Find the area of a trapezoid whose diagonals are 15 and 7 and whose midline (median) is 10.

Step-by-step solution

1. 1

   Translate the data into the standard area formula. For a trapezoid with parallel sides $a$ and $b$ and height $h$,

   $S=\frac{a+b}{2}\,h=m\,h$

   where $m$ is the midline. Here $m=10$, so $S=10h$ and $a+b=2m=20$. Everything now hinges on the height, which the problem never gives directly.

2. 2

   Slide one diagonal sideways to collect the data into one triangle. Translate the diagonal $BD$ by the vector $\vec{DC}$ (i.e. along the direction of the parallel sides). Its endpoint lands on the extension of the longer base, producing a triangle whose three sides are the two diagonals $15$ and $7$ and a base equal to $a+b=20$. This is the key move: the translation changes neither the length of the diagonal nor the height of the figure.

3. 3

   Explain why that triangle has the same area as the trapezoid. The new triangle has base $a+b=20$ and the same height $h$ as the trapezoid, so its area is $\tfrac12(a+b)h=\tfrac{a+b}{2}h$ — exactly the trapezoid's area. So computing the triangle's area is computing the trapezoid's area.

4. 4

   Apply Heron's formula to the triangle with sides 15, 7, 20. Check first that it exists: $15+7=22>20$. The semiperimeter is

   $s=\frac{15+7+20}{2}=21$

   and

   $S=\sqrt{21(21-15)(21-7)(21-20)}=\sqrt{21\cdot 6\cdot 14\cdot 1}=\sqrt{1764}=42$

5. 5

   Cross-check by finding the height directly. Drop both diagonals onto the base direction. Their horizontal projections $x$ and $20-x$ satisfy

   $15^2=x^2+h^2,\qquad 7^2=(20-x)^2+h^2$

   Subtracting removes $h^2$:

   $225-49=x^2-(20-x)^2=40x-400\;\Longrightarrow\;40x=576\;\Longrightarrow\;x=14.4$

   Then $h^2=225-14.4^2=225-207.36=17.64$, so $h=4.2$ and

   $S=10\cdot 4.2=42$

   Two independent methods agree, so the area is $42$.