$\Delta ABC$ is inscribed in a rectangle as shown ($B$ and $C$ are midpoints of two of the rectangle’s sides). What fraction of the rectangle’s area does the triangle cover? Show assumed knowledge

The area of a triangle is half of the base times height. Areas may be added or subtracted to find other areas (Section 10.8: Area). Fractions can be added or subtracted by first making their denominators the same (Section 4.2.5: Equivalent fractions).

Focus on the three other triangles in the rectangle.

Finding the area of $\Delta ABC$ directly is rather difficult – we don’t know its base length or height. (We’ll learn the techniques necessary for that in Section 11.2: Pythagoras' Theorem and Question 11.3.7.) A better idea might be to focus on the other triangles, which are all right-angled so their areas are easily worked out.

If you prefer to use algebra for this then you could call the length of the rectangle $x$, and the height $y$. Then the rectangle’s area is $xy$. The top left triangle measures $x/2$ by $y$, so its area is $xy/4$. The top right triangle measures $x/2$ by $y/2$, so its area is $xy/8$, and the bottom right triangle measures $x$ by $y/2$, so its area is $xy/4$. Then the area of $\Delta ABC$ is \begin{align*} xy - \frac{xy}{4} - \frac{xy}{8} - \frac{xy}{4} &= xy\left(1 - \frac{1}{4} - \frac{1}{8} - \frac{1}{4}\right)\\ &= xy\frac{8 - 2 - 1 - 2}{8}\\ &= \frac{3}{8}xy \end{align*} so the triangle covers $3/8$ of the rectangle.

Alternatively, we could simply reason that the top left triangle could fit four times into the rectangle like so: Similarly, the top right triangle fits eight times and the bottom right triangle fits four times. Therefore together they cover $1/4 + 1/8 + 1/4 = 5/8$ of the rectangle, and the remaining area must be $3/8$ of the rectangle.