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Problem 7

Find the area of the below trapezium:

Trapezium

Hint

You can split the trapezium up like this:

Sectioned trapezium

Hint

If you apply Pythagoras' Theorem to the triangle at the top, you can find $x .$ Once you know $x,$ you can substitute to find the length of each side, and then find the area.

Solution

Sectioned trapezium

Applying Pythgoras' Theorem to the triangle at the top, we get

\begin{aligned} (2 x)^{2} + [2 x - (x + 1)]^{2} & = (2 x + 1)^{2} \\ 4 x^{2} + (x - 1)^{2} & = 4 x^{2} + 4 x + 1 \\ x^{2} - 2 x + 1 & = 4 x + 1 \\ x^{2} - 6 x & = 0 \\ x (x - 6) & = 0 \\ x & \in {0, 6} \end{aligned}

Clearly, $x \neq 0$ so $x = 6$ is the only solution.

The area of the trapezium is given by the formula

\begin{aligned} \frac{a + b}{2} h & = \frac{12 + 7}{2} \times 12 \\ = 114 \end{aligned}

Problem 7 ​

Problem 7