Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 617.

## Thursday, June 9, 2011

### Problem 617: Right Triangle Area, Incircle, Incenter, Tangent, 90 Degrees

Labels:
90,
area,
degree,
hypotenuse,
incircle,
right triangle,
tangent

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s-a=AD=2

ReplyDeletes-c=CD=3

a-c=(s-c)-(s-a)=1

b=2+3=5

2ac=(a^2+c^2)-(a-c)^2=b^2-(a-c)^2=25-1=24,

ac=12

Follows

Area of triangle ABC=(1/2)ac=6 sq units

Let BF=BE=x; then AB=AE+BE=AD+BE=2+x and BC=CE+BF=CD+BF=3+x. By Pythageorean Theorem, we have AB^2+BC^2=AC^2, or (2+x)^2+(x+3)^2=5^2, which gives the quadratic equation x^2+5x=6. Thus, discarding negative solutions, we have x=1. Then, AB=3, BC=4 and the area of triangle ABC is (3*4)/2=6.

ReplyDeleteI have posted my solution, which is the same as above, on my blog.

ReplyDeleteWe write the area in 2 ways

ReplyDelete(2+r)(3+r) = r(2+r+3+r+5)

Simplifying the algebra

r^2 + 5r-6 = 0 from which r= 1 after solving the quadratic and disregarding the negative root

So area = 3 X 4 /2 = 6

Sumith Peiris

Moratuwa

Sri Lanka