SOLUTION: In triangle PQR, let X be the intersection of the angle bisector of angle P with side QR, and let Y be the foot of the perpendicular from X to side PR. If PQ = 9, QR = 10, and PR =
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Question 1202263: In triangle PQR, let X be the intersection of the angle bisector of angle P with side QR, and let Y be the foot of the perpendicular from X to side PR. If PQ = 9, QR = 10, and PR = 17, then compute the length of XY. Found 2 solutions by greenestamps, math_tutor2020:Answer by greenestamps(13195) (Show Source):
There are at least two ways to solve this problem.
Method 1
This is one way to draw things out
I used GeoGebra to make the diagram.
c is a real number such that 0 < c < 10.
As mentioned by the other tutor @greenestamps, we can use the angle bisector theorem to set up this proportion
PQ/QX = PR/XR
So,
PQ/QX = PR/XR
9/c = 17/(10-c)
9(10-c) = 17c
90-9c = 17c
90 = 17c+9c
90 = 26c
c = 90/26
c = 45/13
Therefore,
QX = c = 45/13
XR = 10-c = 85/13
Also mentioned by @greenestamps is Herons Formula to get the area of triangle PQR to be exactly 36 square units.
Notice that triangles PQX and PXR have the same height.
QR is horizontally flat on the ground.
Let h be the unknown height of each smaller triangular piece.
I'll then let A and B represent the areas of triangles PQX and PXR respectively.
A = area of triangle PQX
A = (1/2)*base*height
A = (1/2)*QX*h
A = (1/2)*(45/13)*h
A = (45/26)*h
B = area of triangle PXR
B = (1/2)*base*height
B = (1/2)*XR*h
B = (1/2)*(85/13)*h
B = (85/26)*h
Areas A and B must add up to 36 which is the area of triangle PQR.
A+B = area of triangle PQR
(45/26)*h + (85/26)*h = 36
(45/26+85/26)*h = 36
((45+85)/26)*h = 36
(130/26)*h = 36
5*h = 36
h = 36/5
So,
area of triangle PXR = (1/2)*base*height
area of triangle PXR = (1/2)*XR*h
area of triangle PXR = (1/2)*(85/13)*(36/5)
area of triangle PXR = 306/13
Rotate triangle PXR so that PR is horizontal and acts as the base.
Segment XY is the height of triangle PXR since it is perpendicular to the base PR.
area of triangle PXR = (1/2)*base*height
306/13 = (1/2)*PR*XY
306/13 = (1/2)*17*XY
XY = (306/13)*2*(1/17)
XY = 36/13
XY = 2.7692 approximately
Here's another way to draw things out.
The sides of the triangle are
PQ = 9
QR = 10
PR = 17
I've split side PR into two pieces
PY = m and YR = 17-m
where 0 < m < 17.
Focus on triangle PQR.
Use the law of cosines to determine angle R.
c^2 = a^2 + b^2 - 2*a*b*cos(C)
r^2 = p^2 + q^2 - 2*p*q*cos(R)
(PQ)^2 = (QR)^2 + (PR)^2 - 2*(QR)*(PR)*cos(R)
(9)^2 = (10)^2 + (17)^2 - 2*(10)*(17)*cos(R)
81 = 389 - 340*cos(R)
81-389 = -340*cos(R)
-308 = -340*cos(R)
cos(R) = -308/(-340)
cos(R) = 77/85
R = arccos(77/85)
R = 25.057615418303
R = 25.0576
This value is approximate.
Your calculator needs to be in degree mode.
Follow similar steps to find angle QPR = 28.0725 degrees approximately.
Angle bisector PX will split angle QPR into two equal pieces angle QPX and angle XPY
angle QPX = angle XPY = 28.0725/2 = 14.03625
Let's focus our attention on triangle XPY.
This is a right triangle, so we can use one of the SOH CAH TOA trig ratios.
I'll use tangent so we can connect PY and XY.
The conclusions of the previous two paragraphs found that
XY = 0.25000012m
and
XY = (17-m)*0.46753214
Set them equal to one another and solve for m
I'll skip the steps, but the solution to 0.25000012m = (17-m)*0.46753214 is roughly m = 11.0769
This means:
PY = m = 11.0769
YR = 17-m = 17-11.0769 = 5.9231
and furthermore
XY = 0.25000012*m
XY = 0.25000012*11.0769
XY = 2.769226329228
XY = 2.7692
Or you could compute it like this
XY = (17-m)*0.46753214
XY = (17-11.0769)*0.46753214
XY = 2.769239618434
XY = 2.7692
The drawback of this second method is that we cannot arrive at a nice fraction we did with method 1.
But it is often good practice to be able to solve a problem multiple ways.
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