Finding the Possible Values of k for a Trapezium with a Given Area

In this article, we will explore a mathematical geometry problem involving a trapezium and a given area. The trapezium is enclosed by the lines defined by y0, y6, y8-2x, and yxk, where k is a constant. We will derive the possible values of k given that the area of the trapezium is 66 square units.

Understanding the Geometry

The trapezium is bounded by the straight lines defined as follows:

y0: The x-axis. y6: A horizontal line parallel to the x-axis, 6 units above it. y8-2x: A line with a negative slope, intersecting the y-axis at 8 and the x-axis at 4. yxk: A line with a slope of k, passing through the origin.

Area of the Trapezium

The area of a trapezium can be calculated using the formula:

A frac{1}{2} (B b) cdot h

where A is the area, B and b are the lengths of the parallel sides, and h is the height (altitude).

Step-by-Step Calculation

We start by determining the coordinates of the points where the given lines intersect:

Intersection with y0 and y8-2x

y0: The line y8-2x intersects the x-axis at x4. So, the point is (4, 0). y0: The line yxk intersects the x-axis at x-k. So, the point is (-k, 0).

Intersection with y6 and y8-2x

y6: The line y8-2x intersects the line y6 at x1. So, the point is (1, 6). y6: The line yxk intersects the line y6 at x6-k. So, the point is ((6-k), 6).

Calculating the Area

Let's calculate the area of the trapezium using these points:

B: The length of the side along the x-axis is -k - 4. b: The length of the side along the line y6 is 5 - k.

Substitute these values into the area formula:

A frac{1}{2} cdot (-k - 4 5 - k) cdot 6 66

Simplify and solve for k:

frac{1}{2} cdot (1 - 2k) cdot 6 66

3 cdot (1 - 2k) 66

1 - 2k 22

-2k 21

k -10.5

Let's verify by checking the possible values of k in different intervals:

Case 1: k -4

-k - 4 22

-k 26

k -26

Case 2: -4 k 5

5 - k 22

k -17

This case is not possible.

Case 3: k 5