In trapezoid ABCD with AB parallel to CD, the leg BC is 7 cm and the distance from I, the midpoint of leg AD, to line BC is 4 cm. Find the area of the trapezoid.

Step-by-step solution

1. 1

   Notice what is missing, and stop looking for it. Neither base length nor the height of the trapezoid is given, and they cannot be recovered — infinitely many trapezoids fit the data. So the area must be computable from $BC$ and that one distance alone, which points to an area-splitting argument rather than the usual base-times-height formula.

2. 2

   Split the trapezoid at $I$. Joining $I$ to $B$ and to $C$ cuts the trapezoid into three triangles: $IAB$, $IBC$ and $ICD$. Their areas add up to the whole:

   $S_{ABCD} = S_{IAB} + S_{IBC} + S_{ICD}$

3. 3

   Use the midpoint to fix two of the three heights. Let $h$ be the height of the trapezoid, i.e. the distance between the parallel lines $AB$ and $CD$. Because $I$ is the midpoint of $AD$, with $A$ on line $AB$ and $D$ on line $CD$, its distance to each of those parallel lines is $h/2$. Hence

   $S_{IAB} = \tfrac12 \cdot AB \cdot \tfrac{h}{2}, \qquad S_{ICD} = \tfrac12 \cdot CD \cdot \tfrac{h}{2}$

4. 4

   Add them and compare with the trapezoid. Summing the two outer triangles:

   $S_{IAB} + S_{ICD} = \frac{(AB+CD)h}{4} = \frac{1}{2}\cdot\frac{(AB+CD)h}{2} = \frac{1}{2}S_{ABCD}$

   The two outer triangles take exactly half the trapezoid, so the middle triangle takes the other half:

   $S_{IBC} = \tfrac12 S_{ABCD}$

5. 5

   Compute the middle triangle from the given data. Triangle $IBC$ has base $BC = 7$ and height equal to the distance from $I$ to line $BC$, which is $4$:

   $S_{IBC} = \tfrac12 \cdot 7 \cdot 4 = 14\ \text{cm}^2$

6. 6

   Double it to get the trapezoid. Since $S_{IBC}$ is half the trapezoid,

   $S_{ABCD} = 2 \cdot 14 = 28\ \text{cm}^2$

   A numerical check confirms this is base-independent: building trapezoids with wildly different bases and heights that still satisfy $BC = 7$ and $\operatorname{dist}(I, BC) = 4$ always yields an area of $28\ \text{cm}^2$.