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Question 1205456: Triangle ABC has sides AB = 7, BC = 8, and AC = 9. Point C is reflected over line AB to create point C'. Next, point B is reflected over line AC' to create point B'. Find the area of triangle B'C'C.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52806)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Diagram
Triangle ABC has sides AB = 7, BC = 8, and AC = 9.
That's equivalent to saying the sides are: c = 7, a = 8, b = 9
Each lowercase letter represents the side opposite its uppercase letter (eg: side b is opposite angle B)
I'll use Heron's formula to compute the area of triangle ABC.
First we need the semiperimeter
s = (a+b+c)/2
s = (8+9+7)/2
s = 12
Then we can compute the area
area = sqrt(s*(s-a)*(s-b)*(s-c))
area = sqrt(12*(12-8)*(12-9)*(12-7))
area = 26.83281573 square units approximately
C reflects over line AB to land on C'
Line segment CC' is perpendicular to AB.
Let D be the intersection of segments CC' and AB.
Segment DC is the altitude of triangle ABC when AB is the base.
We know AB = 7 but we don't know DC. But we can easily find it like so
area = 0.5*base*height
26.83281573 = 0.5*7*height
26.83281573 = 3.5*height
height = 26.83281573/3.5
height = 26.83281573/3.5
height = 7.66651878
This is the approximate length of segment DC.
DC' is also this length because of the reflection.
Then,
CC' = DC + DC'
CC' = DC + DC
CC' = 2*DC
CC' = 2*7.66651878
CC' = 15.33303756
Also because of mirroring, B'C' = 8 since BC = 8.
Triangle B'C'C has these sides
BC = 8
CC' = 15.33303756 (approximate)
If we could find side B'C, then we'd be able to use Herons formula to get the final answer.
Technically it is possible to find this missing side length. But it would take too much work I think. Luckily there's another way.
Focus entirely on triangle ABC.
Since we know all three sides, we can use the law of cosines to find each of the missing angles. We'll focus on angle A.
a^2 = b^2 + c^2 - 2*b*c*cos(A)
cos(A) = (b^2 + c^2 - a^2)/(2bc)
cos(A) = (9^2 + 7^2 - 8^2)/(2*9*7)
cos(A) = 0.5238095238095
A = arccos(0.5238095238095)
A = 58.4118644948
This is the approximate measure of angle BAC.
It's also the approximate measure of angle BAC' because of the reflection.
Focus on right triangle C'DA.
We found acute angle C'AD = 58.4118644948 that was computed above.
The angle AC'D is the complement of this
90 - 58.4118644948 = 31.5881355052
I'll skip showing the steps, but using the law of cosines will find that angle B'C'A is roughly 48.18968510421 degrees.
So,
angleB'C'C = (angleB'C'A)+(angleAC'D)
angleB'C'C = (48.18968510421)+(31.5881355052)
angleB'C'C = 79.77782060941
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I apologize if this solution seems a bit cluttered. I'll try to go back and rewrite things when I get the chance. But for now let's just move on.
To summarize so far we found the following:- B'C' = 8
- angle B'C'C = 79.77782060941 degrees (approximate)
- CC' = 15.33303756 (approximate)
We have the SAS (side angle side) case going on here. We know two sides of a triangle and the angle between.
Use the SAS triangle area formula to wrap things up.
area(triangle) = 0.5*side1*side2*sin(angle between those sides)
area(B'C'C) = 0.5*(B'C')*(CC')*sin(angle B'C'C)
area(B'C'C) = 0.5*(8)*(15.33303756)*sin(79.77782060941)
area(B'C'C) = 60.35862404571
Answer: approximately 60.35862404571 square units
Round that value however needed.
GeoGebra can be used to verify each claim that I have made above, and can also be used to verify the final answer.
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