A plot of land is an isosceles trapezoid with parallel sides 12 m and 18 m and area 60 m². Find the length of fence needed to enclose it.

Step-by-step solution

1. 1

   Recover the height from the area. For a trapezoid,

   $A=\frac{b_1+b_2}{2}\cdot h\quad\Longrightarrow\quad 60=\frac{12+18}{2}\cdot h=15h\quad\Longrightarrow\quad h=4\ \text{m}$

   The height is the bridge between the given area and the unknown slanted sides.

2. 2

   Find the horizontal overhang of each leg. Dropping perpendiculars from the ends of the shorter base to the longer base cuts off two congruent right triangles (this is where "isosceles" is used). Together they account for the excess length of the long base:

   $\frac{18-12}{2}=3\ \text{m each}$

3. 3

   Find a leg with the Pythagorean theorem. Each right triangle has legs $3$ m and $4$ m:

   $\text{leg}=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5\ \text{m}$

   (the familiar $3$–$4$–$5$ triangle).

4. 4

   Add up the four sides.

   $P=12+18+5+5=40\ \text{m}$

   $\boxed{40\ \text{m of fence}}$

5. 5

   Check the area from the reconstructed shape. The trapezoid is a $12\times 4$ rectangle in the middle plus two right triangles of area $\tfrac12(3)(4)=6$ each: $48+6+6=60\ \text{m}^2$ ✓, matching the given area. Note also the unit change — the data was an area in $\text{m}^2$ but the answer is a length in m.