In rectangle ABCD the triangle BCD has area 8 and the side BC is 4√2. Find the perimeter of the rectangle.

Step-by-step solution

1. 1

   Relate the triangle to the rectangle. $BD$ is a diagonal of the rectangle, and a diagonal cuts a rectangle into two congruent right triangles. So $\triangle BCD$ is exactly half the rectangle:

   $[ABCD]=2\times 8=16$

2. 2

   Find the unknown side. With $BC=4\sqrt2$ as one side,

   $CD=\frac{16}{4\sqrt2}=\frac{4}{\sqrt2}$

3. 3

   Rationalise. Multiply numerator and denominator by $\sqrt2$:

   $CD=\frac{4\sqrt2}{2}=2\sqrt2$

4. 4

   Add up the perimeter.

   $P=2\left(4\sqrt2+2\sqrt2\right)=2\left(6\sqrt2\right)=12\sqrt2$

   $\boxed{12\sqrt2\approx 16.97}$

5. 5

   Check the area from the two sides. $4\sqrt2\times 2\sqrt2=8\cdot 2=16$ ✓, so the triangle is $8$ ✓. Alternatively compute the triangle directly: $\tfrac12(4\sqrt2)(2\sqrt2)=\tfrac12(16)=8$ ✓ — confirming both the halving step and the rationalisation.