In rectangle $ABCD$, $AD=1$, $P$ is on $\overline{AB}$, and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$. Write the

Question

3 years ago

14

In rectangle $ABCD$, $AD=1$, $P$ is on $\overline{AB}$, and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$. Write the

perimeter of $\triangle BDP$ in simplest form as: $w + \frac{x \cdot \sqrt{y}}{z}$, where $w, x, y, z$ are nonnegative integers. What is $w + x + y + z$?

Mathematics

1 answer:

Studentka2010 [4] · Answer 1 · 2022-03-07T00:31:50+03:00

<h3>Answer:</h3>

w+x+y+z = 12

<h3>Explanation:</h3>

When right angle ADC is divided into three equal parts (trisected), each of those parts is 90°/3 = 30°. Thus, ∠CDB = ∠PBD = ∠PDB = 30°.

The sides of a 30°-60°-90° triangle are in the proportion 1 : √3 : 2. Hence BD = 2, and PD = PB = 2/√3 = (2/3)√3.

Then the perimeter of ∆BDP is ...

... perimeter ∆BDP = BD + DP + PB

... = 2 + (2/3)√3 + (2/3)√3

... = 2 + (4√3)/3 . . . . . . . . w=2, x=4, y=3, z=3

w + x + y + z = 2+4+3+3 = 12