Answer:
Explanation:
Let's break down the solution step by
step:
+
1. The problem involves an isosceles triangle with a base of length 2 and a height of length k. We want to find the maximum area of a rectangle that can be inscribed
within this triangle. 2. To start, we consider the equation of the line that contains the right side of the triangle. In this coordinate system, where the base is on the x-axis and the apex is on
the y-axis, the equation of this line is y = -
k(x-1).
3. Now, we consider a rectangle with a width (x-dimension) of w. The area of this rectangle is given by the product of its width and the corresponding height, which is determined by the equation of the line. So, the area of the rectangle is given by: area = w* (-k((w/2) - 1)) = (-k/2)2 + kw.
4. To find the maximum area, we need to find the value of w where the derivative of the area with respect to w is zero. Taking the derivative and setting it equal to zero gives us: d(area)/dw=0=-kw+k. Solving this equation for w, we get w = 1.
5. Using the formula for the area, we substitute w = 1 to find the size of the rectangle: area = (-k(1/2-1)) = k/2. This value represents half the total area of the triangle we started with.
6. Now, let's consider a specific example where the side lengths of the triangle are 8, 80, and v80. Using the Pythagorean theorem, we can find the height of the triangle: h = √(√80)-(1/2-8)) (80-16) = 8.
7. The area of the triangle is given by the formula: triangle area = (1/2)*8*8-32. 8. Finally, the maximum area of the inscribed rectangle is half the area of the triangle, which is 16 square units.
So, in summary, for an isosceles triangle with a base of 2 and height k, the maximum area of an inscribed rectangle is half the area of the triangle, which is k/2. In the specific example provided, the maximum area of the
the inscribed rectangle is 16 square units.