An isosceles trapezoid has a top base of 4, a bottom base of 4 + 2r and two legs each of length 2r. Find its area and its perimeter in terms of r.

Step-by-step solution

1. 1

   Add the four sides for the perimeter. No height is needed here:

   $P=4+(4+2r)+2r+2r=8+6r.$

2. 2

   Split the overhang to set up a right triangle. The lower base exceeds the upper by $(4+2r)-4=2r$. In an isosceles trapezoid that excess is shared equally by the two ends, so each end overhangs by

   $\frac{2r}{2}=r.$

   Dropping a perpendicular from each upper vertex creates a right triangle with hypotenuse $2r$ (the leg) and horizontal side $r$.

3. 3

   Find the height by the Pythagorean theorem.

   $h^{2}+r^{2}=(2r)^{2}=4r^{2}\;\Longrightarrow\;h^{2}=3r^{2}\;\Longrightarrow\;h=r\sqrt3.$

   The ratio $r:r\sqrt3:2r$ is the 30-60-90 triangle, so each leg meets the long base at exactly $60^{\circ}$ — independent of $r$, which is why the shape is similar for every $r$.

4. 4

   Apply the trapezoid area formula.

   $A=\frac12\bigl(\text{upper}+\text{lower}\bigr)h=\frac12\bigl(4+4+2r\bigr)\bigl(r\sqrt3\bigr)=\frac12\cdot 2(4+r)\cdot r\sqrt3=\sqrt3\,r(r+4).$

5. 5

   Check the formulas on a concrete value and note the constraint. Every $r>0$ gives a genuine trapezoid, since the height $r\sqrt3$ is then positive. Take $r=1$: the bases are $4$ and $6$, the legs are $2$, the height is $\sqrt3=1.7320508$. Perimeter $=4+6+2+2=14=8+6(1)$ ✓, and area $=\tfrac12(10)(1.7320508)=8.660254$, while $\sqrt3(1)(5)=8.660254$ ✓.