l^2
w^2 = 4(4 - l^2)
w = 2√(4 - l^2)
The area of the rectangle, A = l * w = l * 2√(4 - l^2).
To find the largest area, we can take the derivative of A with respect to l and set it equal to zero:
(dA/dl) = 2√(4 - l^2) - 2l*(1/2)(4 - l^2)^(-1/2)
(dA/dl) = 2√(4 - l^2) - l/(√(4 - l^2))
(dA/dl) = 0
Simplifying further, we have:
2√(4 - l^2) = l/(√(4 - l^2))
2(4 - l^2) = l^2
8 - 2l^2 = l^2
3l^2 = 8
l^2 = 8/3
l = √(8/3)
Substituting this value of l back into the equation for w, we have:
w = 2√(4 - (8/3))
w = 2√(12/3 - 8/3)
w = 2√(4/3)
w = 4√3/3
Finally, we can find the largest area of the rectangle by substituting the values of l and w into A = l * w:
A = (√(8/3)) * (4√3/3)
A = (4/3)√(8*3)
A = (4/3)√(24)
A = (4/3) * 2√6
A = (8/3)√6
Therefore, the largest area of the rectangle inscribed in the semicircle is (8/3)√6 square units.
To find the largest possible area of the rectangle inscribed in a semicircle, we need to analyze the given information and establish a relationship between the dimensions of the rectangle.
Let's start by drawing the diagram and labeling the relevant quantities. We have a semicircle with a radius of 2 cm, and an inscribed rectangle within it:
```
A _______ B
| |
|_______|
O C D
```
In the diagram:
- O represents the center of the semicircle.
- A and B are the endpoints of the semicircle.
- CD is the base of the rectangle.
- AC and BD are the lengths of the rectangle.
To establish a relationship, we can observe that the diagonal of the rectangle (AC) is equal to the diameter of the semicircle (AB). The diameter of a circle is twice the radius, so AB = 2 * 2 cm = 4 cm.
Using the Pythagorean theorem, we can find the relationship between the dimensions of the rectangle:
AC^2 + CD^2 = AD^2
Let's calculate the area of the rectangle and express it in terms of CD and AD:
Area of rectangle = CD * AD
Using the relationship derived from the Pythagorean theorem:
Area = CD * (√(AC^2 + CD^2))
However, since we know that AC = 4 cm, we can simplify the equation:
Area = CD * (√(16 + CD^2))
Now, to find the largest possible area, we need to maximize this expression. We can do this by finding the critical points of the function and determining which point yields the maximum value.
To find the critical points, we take the derivative of the area equation with respect to CD and set it equal to zero:
d(Area)/d(CD) = (√(16 + CD^2)) + CD * (1/(2√(16 + CD^2))) * (2CD)
Setting this equal to zero and solving for CD, we get:
(√(16 + CD^2)) + CD * (1/(√(16 + CD^2))) * CD = 0
Simplifying the equation, we have:
(√(16 + CD^2)) + CD^2/(√(16 + CD^2)) = 0
Multiplying both sides by (√(16 + CD^2)), we get:
16 + CD^2 + CD^2 = 0
2CD^2 + 16 = 0
CD^2 = -8
Since we have a negative value for CD^2, it means that no real critical point exists. Therefore, the function does not have a maximum or minimum in the given interval.
However, we can determine the largest possible area by testing the endpoints of the interval (0, 4) and evaluating the area equation:
When CD = 0, the area is zero.
When CD = 4, the area is 4 * (√(16 + 4^2)) = 4 * (√32) ≈ 17.89 cm^2.
Therefore, the largest possible area for the rectangle inscribed in the semicircle is approximately 17.89 cm^2.