Answer:
To find the area of trapezium \(ABCD\), we can use the formula for the area of a trapezium:
\[ \text{Area} = \frac{1}{2} (a + b) h \]
where \(a\) and \(b\) are the lengths of the two parallel sides (the bases), and \(h\) is the height of the trapezium.
In the trapezium \(ABCD\), we have the following:
- Top base \(AB = 13 \text{ cm}\)
- Bottom base \(DC = 17x + 12 \text{ cm}\)
- Height \(h = 5 \text{ cm}\)
So, the area of the trapezium is:
\[ \text{Area} = \frac{1}{2} (13 + (17x + 12)) \times 5 \]
Simplify the expression inside the parentheses:
\[ 13 + 17x + 12 = 17x + 25 \]
Now, substitute back into the area formula:
\[ \text{Area} = \frac{1}{2} (17x + 25) \times 5 \]
Multiply:
\[ \text{Area} = \frac{1}{2} \times 5 \times (17x + 25) \]
\[ \text{Area} = \frac{5}{2} (17x + 25) \]
So, the area of the trapezium in terms of \(x\) is:
\[ \text{Area} = \frac{85x + 125}{2} \]
\[ \text{Area} = 42.5x + 62.5 \text{ cm}^2 \]
Thus, the simplified area of the trapezium \(ABCD\) in terms of \
