In the diagram, A(5 ; 3), B(0 ; 1/2), C and E(6 ; -4) are the vertices of a trapezium having BA || CE - NSC Mathematics - Question 3 - 2022 - Paper 2

First, find the gradient of line CE using points C and E. Since the coordinates of E are (6, -4) and we denote C as (6, yc):

$m_{CE} = \frac{y_{E} - y_{C}}{x_{E} - x_{C}}$

Assuming C lies on the y-axis, the gradient calculation yields:

Essentially, we need to find the y-intercept (c) of the line:

• The equation can be represented as:

$y = mx + c$ Using the gradient we calculated earlier and one of the points, eventually solving gives:

$c = -7\space \text{(substituting values into the equation)}$ The full equation becomes:

$y = \frac{1}{2}(x - 6) - 4$

Given that D is the y-intercept, we find:

$C(6, -10)$ Thus, the coordinates of point C are (6, -10).

The area of quadrilateral ABCD can be calculated as follows:

Using the formula for area:

$Area = \frac{1}{2} \times base \times height$

Base is |7| from the x-coordinates and the height is |5| from the y-coordinates. Hence:

$Area = \frac{1}{2} \times 7 \times 6 = 21.5$

Thus, Area of quadrilateral ABCD = 18.75.

Reflecting E(6, -4) across the y-axis gives:

$K(-6, -4)$

Using the distance formula:

$$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

• Distance KE = 6 units
• Distance EC = 12 units
• Distance CK can be calculated using:

$\sqrt{(6)^2 + (4)^2} = \sqrt{36 + 16} = 14\space units$

Total perimeter:

$Perimeter = KE + EC + CK = 6 + 12 + 14 = 32\space units$

Using trigonometry:

$tan(\angle KCE) = \frac{KE}{KC} = \frac{12}{6} = 2$

Thus:

$\angle KCE = \tan^{-1}(2) = 63.43°$