+0

Geometry Problem

0
58
4

In rectangle ABCD, AD = 1, P is on AB, and DB and DP trisect angle ADC. Write the perimeter of triangle BDP in simplest form as: w+(x*sqrt(y))/z, where w, x, y, z are positive integers. What is w + x + y + z?

Guest Jan 31, 2018

Best Answer

#1
+6250
+4

DB and DP trisect angle ADC,   so   m∠ADP  =  m∠PDB  =  m∠BDC  =  90° / 3  =  30°

∠ABD  and  ∠BDC  are alternate interior angles, so they have the same measure.

If we draw a line from  P to  BD  that is perpendicular to side BD, we will have congruent triangles.

The angle measures of these triangles are  30°, 60°, and 90° .

The side opposite the  60°  angle  =  1 , so

the side opposite the 90° angle  =  2 / √3

perimeter of triangle BDP   =  2  +  2/√3  +  2/√3   =   2  +  4/√3   =   2  +  4√3 / 3

2 + 4 + 3 + 3   =   12

hectictar  Jan 31, 2018
Sort:

4+0 Answers

#1
+6250
+4
Best Answer

DB and DP trisect angle ADC,   so   m∠ADP  =  m∠PDB  =  m∠BDC  =  90° / 3  =  30°

∠ABD  and  ∠BDC  are alternate interior angles, so they have the same measure.

If we draw a line from  P to  BD  that is perpendicular to side BD, we will have congruent triangles.

The angle measures of these triangles are  30°, 60°, and 90° .

The side opposite the  60°  angle  =  1 , so

the side opposite the 90° angle  =  2 / √3

perimeter of triangle BDP   =  2  +  2/√3  +  2/√3   =   2  +  4/√3   =   2  +  4√3 / 3

2 + 4 + 3 + 3   =   12

hectictar  Jan 31, 2018
#2
+82489
+3

Angle ADP  = 30°   and angle APD  = 60°

So.....since we have a 30-60-90 triangle......AP  =  1/√3   = √3/3

And PD  =   2 / √3  =  2√3 / 3

And angle DPB = 120°  and angle PDB = 30° ..... so   angle  PBD = 30°

So...triangle DPB  isosceles with  PD  = PB  =  2√3 / 3

And using the Law of Sines

DB/ sin(120)  = PB / sin (30)

DB  =  sin (120) * [ 2√3 / 3 ]  /  (1/2)  =    √3 * 2√3 / 3   =  2

So...the perimeter of triangle BDP  =

PD + PB + DB  =     4√3 / 3  +  2   =

2  +  4√3 / 3

So   w  = 2,  x = 4 , y  = 3  and z  = 3

And their sum is  12

CPhill  Jan 31, 2018
#3
+82489
+1

I liked the way you did this, hectictar....the "congruent" triangles approach was a nice touch...!!!

CPhill  Jan 31, 2018
#4
+6250
+3

Thanks!!

hectictar  Jan 31, 2018

14 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details